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PLANE GEOMETRY

WILLIS

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PREFACE

This text represents the experience and developing ideals
of a quarter of a century of geometry teaching. The
philosophy and the methods set forth in it are those of the
author's own class-room. He believes that a pupil learns
more of the subject when it is presented as in these pages,
and he is certain that every pupil who is taught by this
method under the guidance of a teacher who is himself
interested in his subject, becomes inspired to learn and is
willing to work to learn. An awakening of a love of the
subject is a service that may be rendered to our pupils
of every intellectual grade. The text is so arranged that it
may be variously used.

A SHORT COURSE. — From the complete material pre-
sented in this text, what may be considered as the funda-
mentals, or as constituting a minimum requirement, has
been designated by a BLACK FACE paragraph numeral.
It constitutes a Short Course.

As much as possible of the remaining material should
be included. This remaining material, which completes and
rounds out the subject, is designated by a LIGHT FACE
paragraph numeral.

A LABORATORY COURSE. — Chapter 1 and the chapters
which contain the Experimental Determination of Principles
may be used as a laboratory course, omitting all of the
chapters containing the Classification and Explanation of
Principles except the most elementary portions.

A COMPLETE COURSE. — The text is primarily intended to
be used as a whole, without many omissions. It contains
all that need be required of any class. The experimental
chapters should be rigorously required of all, and the
accurate or free-hand construction of all figures should be
insisted upon throughout the portions devoted to classifica-
tion and explanation of principles.

541281

VI PREFACE

EXERCISES. — A sufficient number of exercises, carefully
selected and graded, will be found in the list of Additional
Theorems, Exercises and Review Exercises to enable the
teacher to vary his class assignments from term to term for
many years. No class should be expected to solve all.

FOR EXCEPTIONAL PUPILS. — The wealth of material
presented affords a splendid means of developing the pupil
of exceptional ability. Extra work for additional credit
may be allowed every pupil. It is better to stimulate a
pupil to greater interest and accomplishment, especially, in a
subject so vast and offering so great rewards as geometry,
than to hurry an able pupil through the subject, almost
always to his loss rather than to his profit.

DEDICATION. — The conception of this work runs far back
in the years. It is the author's contribution in the service
of his profession. It is dedicated to all those many students
of the grand old science, who perhaps unknowingly, and to
all those teachers who have keenly felt the need of a more
sympathetic interpretation of its immortal truths.

In the hope that it will influence better teaching and
learning, and will help to upbuild and uphold a generation
which will reverence our beloved geometry for its own sake,
as well as for its uses inseparable from a great constructive
civilization — the author sends this little volume forth upon
its mission.

C. ADDISON WILLIS.

GIRARD COLLEGE, PHILADELPHIA.

CONTENTS

CHAPTER I

PAGE

FIRST PRINCIPLES 1

CHAPTER II

STRAIGHT LINES, CIRCLES, SYMMETRY 13

Principles Determined Experimentally.

CHAPTER III

PARALLELS, PERPENDICULARS, ANGLES 30

Classification and Explanation of Principles.

CHAPTER IV

TRIANGLES

Principles Determined Experimentally.

CHAPTER V

TRIANGLES 63

Classification and Explanation of Principles.

CHAPTER VI

POLYGONS 90

Principles Determined Experimentally.

CHAPTER VII

POLYGONS 97

Classification and Explanation of Principles.

CHAPTER VIII

Loci 116

Principles Determined Experimentally.

CHAPTER IX

Loci 123

Classification and Explanation of Principles.

CHAPTER X

THE MEASUREMENT OF SECTS, SIMILAR TRIANGLES 129

Principles Determined Experimentally.

vii

Vlll CONTENTS

CHAPTER XI

PAGE

SIMILAR TRIANGLES, SIMILAR POLYGONS 143

Classification and Explanation of Principles.

CHAPTER XII

CIRCLES 178

Principles Determined Experimentally.

CHAPTER XIII

CIRCLES 194

Classification and Explanation of Principles.

CHAPTER XIV

REGULAR POLYGONS, CIRCLES ASSOCIATED WITH POLYGONS. . . 236
Principles Determined Experimentally.

CHAPTER XV

REGULAR POLYGONS AND CIRCLES 243

Classification and Explanation of Principles.

CHAPTER XVI

AREAS. 260

Principles Determined Experimentally.

CHAPTER XVII

AREAS 269

Classification and Explanation of Principles.

INDEX . . 299

CHAPTER I
FIRST PRINCIPLES

1. Note-book. — Preserve all drawings and written work
in a loose leaf note-book with unruled pages. Number to
correspond with the text. Draw large, accurate figures
with distinct lettering.

2. Drawing Instruments. — Each pupil should be pro-
vided with one six-inch triangle with 30°, 60° and 90°
angles, a six-inch scale divided decimally, a protractor,
and a compass.

3. How Geometry Began. — Many ancient nations, the
Chinese, Babylonians, Egyptians, Greeks, Romans, and
others, constructed wonderful works, as roads, canals,
aqueducts, temples, pyramids, and tombs. They laid
out cities and gardens. These required appreciation of
geometric principles. Herodotus, a Greek historian who
wrote about 450 B.C., -states that the Egyptians developed
many geometric principles because the annual overflow of
the Nile obliterated the boundary marks of the lands, and
it became necessary to re-locate these boundaries. This
was probably the earliest use of surveying for land measure-
ment, and it is from this use of geometric principles that
the name geometry, which means "earth measurement,"
is derived.

EXERCISES

1. What ru:er first appointed surveyors in Egypt? Refer to an
article on geometry in an encyclopedia.

2. Look up the derivation of the word geometry. Name some other
familiar English words containing one or the other of these
Greek root-words.

4. The Modern Value of Geometry. — Geometry forms the
structure of many modern sciences and arts. Artists

PLANE GEOMETRY

apply its principles in pictures and designs; architects use
them in drawing plans; civil engineers and surveyors use
geometry now as they did centuries ago; astronomers could
not measure nor describe the motions and positions of the
stars or planets without it; nor could navigators find their
place at sea. Many artisans, also, and workers in machine
shops, carpenters, stone cutters, and others, make constant
use of geometric principles.

5. Familiar Use of Geometric Terms. — Every word
which has to do with position, form or size, is geometric.

EXERCISES

1. How far is a point on the surface of the earth from the center
of the earth? How tall are you?

2. What is the form of the earth? Of a piece of crayon?

3. What is the area of the floor of the class-room in square feet?

4. What is the volume of the class-room in cubic feet?

5. What kind of surface are the floor and walls?

6. What is the form of the uncut end of a lead pencil?

7. What is the form of a base-ball diamond?

8. What is the form of a tennis court?

9. What is the form of a Rugby football?

10. Write such formulas of mensuration (geometric measurement)
as you can recall. Have these formulas practical use? Many
of them are derived and explained in this text.

6. The Subject Matter. — Geometry considers the prop-
erties of solids, surfaces, lines and points. A solid is a
definite portion of space. Thus the properties of the space,
or room, occupied by an object belong to geometry, and
not the physical properties of the object.

A surface is the boundary of a solid.

A line is the boundary of a surface.

A point is the end of a line, or divides a line.

7. Dimensions. — A dimension is extent of some definite
kind.

A point has no dimensions; that is, it has no extent. It
has position only.

A line has one dimension, length.

FIRST PRINCIPLES O

A surface has two dimensions, length and width (or
breadth).

A solid has three dimensions, length, width, and height.

EXERCISES

1. Refer to a point in the room; where is it?

2. Refer to a line; how long is it?

3. Refer to a surface; how long and how wide is it?

4. Refer to a solid; estimate its three dimensions.

8. Points. — A point is named by a capital letter.

EXERCISES

1. Draw a point • or X. Name it.

2. Draw a straight line. Mark a point on it 1

9. Lines. — A straight line is the shortest line that can
be drawn, or imagined, between two points.

A curved line is one no part of which is straight.

A sect is a straight line of definite length.

A unit sect is a sect of generally known or standard
length, as the inch, meter, mile.

A sect is measured in terms of some unit sect. The
measurement is a number which expresses how many times
the given sect contains the unit sect.

A line is named by two capital letters, or by a single
letter.

A B_ t x [ y [

A sect is bisected when it is divided by a point or inter-
secting line into two equal parts.

EXERCISES

1. From what word is straight derived?

2. What instrument is used to draw a straight line?

3. What instrument contains a unit sect?

4. What instruments may be used to copy a given sect?

5. Draw a sect of 3 inches; one of 3.4 inches.

6. Draw a sect as long as the ^ *. B
sect AB, or x. -* ^—

4 PLANE GEOMETRY

7. Measure the sect AB.

8. What drawing instrument is used to draw a curved line?
What kind of curved line may be drawn with it?

9. May this curved line be measured in the same units as a
straight line?

10. Draw such a curved line; measure its length in inches.

11. Bisect (by judgment) each of the sects of Example 5 by a point.

12. Bisect the sect of Example 6 by a line.

13. Draw and trisect a sect.

14. Can an indefinite line be bisected?

15. Draw and define a broken line.

16. Draw and define a mixed line.

10. Positive and Negative Sects. — These terms refer
to the direction in which a sect is drawn or measured. In
general, sects are considered positive. Sometimes it is-
necessary, or convenient, to distinguish between a sect
drawn, or measured, from left to right and one drawn, or
measured, from right to left. The former direction is then
generally considered positive, and the latter negative.

EXERCISES

1. Draw a sect and read it as a positive sect. Read it as a negative
sect.

2. If the distance and direction of a point A from a point B are
indicated on a surveyor's plot by + 785.46 feet; how are the dis-
tance and direction of point B from point A indicated?

3. Read sect AC as the sum A B C

of two sects. Read AB as H ' '

the (algebraic) sum of two sects, one positive and the other
negative.

4. A man walks from point A, —250 feet to point B; and from point
B, +300 feet to point C. Make a figure to show where point C
is located with reference to point A.

11. Planes. — A plane, or a plane surface, is such a surface
that a straight line between any two points of the surface
lies wholly within it.

Plane geometry is the geometry of figures which are, or
could be, drawn on a plane.

EXERCISES

1. How does a carpenter test the planeness of a board? Does the
method agree with the definition of a plane?

FIRST PRINCIPLES 5

2. Why is a carpenter's plane so called? Why is a certain machine
tool called a planer?

3. Define a plane curve. Refer to, or form with a string or wire,
a curve that is not plane.

4. Are two pencils with points in contact, in the same plane?

5. What kind of surfaces are cylindrical and conical surfaces?

12. Straight Lines in Relative Position. — Two straight
lines in the same plane may be intersecting, parallel or
coinciding. A system of parallels consists of three or more
parallel lines.

Three or more lines are concurrent which have a common
point of intersection. They form a pencil of lines.

EXERCISES

1. Draw two parallels; a system of parallels; two intersecting lines;
four concurrent lines; two coinciding lines; a pencil of three lines.

2. Define parallel lines.

3. Hold two pencils so that they are not intersecting, parallel or
coinciding. Are they in the same plane? Refer to two such
lines in the room.

13. Generation by Motion. — The line, surface and solid
may be regarded as generated by motion.

EXERCISES

1. When a point moves, what does it generate; that is, what does
the path of the point represent?

2. If a line moves, what does it generate?

3. If a surface moves, what does it generate?

14. Circles. — A circle is a portion of a plane bounded by
a curved line every point of which is equidistant from a
fixed point within it, called the center.

The circumference is the bounding line;
which is also called a circle.

A radius is a straight line joining the
center with a point on the circum-
ference.

A chord is a straight line joining two
points of the circumference.

A diameter is a chord which passes through the center.

6 PLANE GEOMETRY

A tangent is a straight line which touches the circumfer-
ence in only one point however far it is extended.
An arc is a portion of a circumference.
A unit arc is an arc of standard length.

A degree of arc (1°) is of a circumference.

The degree is the unit arc which is most used.

The measurement of an arc is a number which expresses
how many times the given arc contains the unit arc of the
same circle.

A minute of arc is ™ of a degree.

A second of arc is ^ of a minute.

A semicircle is one-half of a circle.

A quadrant is one-quarter of a circle, or of a circumference.
Every circle contains 360° of arc, but the size of a circle
is given by the length of its radius or diameter.
A circle is named by a capital letter at the center.

EXERCISES

1. Draw a circle. Draw a radius, diameter, chord, tangent, degree
of arc, quadrant, semicircle, semicircumference.

2. Draw a curve which is not a circle.

3. What instrument is used to measure arcs of circles?

4. How many degrees of arc are there in a quadrant? In a semi-
circumference? How many quadrants are in a semicircumfer-
ence or semicircle?

5. Draw a circle and mark arcs of approximately 60°; 120°; 45°;
30°; 1°.

6. On a circle, lay off an arc AB = +160°; from point B, lay off
an arc BC = —100°. Read the value of arc CA.

15. Relative Position of Two Circles. — Two circles of
the same, or different, radii may be situated in relation to
each other in three ways: (1) out of contact; (2) tangent; (3)
intersecting.

Concentric circles have the same center and different
radii.

FIRST PRINCIPLES 7

EXERCISES

1. Draw two circles out of contact in two different ways (internally
and externally out of contact).

2. Draw two circles tangent in two different ways (internally and
externally).

3. Draw two intersecting circles.

4. Draw two concentric circles; three concentric circles.

5. What class includes concentric circles?

16. Angles. — An angle is a figure formed by two straight
lines drawn from the same point.

The parts of an angle are the sides and the vertex.

An angle may be considered as having been generated by
the rotation of one of its sides about the vertex from a posi-
tion of coincidence with the other side. An angle may be
positive or negative according to the direction of this rota-
tion. The counter-clockwise direction is generally regarded
as positive.

A measurement arc is often used to indicate the size or
direction, or both, of an angle.

An angle is named (1) by three capital letters, reading
from one side to the other with the vertex letter in the
middle ; (2) when no confusion can occur, by the vertex let-
ter alone; (3) by a small letter, or number, or capital letter
placed within the angle.

EXERCISES

1. Read the angles of Figures 1 and 2 in all possible ways.

2. Define the vertex and sides of an angle.

3. Draw two lines, A B and CD, intersecting at point 0. Read all
the angles which are formed.

8 PLANE GEOMETRY

17. Angles Named According to Size.

A zero angle is one whose sides coincide before
rotation.

A straight angle is one whose sides lie in the same straight
line in opposite directions from the
vertex.

A right angle is one-half of a straight angle.

A perigon is an angle whose sides coincide ^\

after rotation.

An acute angle is less than a right angle.

An obtuse angle is greater than a right angle and less than
a straight angle.

A reflex angle is greater than a straight angle and generally
is not more than a perigon.

An oblique angle is any angle other than a multiple of a
right angle.

Two angles whose sum is a
right angle are complementary.

Two angles whose sum is a straight angle are supplement-
ary.

Two angles whose sum is a perigon are explementary.

The angles are complements, supplements, explements of
each other.

A unit angle is an angle of standard size.

A degree of angle is §V of a right angle.

The measurement of an angle is a number which expresses
how many times the given angle contains the unit angle.

EXERCISES

1. Draw angles of each kind named above, with their measurement
arcs.

2. Calculate the complement and supplement of 72°.

3. What angle has a negative complement and a positive supple-
ment?

4. Estimate the angles x, y and DEG of paragraph 16; in right
angles, degrees and straight angles.

5. Draw angles with the protractor to the following measures:
1 degree, or 1°; 45°; 130°; 2£ right angles; 180°; If straight angles.

FIRST PRINCIPLES

6. What angle occurs most frequently in familiar objects?

7. Is the size of an angle determined by the length or by the posi-
tion of its sides?

18. Angles Named According to Relative Position. —
Unless otherwise stated, these names refer only to two
angles.

Adjacent angles have a common side and vertex.

Opposite angles (or vertical angles) are the non-adjacent
angles formed by two intersecting lines.

Supplementary-adjacent angles are both supplementary
and adjacent.

A transversal is a line which intersects two or more
other lines.

The names given to pairs of angles, one at each point of
intersection, are:

Corresponding; r and w.
Alternate-interior; u and

Alternate-exterior; rand

Consecutive-interior; I
and x.

Consecutive-exterior; s
and y.

EXERCISES

1. Draw two externally adjacent angles. Draw two internally
adjacent angles. Draw four successively adjacent angles.

2. Draw two intersecting lines. Name all pairs of opposite angles.

3. Draw two supplementary-adjacent angles. Draw two supple-
mentary angles that are not adjacent.

4. Name all other pairs of corresponding angles formed in the trans-
versal figure; of alternate-interior angles; of alternate-exterior
angles; of consecutive-interior angles; of consecutive-exterior
angles.

5. Draw three lines cut by a transversal. Name a set of corres-
ponding angles.

10 PLANE GEOMETRY

19. Other Angle Names. — A perpendicular is a straight
line which forms a right angle with another straight line.
The foot of the perpendicular is the point of intersection.

A bisector of an angle is a straight line which divides it
into two equal angles.

EXERCISES

1. Draw AB perpendicular to CD at point A. Name the foot of
the perpendicular A B. Is line CD also perpendicular to line A B ?

2. Can an angle bisector be drawn otherwise than through the
vertex?

3. Draw acute, right, obtuse, reflex and straight angles, and a
perigon. Draw by judgment the bisector of each.

4. Draw and trisect an angle?

20. Distance. — The distance between two points is the
length of the straight line joining the points.

The distance from a point to a straight line is the length
of a perpendicular drawn from the point to the line.

The distance between two parallels is measured on a
perpendicular drawn to both parallels.

The distance from a point to a circle is measured on a
radius of the circle, or on a radius prolonged, which is drawn
through the given point.

EXERCISES

1. Mark two points. Draw and measure the distance between
them.

2. Mark a point and draw a straight line. Draw and measure the
distance from the point to the line.

3. Draw two parallels. Draw and measure the distance between
them.

4. Draw a circle and mark a point within it. Draw and measure
the distance from the point to the circumference.

5. Mark a point outside the circle which is two inches from the
circumference.

21. REVIEW EXERCISES

1. Give the geometric names of various objects in the class-room.

2. What kinds of angles are formed by the corners of your draw-
ing triangles? Measure them in degrees.

FIBST PRINCIPLES 11

3. Name two streets that intersect at a right angle; two that inter-
sect at an oblique angle; two streets that will not intersect no
matter how far they are extended in either direction.

4. Can a sect be measured in degrees? Or an angle in inches?
Has a degree of angle any equivalent in inches?

5. Can an arc be measured in inches?

6. If you walk from a given point A, +250 feet to a point B; and
then walk from point B, — 400 feet to a point C; how is
point C situated with respect to point A? Make a drawing

7. If the terminal side of an angle rotates from the initial side A B,
+ 120 degrees to the position AC; and then rotates from the
position AC, —200 degrees to the position AD; how is the
terminal side AD situated with respect to the initial side AB?

8. Judge the measure of each of these angles in terms of the right
angle; also in terms of the degree.

At.

9. Reduce f of a right angle to degrees, minutes and seconds.

10. Reduce 22° 30' (22 degrees and 30 minutes) to a fraction of a
right angle.

11. How many degrees of angle has the minute hand of a clock
described (generated) from twelve o'clock to one minute (of
time) past twelve?

12. How many degrees of angle are between a line running from a
certain point northeast, and a line from the same point
due east?

13. Calculate and draw the supplement of 200 degrees; the com-
plement of 100 degrees.

14. Write the complement of x degrees; the supplement. Write
the complement of x right angles; the supplement.

15. Find the angle whose supplement equals three times the com-
plement. Help. 180° - x = 3(90° - »).

16. Express algebraically the relation that x and y are comple-
mentary; supplementary.

17. One of two complementary angles is 10° less than twice the
other angle. Find the angles.

18. The difference between two supplementary angles is 25°.
Find the angles.

19. Upon what does sign depend in sects, angles and arcs?

20. What angle-bisector is a perpendicular to one of the sides of
the angle?

21. Measure the distance from this point to the nearest and
farthest edges of this page.

PLANE GEOMETRY

22. What line is the distance from the center of a circle to the
circumference?

23. On what line is the distance measured between two concentric
circles?

24. On what line is the distance measured between the circumfer-
ences of two circles that are out of contact?

25. Define parallels by means of a property of the distance between
them.

26. Show that in every use of the term distance between two objects,
the shortest distance is referred to.

27. Draw two straight lines which do not coincide, which form a zero
angle at infinity.

22. ABBREVATIONS AND SYMBOLS

adjacent adj.

angle, angles Z, Zs

alternate-interior alt. -int.

alternate-exterior alt.-ext.

circle, circles O, ©

consecutive cons.

corresponding corres.

degree, minute, second . °

equal, is equal to =

equals =s

is unequal to 7^

unequals F^S

exterior ext.

interior. . . int.

feet, inches ' "

is greater than >

is less than <

opposite opp.

parallel, is parallel to ... 1 1

parallels ||s

perpendicular, is perpen-
dicular to _L

perpendiculars _l_s

point, points pt, pts.

right rt.

straight st.

supplementary-adjacent sup. -adj.
paragraph reference. . . . l,26,etc.

CHAPTER II
STRAIGHT LINES, CIRCLES, SYMMETRY

EXPERIMENTAL INVESTIGATION OF PRINCIPLES

23. Geometrical Experiments. — The facts of geometry
are all about us. We intend to inquire into nature's laws
and to discover these facts through our own powers of
observation. For this purpose a series of experiments will
be undertaken; an experiment being an investigation made
for the purpose of ascertaining a fact.

We shall make a drawing according to some assigned
(or given) property, and shall observe some new property
which is invariably associated with the given property.
The finding of this new property is the result of the
experiment.

Make all experiments for the general case, so that results
shall not be limited in their application. State the result
in the shortest and clearest expressions possible. Intro-
duce nothing outside of the experiment. If an experiment
is inconclusive, do not hesitate to so state the result. You
are not to state what you think ought to be true, but exactly
what the experiment shows to be true. Sometimes it is
well to repeat an experiment one or several times.

EXERCISES

1. Are other sciences based upon experiment?

2. How may the equality or difference of two sects be tested by the
use of instruments? Of two angles?

3. How may the perpendicularity, or lack of it, of two lines be
tested?

24. Experiment I. — The number of straight lines that
can be drawn through given points.

(a) One given point.

(b) Two given points.

14 PLANE GEOMETRY

(a) Mark a point A. Draw a straight line through it.
If possible, draw other straight lines through point A.
Observe the number of straight lines which can be drawn
through point A, and their directions.

RESULT. — Any number of straight lines can be drawn
through a given point, and in any direction.

(6) Mark two points A and B. Draw a straight line
through (containing) both points. If possible, draw other
straight lines containing both points, in different positions
from the line first drawn. Observe the number of straight
lines which can be so drawn. State result.

(c) Mark three points at random. Observe the number
of straight lines which can be drawn containing all the
points. Can a straight line be drawn containing any three
marked points? Make the experiment for four points.

RESULT. — In general, no straight line can be drawn
through three or more given points. But the points may
be so situated that one straight line shall contain them.

25. Experiment II. — Prolongation of a straight line.
Draw a (straight) line. Prolong (extend) it in both

directions, using dotted lines. Observe how far it can be
thus extended if the drawing is not limited by the extent of
the paper.

RESULT. —

26. EXPERIMENT III. — The number of points of intersec-
tion of straight lines and circles.

(a) Two straight lines.

(b) One straight line and one circle.

In stating the results, observe that in certain positions
of the straight lines and circles there may be no points of
intersection, and in other positions there may be one or
more points of intersection or of contact. State the com-
plete result for each part.

27. Constructions. — A construction is a drawing made
for the purpose of producing a figure with some required

STRAIGHT LINES, CIRCLES, SYMMETRY 15

property. The method of obtaining this required figure is
the subject of experiment.

The different construction methods are:

(1) Rule and compass.

(2) Measurement.

(3) Trial and error.

(4) Copying a standard.

28. Experiment IV. — To draw (or construct) the bisector
of an angle. That is: An experiment to find a method of
drawing the bisector of an angle.

/.ABC is the given angle;
Required to construct its bisector.
METHOD (a). — (1) Draw an arc, center at B, any radius,
intersecting BA and BC in points D and E.

(2) Draw two arcs, centers at D and E, any equal radii
greater than J DE, intersecting each other at F.

(3) Draw BF, which is the required bisector of /.ABC.
METHOD (b). — (1) Measure /.ABC with the protractor.

(2) Divide the measurement by 2.

(3) Lay out /.ABG equal to the quotient.

EXERCISES

1. After obtaining the bisector BG, how may /sABG and GBC
be compared in order to test their equality?

2. Can any angle, straight, obtuse or reflex, be bisected by these
methods?

3. Can another different bisector of /.ABC be drawn?

4. What methods as defined in 27, have been employed in the
preceding experiment?

5. Can an angle be divided into three, four, five, etc., equal angles

by either of these methods?

16 PLANE GEOMETRY

29. Experiment V.— To bisect a sect.

Make the construction: (a) using the rule and compass
method; (6) using the measurement method; (c) using the
trial and error method.

A CED C BD

-H h-fh 1 M

AB is the given sect;
Required to bisect it.

METHOD (c). — (1) Lay off with the compass any length
AC judged = JA£, and lay off CC' = AC.

(2) If C' does not coincide with B, set the compass to a
different length as AD, and make DD' = AD.

(3) Repeat until the middle point of A B is found, as at E.

EXERCISES

1. At how many points can a sect be bisected?

2. Can a sect be divided into any number of equal parts?

3. Which of the methods of the preceding experiment can be used
to divide a sect into three, four, five, etc., equal parts?

30. Experiment VI. — To construct a perpendicular to a
line at a point in the line.

Find (a) a method by rule and compass; (b) a method
by measurement; (c) a method by copying a standard.

AB is the given line, and C is a point in AB;
Required to draw a perpendicular to A B at point C.
METHOD (c). — (1) Place a standard right angle with
vertex at C, and one side coinciding with CB.

(2) Draw CD, which is the required perpendicular.

STRAIGHT LINES, CIRCLES, SYMMETRY 17

EXERCISES

1. Can a perpendicular be drawn (erected) at any point in the line?

2. How many perpendiculars can be erected at a given point as C,
on one side of line AB1

3. Methods (a) and (6) in the preceding experiment are equivalent
to bisecting what kind of angle?

31. Experiment VII. — To construct a perpendicular to a
line from a point without (outside of) it.

Draw and describe methods (a) by rule and compass;
(6) by copying a standard. .

32. Experiment VIII. — To construct an angle equal to a
given angle ; or, to copy a given angle.

Draw and describe methods (a) by rule and compass;
(b) by measurement.

EXERCISES

1. Use these methods to draw an angle
equal to a given obtuse angle.

2. Use the method indicated in the figure
to copy a given angle. This is a
modification of the measurement
method.

33. EXPERIMENT IX. — To copy a sect; addition and sub-
traction of sects.

Draw two sects. Describe different methods of copying
them, of drawing a sect equal to their sum, and of drawing
a sect equal to their difference.

34. EXPERIMENT X. — Addition and subtraction of angles.
Draw two angles; etc.

35. EXPERIMENT XL — To draw the complement and the
supplement of a given angle.

Draw an angle. Find (a) a method by rule and com-
pass; (6) a method by measurement; of drawing the re-
quired angles.

36. Experiment XII. — The relative size of opposite
angles.

Draw two intersecting lines. Observe the size of opposite
angles. State result.

18 PLANE GEOMETRY

37. Experiment XIII. — The relative position of two
perpendiculars to the same straight line.

Draw the figure. State result.

38. EXBERIMENT XIV. — To construct a line through a
given point parallel to a given line.

Draw the given line and the given point. Use the
principle of 37 to complete the required figure.

EXERCISES

1. Can a line be drawn through any given point parallel to any
given line?

2. How many lines, in different positions, can be drawn through
a given point parallel to a given line?

39. Experiment XV. — The position of two lines cut by
a transversal making a pair of corresponding angles equal.

To Draw the Required Figure. — (1) Draw AB cut by
transversal CD, forming Zz.

(2) Mark point F on CD; construct Z.y = Z.x, and in
such a position that it forms with Z.x a pair of correspond-
ing angles; using either the rule and compass method of
copying the angle, as shown, or copying the angle with a
protractor.

(3) Draw GH, the side of Z.y.

Observe the position of A B and GH. The result should
be a complete statement of the property of the figure, thus :

STRAIGHT LINES, CIRCLES, SYMMETRY 19

RESULT. — Two straight lines which are cut by a trans-
versal so as to make a pair of corresponding angles equal,
are parallel.

EXERCISES

1. May any pair of corresponding angles be made equal with the
same result?

2. Draw a figure in which the protractor is used to make Z.y = Z.x.

40. Experiment XVI. — The position of two lines cut
by a transversal making a pair of alternate -interior angles
equal.

Draw the figure. State result.

41. EXPERIMENT XVII. — The position of two lines cut by
a transversal making a pair of alternate -exterior angles
equal.

Draw the figure. State result.

42. Experiment XVIII. — The position of two lines cut
by a transversal making a pair of consecutive -interior
angles supplementary.

To Draw the Figure. — (1) Draw supplementary angles
x and y.

(2) Draw AB, mark points C and D.

(3) Draw Zs x' and y' equal to Zs x and y respectively,
and EF and GHy the sides of Zs x' and yf. Or measure

Zs x' and y' with a protractor, so that their sum equals
180°. State result.

20 PLANE GEOMETRY

43. EXPERIMENT XIX. — The position of two lines cut by
a transversal making a pair of consecutive -exterior angles
supplementary.

Draw the figure. State result.

44. Experiment XX. — To construct a line through a given
point parallel to a given line.

This figure has been constructed c

by means of two perpendiculars to
the same line; 38. Employ in the
present construction the principles
of 39, 40 and 41.

45. EXPERIMENT XXI. — To construct a line through a
given point parallel to a given line, by using rule and
triangle.

This construction can be made by placing the rule and

triangle in certain positions,
and by then sliding the triangle
along the rule. Find the
best positions for the rule and
triangle. This is an accurate
and rapid method of drawing
parallels and is the best
method for general use. It is
much used by draughtsmen.

EXERCISES

1. What represents the transversal in this construction? What
pair of angles is made equal?

2. To which of the methods of 44 is this method equivalent?

3. The rule may be considered as a pair of standard parallels.
Draw two parallel lines by this instrument alone. May a line
be drawn through any given point parallel to any given line
by this method?

46. Experiment XXII. — Relations between pairs of angles
when two parallels are cut by a transversal.

Draw two parallels by any method.

(a) Draw a new transversal. Observe the relative size of
each pair of corresponding, alternate-interior, etc. angles.

' STRAIGHT LINES, CIRCLES, SYMMETRY 21

(6) Draw a perpendicular to one of the parallels and
observe its position with respect to the other parallel.
State results for both (a) and (6) .

47. EXPERIMENT XXIII. — The position of two parallels
to the same line.

Draw the figure. State result.

48. Experiment XXIV. — Relation of angles whose sides
are parallel, each to each.

To Draw the Figure. — (1) Draw an oblique angle.
(2) Draw other angles ^«C^ —

whose sides are parallel, each **£*. ^ c

GIVEN ANGLE

to each, to those of the given .^

angle. There are four posi-
tions in which such angles can be drawn, two of these
positions being shown in the figure at B and C.

Observe the relation between the given angle and the
constructed angles. State result.

49. Experiment XXV. — Relation of angles whose sides
are perpendicular, each to each.

Draw the figure and state result.

50. EXPERIMENT XXVI. — Properties of a perpendicular.
(a) Observe the relative lengths of a perpendicular and

an oblique line drawn from the same given point to the
same given line. State result.

(6) Draw a line and a perpendicular to it; mark a point
in the perpendicular; measure two equal distances on the
given line from the foot of the perpendicular (one on each
side of it); draw two oblique lines from the point in the
perpendicular to the ends of the equal distances.

Observe how the lengths of the two oblique lines compare.
State result.

(c) Make an experiment similar to part (6), in which the
two distances measured on the given line from the foot of
the perpendicular are unequal. State result.

SYMMETRY

51. THE SYMMETRY OF PLANE FIGURES is OF Two
KINDS: — (1) A figure possesses symmetry with respect to an

PLANE GEOMETRY

axis, which is a straight line, when it can be folded upon
the axis so that the part of the figure on one side of the axis
coincides with the part on the other side. The axis is called
the axis of symmetry.

(2) A figure possesses symmetry with respect to a center,
which is a point, when, after rotation about the center
through any integral part of a perigon, it coincides with its
original position. The point is called the center of symmetry.

Symmetry with respect to a center is two-fold, .three-
fold, etc., according to the integral division of a perigon
through which it is rotated into coincidence with the
original position.

Any one element, or symmetrical division, of a figure
is the symmetrical impression of any other element.

EXERCISES

1. Name the kind of symmetry possessed by each of these figures.

2. Draw a square. Draw all possible axes of symmetry. Show
that it possesses two-fold and four-fold symmetry with respect
to a center.

3. Go over your note-book and select all the figures possessing
symmetry of any kind.

4. How many axes of symmetry has a circle? How many kinds
of symmetry with respect to a center has it?

52. EXPERIMENT XXVII. — The law of symmetry of a
point with respect to an axis.

Draw a line AB and a point C without it. Fold the
paper on AB and trace the symmetrical impression C", of
point 0. Unfold the paper; draw CC'.

Observe the relation of CC' to the axis AB.

State this relation as the result of the experiment.

STRAIGHT LINES, CIRCLES, SYMMETRY 23

53. EXPERIMENT XXVIII. — To construct the symmetrical
impression with respect to an axis :

(a) Of a sect.

(b) Of an indefinite straight line.

(a) Find the symmetrical impression of the two end
points of the sect; join them by a straight line.

(b) Find the symmetrical impression of any two points
of the given line, etc.

(c) Find the symmetrical impression of an indefinite
number of points of the given curve; draw a curve con-
taining them.

Test the accuracy of each construction by folding the
paper on the axis.

EXERCISES

1. Construct the symmetrical impression of a perpendicular with
respect to the line on which it stands, as an axis.

2. Construct the symmetrical impression of a line parallel to the
axis.

3. Construct the symmetrical impression of a line intersecting the
axis. Show also how this can be done by copying an angle.

4. Construct the symmetrical impression of a circle with respect to
an axis outside the circle. What is the least number of points
which will determine the symmetrical circle?

54. EXPERIMENT XXIX. — The law of symmetry of a point
with respect to a center :

Mark two points A and B. Fix a piece of transparent
paper with a pin at A ; trace point B upon the transparent
paper. Revolve the transparent paper through a straight
angle; trace the new position of point B on the drawing at
B'. Remove the transparent paper and draw BB'.

Observe the relation of BB' to the center of symmetry A .
State result.

55. EXPERIMENT XXX. — The two-fold symmetrical im-
pression with respect to a center :

(a)0fasect.

\ (b) Of an indefinite straight line .
(c) Of an irregular curve. A

PLANE GEOMETRY

Test the accuracy of each con-
struction by tracing the figure
on transparent paper and re-
volving on the center through
a straight angle.

EXERCISES

1. Construct the symmetrical im-
pression of a sect with respect to
a center at one of its extremities.

2. Construct a design with three-
fold symmetry from the given
element xyz. What must be the

measure of angle a?

3. Are doorways, windows, mantlepieces,
building outlines, symmetrical?

4. Are most pieces of furniture symmetri-
cal? Name some with one or the other
kind of symmetry.

5. Examine any pleasing outline of design, landscape gardening,
architecture, etc. Do all possess symmetry or a modified
symmetry?

6. Examine some good group pictures or statuary. Do they
possess a certain balance of proportion and grouping that may
be regarded as symmetry?

7. Are flowers, leaves, entire plants, symmetrical?

56. EXPERIMENT XXXI. — Geometrical illusions.

The properties of some geometric figures are always
judged incorrectly. Draw the following figures much
enlarged. It is well to make blackboard drawings of some of
them.

(a) Draw four vertical lines, equally spaced.
Rule an oblique line as shown. Do the parts
of the oblique line appear to lie in the same
straight line?

(6) Measure a perpendicular equal in length to
the horizontal line. Do they appear equal?

STRAIGHT LINES, CIRCLES, SYMMETRY

(d!) Draw two equal parallel lines. Attach the
"arrow-head" and " feather" marks at the ends.
State result.

(e) Draw two horizontal parallel lines. From
a point A midway between them, draw radial lines / \
as shown at A', at about equal angles.
State result.

(/) Draw four parallel lines. Draw
sets of short oblique parallel lines
across them. State result.

GENERAL RESULTS. — Our judgment
of geometrical figures may lead us into
errors. The results of the experi-
ments performed in this chapter have been
stated on the assumption, which is probably
a safe one, that nature prefers parallels, per-
pendiculars, equalities, universal laws and
general simplicity.

57. REVIEW EXERCISES

1. Draw opposite angles; bisect one of them; prolong the bisector
through the other angle of the pair. How does it divide the
other angle? State result.

2. Divide a given sect into four equal parts.

3. Construct accurate angles of 90° and 45°, using rule and com-
pass. Test these angle values of your protractor and triangles.

4. Divide a given obtuse angle into four equal parts.

5. Draw with the protractor an angle of 25°. Extend both sides
through the vertex. Calculate and measure the other three
angles thus formed.

6. Draw a triangle ABC and a point D without it. Construct the
two-fold symmetrical impression of ABC with respect to the
center D,

7. Draw a circle with a chord AB. Construct the symmetrical
impression of the circle about the chord as an axis.

8. Which of the printed Gothic capitals, A, B, C, etc., possess

26 PLANE GEOMETRY

symmetry with respect to: (a) a vertical axis; (6) a horizontal
axis; (c) both vertical and horizontal axes; (d) a center and not
to any axes; (e) a center and also one or two axes; (/) oblique
axes? (g) Which letters are asymmetrical (not symmetrical) ?
9. State all the methods that may be used to draw a line through
a given point parallel to a given line. ~

10. Construct a design from the given element xy. B
Use: (1) AB as an axis; (2) AC as an axis; (3) ^"/V
DE as an axis. What kinds of symmetry will

the complete design possess? ^

11. Mark a point on the floor. With a piece of
string find the shortest length from the point to
the wall. Draw this position of the string.
What line will this position be?

12. Draw two sects AB and CD: (a) perpendicular bisectors of
each other, but not equal; (6) so that AB is the perpendicular
bisector of CD, but AB is not bisected; (c) so that each bisects
the other; but not perpendicular; (d) so that they intersect but
neither sect is bisected.

A-

B-

13. Draw two parallel lines as shown. Draw radial lines from
points A and B, terminating in line CD. State the illusion
result.

14. When a draughtsman draws parallels by sliding his T-square
along the edge of his board, what construction principle is
employed?

APPLICATIONS

58. Surveying, Building and Designing. — A surveyor
uses a tape to measure or lay out lines (sects), and a transit
to measure or lay out angles. A carpenter or mason uses a
tape or rule to measure or lay out lines.

The tape is a flexible scale, usually 50 or 100 feet long.
The transit is an accurate portable protractor. The transit
consists of an accurately graduated horizontal circle,
sometimes also a vertical circle or part of such a circle, a

STRAIGHT LINES, CIRCLES, SYMMETRY

telescope provided with two intersecting cross-hairs at the
focal point within the tube of the telescope, spirit levels and
levelling screws for levelling the graduated circle, clamps
and slow motion screws for clamping or rotating the gradu-

ated circle or the telescope, either separately or together,
and a tripod for mounting the instrument.

The transit is set up over the vertex of the angle which
is to be measured or laid out. The graduated circle is

28 PLANE GEOMETRY

levelled and the telescope focussed for cross-hairs and dis-
tance. The telescope is then set with the " upper motion"
so that the index reading of the circle is 0°. While clamped
in this position, the circle and telescope are rotated together
with the " lower motion" until the vertical cross-hair of
the telescope cuts a point which marks one side of the
angle to be measured or laid out. The upper clamp is
loosened and the telescope alone is rotated into the position
of the other side of the angle. The reading of the index
on the graduated circle then gives the value of the angle.
A text on surveying should be consulted for more complete
instructions in the use of a transit, and for the best methods
of measuring and laying out lines on the ground.

Some of the construction principles of the preceding
experiments may be modified, or used without modification
in practical problems.

EXERCISES

1. Show how the angle at a corner of the room may be bisected
by using a piece of "string about ten feet long.

2. Show how the angle of a field may be bisected by using

a tape. The tape must be so used that it is unneces-
sary to draw arcs on the ground. Points may be marked
by stout wire marking pins, stuck in the ground. These
should be made of about &" wire and be about 12"
long.

3. How may a surveyor bisect the length of a curb between two
street corners?

4. How may a surveyor erect a perpendicular to a line by using
a tape?

5. How may a mason erect a perpendicular to a wall by using a
piece of string?

6. Adapt the construction of 31 to the use of a tape; the point
being nearer the line than the length of the tape.

7. How may a surveyor bisect an angle by means of a transit?

8. How may a surveyor erect a perpendicular by means of a
transit ?

9. How may a surveyor lay out in another place an angle equal
to a given angle by using a transit?

10. How may a surveyor lay out a street parallel to another exist-
ing street by making use of the principles of 38 or 44? Explain

STRAIGHT LINES, CIRCLES, SYMMETRY

at what point an angle must be measured with a transit, and at
what other point an equal angle must be laid out.

11. A border for a hardwood floor is to
be laid around a room as shown.
How may the angle bisectors be
obtained where the strips of wood
are fitted together?

12. Sketch a symmetrical design for a
colored glass window for a bath-
room or stairway.

13. Sketch a symmetrical design for a

decorated plate, book cover, lamp shade, rug, wall-paper
pattern, or some other familiar object.

59. STANDARD FORM SHEET FOR RECORDING FIELD WORK.
FIELD WORK

DEPARTMENT OF MATHEMATICS.-

-SCHOOL

•---. c

To lay out a line parallel to the curb of a road or path, and
passing through a given point (at a distance of several hundred
feet from the curb) .

Geometrical Principle Employed. — If two straight lines

are cut by a transversal making a pair of alternate interior

£ angles equal, the lines are

parallel.

AB is the curb and C is
the given point.

(1) Mark any convenient
point D, on the curb, and
set a transit over point D;
measure the angle BDC.

(2) Move the transit to
point C; lay out the angle

DCE equal to the angle BDC', drive stake E.

(3) CE is the required parallel.

NOTE. — If a building or other obstacle is in the way of the
line CE, the angle DCF may be laid out equal to 180°-
angle BDC. The work may be checked by repeating with
another point G on AB.

CHAPTER III
PARALLELS, PERPENDICULARS AND ANGLES

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

60. The Importance of Experimental Investigation. —
The natural world about us is full of facts. We learn these
facts from observation, either at large or in a laboratory.
Many of the broader principles of Chapter I may be con-
sidered as being derived from general observation. We see
types of lines, surfaces and solids, symmetry and motion,
indefinitely repeated in natural objects. The experimental
results which were obtained in Chapter II were those of a
geometrical laboratory. Every science passes through a
certain period of inception during which facts are accumu-
lated, and this is true of chemistry, physics, psychology,
biology, physical geography, astronomy, or any other
science. A fact, or principle is proved by indefinitely
repeating the experiment or observation.

61. The Importance of Analysis and Classification.—
Facts alone are not sufficient to constitute a science. Ex-
perimental results may be incorrectly or imperfectly
stated, due to errors of measurement or of judgment. If,
however, a system of classification can be built up in which
all experimental results are shown to agree with some gen-
eral principles, and thus to agree among themselves, this
very agreement is a further proof that these principles
have been correctly determined by experiment. The
greater the number of independent principles thus success-
fully analysed, the greater will be the certainty that the
system of classification adopted is a true expression of
nature's laws. We will now proceed with the analysis
and classification of the facts or principles which we have
noted in Chapter II.

PARALLELS, PERPENDICULARS AND ANGLES 31

62. The First Step in Classification. — This consists in
the selection of the simplest, and therefore most reliable,
principles as indicated by the experiments. These princi-
ples are either incapable of analysis in terms of still more
elementary principles, or are not readily analyzed, or are
so elementary that it is not worth while to analyze them.

63. Postulates. — These are the elementary principles
selected as the basis of classification. Just what postulates
are sufficient and necessary can only be determined as the
analysis proceeds. It may be found necessary to increase
the number originally selected or some of those selected
may be found not to be required.

POSTULATES

1. Only one sect can be drawn between two given points
(24(6)).

2. Two straight lines can intersect in only one point
(26(a)).

3. A straight line and a circle, or two circles, can inter-
sect in two and not more than two points (26(6) and (c)).

4. A straight line is the shortest distance between two
points.

5. A sect or an angle can be divided into any number of
equal parts, and in only one way (28 and 29) .

6. All straight angles are equal.

7. All right angles are equal.

8. A right angle is one-half of a straight angle.

9. A perpendicular forms a right angle.

10. Only one perpendicular can be drawn to a straight
line at a given point in the line, or from a given point with-
out the line (30 and 31).

11. All radii of the same circle are equal.

12. Any definite figure can be constructed: as a straight
line through a given point; a line through two given points;
a sect of a given length; an angle equal to a given angle; a
line passing through a given point and perpendicular to a
given line; etc. (28 to 35, 38.)

32 PLANE GEOMETRY

64. Axioms. — These are numerical relations which may
also enter into the analysis of geometric principles. They
are general in their application, not being limited to geome-
try, and are derived from reasoning of the most elementary
kind.

Complete the statement of the axioms which are left
incomplete.

AXIOMS

1. Things equal to the same thing are equal to each other.

2. If equals are added to equals, the sums are equal.

3. If equals are subtracted from equals, — — .

4. If equals are multiplied by equals, — — .

5. If equals are divided by equals, — — .

6. If equals are added to unequals, the results are unequal
in the same order.

7. If equals are subtracted from unequals, — — .

8. If unequals are subtracted from equals, — — .

9. If unequals are added to unequals, the greater to the
greater and the less to the less,— — .

10. If, of three unequals, the first is greater than the
second, and the second is greater than the third, — — .

11. The whole is equal to the sum of all its parts.

12. The whole is greater than any of its parts.

13. A quantity may be substituted for its equal in any
expression.

65. The second step in classification is the preparation
of a graded list of those experimental results, or principles,
which can be analyzed in terms of other principles, and
which are sufficiently important to require such analysis
and explanation.

66. Theorems. — A theorem is a principle that is analyzed
in more elementary terms.

A corollary is a theorem which is classified .under another
theorem.

The order of arrangement of theorems can be considerably
modified. All that is necessary is to preserve their con-
secutive dependence.

PARALLELS, PERPENDICULARS AND ANGLES

anal.

hypothesis

hyp.

ax.

postulate

post.

con.

theorem

theo.

cor.

67. Geometric Style. — A certain formal method has
been developed as suitable to written geometrical analysis.

The statement of a theorem is divided into two parts :

(1) The hypothesis, which is a statement of the given, or
constructed, properties of the figure.

(2) The conclusion, which is a statement of the new, or
resulting, properties of the figure.

The analysis (often called the proof) explains how the
conclusion is worked out from more elementary principles.
It consists of a statement of each separate logical step
made, and the reason or explanation of the step.

68. Convenient Abbreviations.

analysis
axiom
conclusion
corollary

69. Theorem I. — Opposite an-
gles are equal (36).

Hypothesis. — Straight lines
AB and CD intersecting and
forming pairs of opposite angles,
x and z, y and w;

Conclusion. — x = z, y = w.

Analysis.

STATEMENT REASON

1. x + y = a straight Z Axiom 10

2. y + z = a straight Z Axiom 11

3. x + y = y + z Axiom 1

4. x = z Axiom 3

Show also that the fact that y — w, is dependent upon
these axioms.

EXERCISES

1. Draw the two straight angles separately which are considered
as making up the composite figure of the theorem.

2. Does the analysis show that the fact that "opposite angles are
equal" is not an independent principle?

34 PLANE GEOMETRY

3. Work out the same kind of analysis for x = z, in which Zs
x and z are combined with Z.w instead of with /.y.

4. In analyzing the second part, y = w, we may write y + x =
a straight Z, w + z = a straight /., y + x = w + z, y= w.
Why may y and w be combined with different angles hi the
first two statements?

5. Upon what must the analysis of the first theorem depend?
Upon what may that of the second theorem depend? Of other
theorems ?

ADDITIONAL THEOREMS

70. These are theorems of minor importance, not assigned
a theorem number.

(1) Complements of equal angles are equal.

Helps. — Draw two equal angles; construct their comple-
ments. State the hypothesis and conclusion. Write
the analysis to show how the equality of the complements
depends upon Postulate 7 and Axioms 1, 3 and 11.

(2) Supplements of equal angles are equal.

71. Theory of Parallels. — The analysis of the principles
of 37, and 39 to 46, cannot be made entirely in terms
of the selected postulates and previous theorems. Two
additional postulates will therefore be stated:

Postulate 13. — Two lines that are cut by a trans-
versal making a pair of corresponding angles equal, are
parallel (39).

Postulate 14. — If two par-
allel lines are cut by a trans-
versal, any pair of correspond-
ing angles is equal (46).

72. Theorem II. — Two lines
that are cut by a transversal
making a pair of alternate-
interior angles equal, are paral-
lel (40).

Hypothesis. — AB and EF
cut by transversal CD making
alternate-interior Zs x and y
equal :

PARALLELS, PERPENDICULARS AND ANGLES 35

Conclusion.— AB\\EF.
Analysis.
STATEMENT REASON

1. x = y Hypothesis (i.e., made equal in

drawing the figure.)

2. x = z Theorem I (a principle previously

analyzed.)

3. y = z Axiom 1.

4. AB\[CD Postulate 13.

EXERCISES

1. Draw separate figures showing the opposite angles and the cor-
responding angles which are considered as making up the figure
of the theorem.

2. Make the analysis using the corresponding Zs x and w.

3. Construct a figure with the other pair of alternate-ulterior angles
equal, and show that a pair of corresponding angles is also equal,
and hence the lines are parallel.

73. Theorem III. — Two lines that are cut by a transversal
making a pair of alternate-exterior angles equal, are parallel
(41).

Construct the figure; state the hypothesis and conclu-
sion. Analyze (a) by using Postulate 13 ; and (6) by using
Theorem II.

ADDITIONAL THEOREMS

74. (1) State the result of the experiment of 42 as a
theorem. Analyze it in more than one way.

Helps. — (1) x -j- y = a straight \ r

Z; why?— (2) x + z = a straight -
Z; Ax— (3) z + y = x + z; Ax —
(4) y = z; — (5) lines are ||.

(2) State the result of the ex-
periment of 43 as a theorem. \
Analyze it in more than one way.

75. Theorem IV. — Two lines perpendicular to the same
line are parallel (37).

Helps. — (1) The right angles are equal. What postulate?

36 PLANE GEOMETRY

(2) These form a pair of angles with the third line considered
as a transversal.

76. A converse theorem is one in which the hypothesis
and conclusion are respectively the conclusion and hypo-
thesis of a previous theorem. The previous theorem is
the direct theorem.

77. Theorem V. Converse of Theorem II.— // two
parallel lines are cut by a transversal, any pair of alternate-
interior angles is equal (46).

State the hypothesis and conclusion.
Analysis.

STATEMENT REASON

1. y = z Postulate 14.

2. x = z What theorem?

3. x = y What axiom?

EXERCISES

1. Draw separate figures showing how the figure of the theorem is
made up of more elementary parts.

2. Make the analysis of Theorem V with some other pairs of alter-
nate-interior and corresponding angles.

78. Theorem VI. Converse of Theorem III (46).

State the theorem. Draw the figure; write out the
analysis in more than one way.

PARALLELS, PERPENDICULARS AND ANGLES

ADDITIONAL THEOREMS

79. (1) State the converse of Theorem 1, 74. Analyze
by the method of Theorem V.

(2) State the converse of Theorem 2, 74. Analyze by
any method.

80. Theorem VII. Converse of Theorem IV.— A line
which is perpendicular to one of two parallels is perpendicular
to the other also (46).

Helps. — Consider the perpendicular as a transversal.

81. Theorem VIII. — Two angles whose sides are parallel,
each to each, are equal if both are acute or both obtuse; and are
supplementary if one is acute and one obtuse (48) .

Hypothesis. — An acute /.x, and acute Zs y and z, with
sides parallel to the sides of Zx.

Conclusion. — x = y and x = z.

Helps. — (a) Prolong BA and EF to form /.w. Notice
that the figure now contains two
pairs of parallel lines cut by trans-
versals, and hence Zs x and y are
both equal to /.w. Why? .'. x = y.
Show also that x = z.

(b) Draw an obtuse angle, and
two other obtuse angles whose sides

are parallel to the sides of the given angle. Show that
these angles are each equal to the given angle.

PLANE GEOMETRY

sides are parallel to the sides of the given angle. Show
that these angles are supplementary to the given angle.

82. Theorem IX. — Two angles whose sides are perpendicu-
lar, each to each, are equal if both are acute or both obtuse,
and are supplementary if one is acute and the other obtuse
(49).

State the hypothesis and conclusion.
Analysis.

REASON
What postulate?

What theorem?
What theorem?
70. Theorem (1)
What axiom?

STATEMENT

1. Draw BG1.BA and BE
±BC

2. BG\\ED and BH\\EF

3. w = y

4. w = x

5. .'. x = y

Analyze also for other angles as in Theorem
VIII (81).

83. General directions to be observed in the analysis of
theorems.

1. Draw large and distinct figures, with all necessary
and no unnecessary lettering.

2. Construct the figure from its given properties.

PARALLELS, PERPENDICULARS AND ANGLES

3. Eliminate all useless statements.

4. Cultivate a correct style.

5. Auxiliary lines may be drawn with any assigned
property. If the line possesses additional properties, it is
part of the analysis to show that these exist. Such lines
are drawn dotted.

ADDITIONAL THEOREMS

84. (1) The bisectors of two supplementary-adjacent angles
are perpendicular.

(2) An angle ABC is bisected by BD; through the vertex
B a line EF is drawn perpendicular to BD. Line EF makes
equal angles with the sides of angle ABC.

State this theorem in general terms.

(3) In AABC, AB>AC; point D is taken on AB so that
AD = AC. ThenBOBD.

(4) The bisectors of alternate-interior angles of parallel
lines are parallel.

(5) The bisectors of corresponding angles of a system of
parallels form another system of parallels.

(6) The bisectors of two pairs of opposite angles formed by
two intersecting straight lines are perpendicular.

(7) The bisectors of a pair of consecutive-interior angles of
parallels are perpendicular.

Helps.— (1) Draw MN\\EF.—(2) x + y = a straight Z;
—(3) s + t = a, right /;— (4) s = v, t = w, etc.

PLANE GEOMETRY

(8) Two lines parallel to the same
line are parallel to each other (47).

Helps.— Hyp.: AB and BF\\CD.
Analysis. — (1) DrawG#; (2) x =
y,y=z; (3) x = z\ (4) AB\\EF.

(9) // one straight line meets
another straight line so that adjacent
angles are equal, the angles are
right angles, and the lines are per-
pendicular.

Helps.— (1) x + y = 180°;—
(2) x = y;—(3)2x = 180°; — (4> x
= 90°;— (5) CD.LAB.

85. REVIEW EXERCISES

1. Are we compelled to follow any certain order in making experi-
ments? Or are the experiments, for the most part, completely
independent of each other? Are the analyses of the results of
these experiments also independent?

2. Mark two points on the surface of a globe, or ball. Will the
straight line containing them lie wholly in the surface?

3. Draw a line on the paper by marking along an edge of your
triangle. Turn the triangle over on this line as an axis. Does
the drawn line still coincide with the edge of the triangle? If
so, what kind of line is it? Has any other line this property?

4. How many straight lines are determined by three points?

5. How many points are determined by three straight lines?

6. Do two points in space determine a straight line?

7. Of what angle is the complement equal to the angle?

8. Of what angle is the supplement equal to the angle?

9. Of what angle is the supplement double the angle?

10. The difference of two supplementary angles is d degrees.
Find the angles.

11. Given a line AB, a point C without it, and an angle x. Find
a point D on AB such that CD makes an angle CD A = x.
Use the rule and compass method.

12. x and 2 x represent any pair of adjacent angles on a figure
formed by a transversal cutting two parallels. Draw the figure
and write on it the measures in degrees of the eight angles.

13. Two parallels are cut by a transversal so that x and y are

PARALLELS, PERPENDICULARS AND ANGLES

adjacent angles and x and 2y are corresponding angles. Sketch
the figure and calculate all the angles in degrees.

14. Draw the right angle of your triangle. Turn the triangle over
using a perpendicular side as an axis. Draw the right angle
again adjacent to the angle first drawn. How does this draw-
ing test the accuracy of the right angle of the triangle?

15. The sum of two adjacent angles is 85°. Find the angle between
their bisectors?

16. The angle between the bisectors of two adjacent angles is 80°
30'. What is the sum of the angles?

17. Reproduce this figure on a larger scale.
In what does the illusion consist?

18. Measure the distance from the vertex of the
right angle of your triangle to the hypotenuse.

19. A country road runs from southeast to northwest. In what
direction would you cross the fields from a point to the west of
the road, to reach it by the shortest way?

20. Would surveyors, architects and navigators be able to make
use of the principles of geometry even if the analysis of experi-
mental principles had never been developed?

21. In what two ways are axioms distinguished from postulates?

22. John and James each received two dollars for spending money
during a holiday. John spent more of his money than James
spent of his. Who has more remaining? What axiom is
illustrated?

23. If the greater quantity of one pair of unequals is added to the
less quantity of another pair of unequals, and the less quantity
of the first pair is added to the greater of the second pair,
which sum will be greater? Can such a result be stated as an
axiom ?

24. Express the axioms algebraically. Thus, Axiom 1; a = b,
a = c, . ' . b = c.

25. Which of the axioms have been used in algebra: (1) In the re-
duction of equations, (2) in the processes of elimination in
simultaneous equations, (3) in the reduction of inequalities?

APPLICATIONS

86. LIGHT REFLECTED FROM MIRRORS.
The law of reflected light: It is proved
experimentally that the angle of inci-
dence, x, equals the angle of reflection,
y, or x' = yf. NN is called the normal
to the plane of the mirror MM.

PLANE GEOMETRY

EXERCISE

Show that if two mirrors,
M and Mr, are parallel, the
incident ray A and the finally
reflected ray B, are parallel.
Help. — Show that a = /3.

87. SURVEYING. — Measurement of an inaccessible angle.
An angle in an inaccessible corner of a field may be measured
as shown in the sketches. Ex-
plain how this is done and the
principle involved.

Bearings. — Point 0 is a point of ob-
servation; SN is a meridian through 0;

the angles x, y, z, w, are the bearings B

of points Ay B, C, D, as seen from 0. A

is northeast of 0, B is northwest, etc.

If x = 35°, the bearing of line OA is read

north 35 degrees east, and is written N

35° E. The length OA is the distance or

course. A point can be located from a

given point when its bearing and distance and also the

direction of the merid-
ian through the given
point are known.

A surveyor may make
a map of a piece of
ground, or of a broken
line, by measuring bear-
ings and distances, first
at point A, next at point
B, etc. .

PARALLELS, PERPENDICULARS AND ANGLES

An angle of elevation is an angle measured in a vertical
plane between a horizontal and an
oblique line, the horizontal line
being below the oblique line.

An angle of depression is an
angle measured in a vertical
plane between a horizontal and
an oblique line, the horizontal
line being above the oblique line

EXERCISES

1. An imaginary line is drawn from the top B of a hill to a point A
in the plain that stretches

away from its base. The
angle of elevation x of the
top of the hill as observed at
A is 35°. What is the angle
of depression y of the point
A as observed at £? Why?

2. Are two vertical lines, determined by plumb-lines
several feet or several hundred feet apart absolutely
parallel? Why not?

3. Draw any meridian SN through a given point X.
Locate a point Y whose bearing from X is S 50° W,
and whose distance from X is 175 feet. Plot on a
scale of 100 feet to the inch.

4. The bank of a river is located by a series of straight lines. Plot
the river bank by drawing an irregular curved line agreeing
closely with the broken line ABCDEF. Such a line is called in
surveying a traverse line.

FIELD-NOTES

Beginning point
of line

Bearing of line from
beginning point

Length of line,
feet

A ...

N 50° E

120

N72°E

S 45° E

150

S 80° W

E..

N25° W

120

PLANE GEOMETRY

88. LATITUDE. — The latitude of a place is determined by
measuring the angle of elevation e, of the sun when it is
directly south, which occurs at, or about, noon.

The angle of elevation of the sun can be most accurately
determined by using an artificial horizon. This is a mirror
accurately levelled, or a shallow disk of mercury; Af,
Figure 2.

In Figure 3, P is the place on the earth of which the
latitude is required, EQ is the equator, NN' is the axis of
the earth, PH is the horizontal plane corresponding to TH
of Figures 1 and 2. The angle QOP is the latitude of P.
This figure shows the position of the sun with respect to
the earth at the instant of equinox, which occurs at some
place on the earth on March 21st and September 23d.

At all times between September 23d and March 21st,

PARALLELS, PERPENDICULARS AND ANGLES

FIG. 5.

the sun is south of the equator, and the angle which a

line OS', or the (approximately)

parallel line PS, makes with the plane

EQ of the earth's equator, is called

the south or minus declination of the

sun. This angle is QOS'.

At all times between March 21st
and September 23d, the sun is north
of the equator, and the angle QOS' A
is called its north or plus declination,
Figure 5.

Commander Peary, Capt. Amund-
sen and Capt. Scott located the poles
of the earth by this or a similar
method.

EXERCISES

1. Show that one-half of the total angle STM, Figure 2, is the
angle of elevation of the sun, STH.

2. How is the angle of elevation E, Figure 3, related to the latitude
angle L, of place P?

3. In Figure 4, draw OA _J_ OS'. Show that the declination angle

D, of the sun, equals angle NO A; and that the angle of elevation

E, of the sun, equals angle AOP. Then L = 90° - (E + D);
and if ZD is given its minus sign, L = (90° — E) + D.

4. In Figure 5, draw OA JL OS'. Show that the declination angle
D = ANOA; and that £E = /.AOP. Then L = 90° +
D - E; and if D is given its plus sign, L = (90° - E} + D, as
in Exercise 3.

5. If the angle of elevation of the sun measured near the Old State
House in Philadelphia, at an instant when the sun was exactly
south, and when its declination was zero, was 50.05°; what is
the latitude of this point?

6. The angle of elevation of the sun when its direction is south,
measured at a place P, was 75° 10'. The declination at this
time was +13° 45'. What is the latitude of the place of
observation?

Another Method of Finding the Latitude of a Place. — The
measuring quadrant illustrated in Figure 6 was first used
by Tycho Brahe, a Danish astronomer (1546-1601), to
measure the angles of elevation of stars. If the angle of

PLANE GEOMETRY

elevation HPS of the pole star (Polaris) is measured at any
time of the year, (Figures 6 and 7) the latitude L of the
place P can be found. The reason of this is that the pole
of the earth extended northwards to its point of intersection
with the " celestial sphere " is very close to Polaris. Polaris,
however, describes a small circle around the pole, and there-
fore the angle of elevation of the star must, in general, be
corrected to obtain the angle of elevation of the pole. This

FIG. 7.

method is troublesome to apply because the crosshairs of
the telescope must be illuminated, since the observation is
made at night. The correction of the angle requires the
use of certain tables published by the Government Printing
Office at Washington, D. C.

EXERCISES

7. Show what arc of the quadrant, Figure 6, is read to give the
value of angle HPS, and show what relation /.HPS has to the
latitude angle L, of Figure 7.

8. The angle of elevation of Polaris is measured at a certain place
= 49° 18'. The correction to be applied to obtain the angle of
elevation of the pole is — 1° 2'. Find the latitude of the place
of observation.

FIELD WORK

89. DEPARTMENT OF MATHEMATICS — — SCHOOL.

To determine the latitude of School.

Geometrical principle employed:

Two acute angles whose sides are perpendicular, each to
each, are equal.

PARALLELS, PERPENDICULARS AND ANGLES 47

METHOD. — (1) Set up and accurately level the transit,
about 15 minutes before noon on March 21st, or September
23d.

(2) Set the horizontal crosshair tangent to the lower
edge of the sun; as the angle of elevation of

the sun increases, follow it up with the
crosshair, until the angle of elevation has
reached its maximum.

(3) Read the vertical angle; level the telescope by the
telescope spirit level and read the index error of the vertical
circle.

(4) Add 0° 15' to the value of the angle of elevation, to
allow for parallax, refraction and the radius

of the sun.

(5) Latitude = 90° — corrected angle of
elevation of sun.

Note: Dark glasses must be fastened in front of the
object glass of the telescope.

To determine the latitude of a place with accuracy within
one or two minutes, the Nautical Almanac must be used.

The latitude of a place can be determined on any day of
the year, if the meridian altitude of the sun is measured on
that day. and the tables found in the Nautical Almanac for
decimation, parallax, refraction and semi-diameter of the
sun, are consulted.

CHAPTER IV
TRIANGLES

PRINCIPLES DETERMINED EXPERIMENTALLY

90. Triangles Defined. — A triangle is a plane figure
bounded by three straight lines.

The sides are the bounding lines.

The perimeter is the sum of the sides.

The interior angles, or the angles, of the triangle are the
angles formed by the sides.

The vertices of the triangle are the vertices of the angles.

An exterior angle is an angle formed by one of the sides
extended and the adjacent side not extended.

The parts of a triangle are the three sides and the three
angles.

EXERCISES

1. Name the triangle of the figure. Name the sides; the angles by
a single letter; the vertices; the exterior angles which are drawn.

2. What angle of the figure is neither an interior nor an exterior
angle? To which angle is it equal?

3. Draw a triangle with all possible exterior angles.

4. What is the relation between an interior and an exterior angle
at the same vertex?

91. Triangles Classified. — The classification is based
upon: (1) the relative length of sides; (2) the size of angles.

TRIANGLES 49

According to sides, a triangle is called:

Equilateral, when the three sides are equal.

Isosceles, when two sides are equal.

Scalene, when no two sides are equal.
According to angles, a triangle is called :

Equiangular, when the three angles are equal.

Right, when one angle is a right angle.

Obtuse, when one angle is obtuse.

Acute, when all angles are acute.

Oblique, when it is not a right triangle.

EXERCISES

1. Draw triangles of each of the eight kinds.

2. Draw triangles of every possible kind, combining a name from
the first list with one from the second; as, obtuse-isosceles; etc.

3. What triangle is the general type!

92. Relations Between Triangles. — Congruent triangles
are triangles which are alike in all respects, and which can
be made to coincide.

Similar triangles have the same shape, or angles, but not
the same size, or lengths of sides. They are considered in
Chapters X and XI.

The homologous parts of congruent or similar triangles are
the correspondingly situated parts.

EXERCISES

1. Draw two congruent triangles. Name all the homologous sides,
angles, vertices, and exterior angles.

2. Draw two similar right triangles. Name the homologous sides
and angles.

93. Position of the Parts of a Triangle.

An angle is included by two sides.
A side is included by two angles.
Two angles are adjacent to each side.
Each side is opposite an angle.
Each angle is opposite a side.

50 PLANE GEOMETRY

EXERCISES

1. Draw a triangle ABC; letter the sides opposite the angles A, B,C,
respectively a, 6, c. This will be called the standard lettering of
a triangle.

2. What angle is included by a and c? What sides include C?
What angles are adjacent to a; to 6? What angle is opposite 6?
What side is opposite C? What side is included by angles A
andC?

Names of Special Parts of Triangles.
In an isosceles triangle:

The base is the unequal side.

The vertex angle is included by the equal sides.

The base angles are adjacent to the base.
In a right triangle :

The perpendicular sides include the right angle.

The hypotenuse is opposite the right angle.

EXERCISES

1. Draw an isosceles triangle. Name all the parts.

2. What angle of the triangle in Exercise 1 is opposite the base?
What angles are opposite the equal sides?

3. Draw a right triangle. Name all the parts.

4. What angles of the right triangle include the hypotenuse?
What angles are opposite the perpendicular sides?

In any triangle:

The base is the side on which it stands; but any side

may be called the base.
• An altitude is a line drawn from a vertex perpendicular

to and meeting the opposite side.
A median is a line drawn from a vertex to the middle
point of (bisecting) the opposite side.'

EXERCISES

5. Draw a triangle with all possible altitudes.

6. In what kind of triangle does one altitude (or more) fall upon a
side produced?

7. In what kind of triangle are some of the sides the altitudes?

8. Draw a triangle with all possible medians. Will the medians of
a triangle ever fall outside the triangle?

TRIANGLES 51

Abbreviations.

altitude alt. isosceles isos.

congruent, is congruent to . . ^ similar ~

homologous horn. triangle, triangles A, A

94. Experiment I. — Properties of the angles of a triangle.

(a) The number of obtuse angles possible in a triangle.
(6) The number of right angles possible in a triangle.

(d) The sum of the angles of a triangle.

(e) The relation between an exterior angle and the
interior angles of a triangle.

Part (a).-^-(l) Draw a triangle with one obtuse angle.
(2) Draw, if possible, a triangle with more than one obtuse
angle. State result.

Part (6). — Make the same experiment for one or more
right angles. Notice that an infinite triangle can be drawn
with two right angles. State result.

Part (c). — How many acute angles may a triangle have?
How many must it have? State result.

Part (d). — Measure the angles of several triangles of diff-
erent shapes with the protractor. State result.

Part (e). — Measure the exterior angles of the triangle of
part (d) and compare their values with the sum of certain
interior angles. State result.

95. EXPERIMENT II. — A method of finding the third angle
of a triangle when two angles are known, without drawing
the triangle.

METHOD (a). — Draw the two given angles, x and y.
Draw the angles x and y adjacent, and find the angle z
which completes a straight angle, z is the third angle
required.

METHOD (6). — Measure the given angles with a pro-
tractor, and find the third angle without drawing it.

52 PLANE GEOMETRY

EXERCISES

1. Two of the angles of a triangle are 75° and 38°. Calculate the
third angle without a drawing.

2. Can a triangle be drawn with the angles 81°, 47°, 52°? With
the angles 35°, 120°, 40°? With the angles 20°, 136°, 12°?

3. Can a triangle be drawn with two (interior) angles 75° and 45°
and the exterior angle at the third vertex equal to 120°.

4. If the interior angles A, B, C of a triangle are 70°, 30° and 80°,
calculate the three exterior angles A', B', Cf, adjacent respectively
to angles A, B, C.

5. If all the angles of a triangle are equal, how large is each?

6. If one angle of a triangle is 126°, and the other two angles are
equal, how large is each exterior angle?

7. If two angles of a triangle are each 72° 30', calculate the third
angle.

8. One angle of a triangle is double another angle, and the third
angle is 20° more than the smallest angle. Calculate the angles
and sketch the triangle.

96. Construction of Triangles from Given Parts. — In
general, three given parts, sides or angles, are sufficient to
construct a triangle. If less than three parts are given,
the triangle is not determined. If more than three parts
are given, the triangle cannot be constructed unless the
additional parts agree by chance or by calculation with the
triangle as determined by three given parts.

Different Combinations of Three Parts of a Triangle.

(1) Three sides.

(2) Two sides and one angle; of which there are three
possible arrangements:

(a) Two sides and the included angle.

(&) Two sides and the angle opposite the greater of the

two sides,
(c) Two sides and the angle opposite the less of the

two sides.

(3) One side and two angles; of which there are three
possible arrangements:

(a) The side included by the two angles.
(6) The greater angle opposite the side and the less
angle adjacent to it.

TRIANGLES 53

(4) Three angles.

Values of the Given Parts. — The lengths of sides may be
given (1) equal to sects which are drawn, or (2) by numerical
measure. The size of angles may be given (1) equal to
drawn angles, or (2) by numerical measure.

97. Experiment III. — A method of drawing a triangle of
which three sides are given.

Given. — Sects a, 6, e, from
which the sides of the triangle
are to be measured. b

METHOD. — (1) Measure CB = a.

(2) Draw two arcs, centers at C and B, with radii equal
respectively to b and c, intersecting each other at A.

(3) Draw AC and AB.

(4) ABC is the required triangle.

Discussion. — (1) If a = b + c, the triangle becomes a
straight line.

(2) If a<b + c, no triangle is possible.

EXERCISES

1. Does the order of placing the sides affect the shape or size of the
triangle?

2. Construct a triangle whose sides are 3.2, 2.4, and 2.7 inches.
Measure the first side from the scale, and measure the two radii
from the scale. Measure the angles of the triangle.

3. Are triangles possible whose sides are (1) 12, 17, 15; (2) 106, 363,
257; (3) 93, 71, 50; (4) 28, 28, 15; (5) 80, 80, 80; (6) 29, 81, 46?

4. Construct an isosceles triangle whose sides are 3, 3 and 4 inches.

5. Construct an equilateral triangle whose sides are each 3.5 inches.

98. Experiment IV.— To draw a triangle of which two
sides and the included angle are given.

Helps. — Draw two sects a and b, and an angle C,
from which the given parts of the triangle are measured.

54 PLANE GEOMETRY

Describe the method as in 97. Are there any special or
impossible cases?

EXERCISES

1. Construct a triangle of which two sides are 2.9 and 1.5 inches,
and the included angle is 57°. Measure the sides from the scale
and the angle directly from the protractor. Measure the remain-
ing side and angles.

2. Are triangles possible with the parts (1) a = 150, b =3, C =
86°; (2) a = 10, b = 10, C = 162°; (3) a = 7, c = 5, B = 215°?

3. Construct an isosceles triangle whose equal sides are each 3.0
inches and whose vertex angle is 70°.

4. Construct a right triangle whose perpendicular sides are 4.2 and
1.7 inches.

99. Experiment V. — To draw a triangle of which two
sides and the angle opposite the greater of the two sides
are given.

Construct the figure, and describe as in 97. Note any
special or impossible cases.

EXERCISES

1. Construct a triangle of which two sides are 2.9 and 1.5 inches,
and the angle opposite the 2.9-inch side is 57°. Measure the
remaining parts.

2. Construct a right triangle of which the hypotenuse is 4.0" and
a perpendicular side is 2. 3".

100. Experiment VI. — To draw a triangle of which two
sides and the angle opposite the less of the two sides are
given. a -i — +-

This is the most com- A"1
plicated construction of
a triangle from three
given parts. Several
cases occur.

(1) Use these given
measurements. Con- FlG- 1-

struct two entirely different triangles (two figures) from
the measurements, the required relative position being
maintained in both figures.

TRIANGLES

(2) Use these given
measurements . What
do you observe in this
case as to the number of
possible triangles and
the kind of triangles ob-
tained?

(3) Use these given
measurements. What

do you observe in this case?

FIG.

FIG. 2.

(4) Draw any two sects
a and 6, with a>6, and a
right angle B. Construct
the triangle, if possible.

(5) Draw any two sects
a and 6, with a>6, and an
obtuse angle B. Construct
the triangle, if possible.

In order to classify these
results, draw a perpendicular in Figures (1), (2) and (3),
from vertex C of the triangle, to side c, as in Figure 4.
Call this the perpendicular distance, p, of the triangle.

RESULT. — When the given angle is acute and (1) when
b>p, two different triangles can be constructed; (2) when
b = pt one right triangle can be constructed; (3) when
b<p, no triangle can be
constructed.

When the given angle is
right or obtuse (the angle
being opposite the less of
the two given sides), no
triangle can be constructed.

Because of the fact that.
two entirely different tri- B
angles can sometimes be
constructed from the same measurements, this construction
is called the ambiguous case in triangles.

FIG. 4.

56 PLANE GEOMETRY

EXERCISES

Construct triangles from the following measurements. Note which
case arises in each construction. Measure the remaining parts in
each figure (in some figures there are two sets of parts).

1. a = 3.2", b = 2.1", B = 25° 4. a = 3.2", b = 1.5", B = 40°

2. a = 3.2", b = 2.1", B = 50° 5. a = 2.7", b = 2.0", B = 90°

3. a = 3.2", b = 1.6", B = 30° 6. a = 2.9", 6 = 1.8", B = 110°

101. Experiment VII. — To draw a triangle of which two
angles and the included side are given.

Construct the figure from given parts and observe special
and impossible cases.

EXERCISES

1. Construct a triangle of which one side = 2.8", and the two angles
including the side are 42° and 70°.

2. Construct an isosceles triangle of which the base is 4.0" and the
base angles are each 35°.

3. Which of the following triangles are (a) possible, (6) special, (c)
impossible? (1) a = 10, B = 50°, C = 75°; (2) a = 75, B =
130°, C = 65°; (3) b = 12, A = 170°, C = 10°.

4. If the given side of a triangle is 2", and the two angles including
the side are supplementary, what is the value of the third angle
of the triangle?

102. Experiment VIII. — To draw a triangle of which one
side and two angles are given, either angle being adjacent
to the given side and the other angle opposite the given side.

Helps. — (1) Find the third angle of the triangle; (2) pro-
ceed as in 101. Notice that two possible arrangements of
the given parts may be specified, but only one triangle
can be drawn for each arrangement.

EXERCISES

1. Construct triangles from the following measurements: (1) a =
3.4", B = 72°, A = 85°; (2) a = 3.4", A = 72°, B = 85°; (3)
b = 1.5", B = 87°, C = 135°.

2. Construct a right triangle with the hypotenuse = 3.5" and an
acute angle = 37°.

TRIANGLES 57

103. EXPERIMENT IX. — To draw a triangle of which the
three angles are given.

Helps. — (1) What relation must exist
between the three given angles? (2)
Observe the number of triangles that
can be constructed with the same given
angles.

EXERCISES

1. Show that when three angles of a triangle are known, no more
is given than when two angles of the triangle are known.

2. Construct an equiangular triangle.

104. Experiment X. — Properties of an isosceles triangle.
Draw an isosceles triangle. Observe:

(1) The relative size of the base angles.

Bisect the vertex angle of the triangle. Observe:

(2) Where the bisector meets the base.

(3) The position of the bisector with respect to the base.

(4) The symmetry of the triangle.

Draw and bisect an exterior angle at the vertex. Ob-
serve :

(5) The position of the bisector.

Extend one of the equal sides of the triangle through the
vertex by a distance equal to the length of a side; join
the extremity of the part extended with the nearer end
of the base. Observe:

(6) The position of the connecting line.

State results, including any other properties of the
triangle which you discover.

105. EXPERIMENT XI. — To draw a perpendicular to a
given line through a given point, (a) when the point is in
the line, and (b) when the point is without the line.

Helps. — Draw a straight line and mark the given point.
Adapt the figure of 104, part (6), to both -constructions.
Thus in (6) , the given line contains the base of the isosceles
triangle, and the given point is at the extremity of the
extended side of the triangle.

58 PLANE GEOMETRY

106. EXPERIMENT XII. — To draw a line parallel to a given
line and passing through a given point.

Helps. — Draw a straight line and mark the given point.
Adapt the figure of 104, part (5). Describe method.

107. Experiment XIII.— Properties of an equilateral
triangle.

108. Experiment XIV. — A triangle with two equal angles.
Helps. — Draw a side of the triangle; and make the two

angles adjacent to the side, equal. Observe the kind of
triangle that is formed.

109. Experiment XV.— A triangle with three unequal
sides.

Helps. — Draw the triangle. Observe the size of the
angles and how they are situated with respect to the sides.

EXERCISES

1. The sides of a triangle are a = 5, b = 7, c = 5. Into what
parts, or segments, does the bisector of angle B divide side 6?

2. The angles of a triangle are 70°, 55°, 55°. What kind of triangle
is it?

3. The sides of a triangle are a = 5, 6 = 6, c = 7; the angles are
about 44°, 78°, 58°. How are these angles situated in the
triangle?

110. Experiment XVI. — A line joining the middle points
of two sides of a triangle.

Helps. — Draw the triangle, mark the middle points of
any two sides; join these points with a line. Observe (a)
the position of this line, and (6) its length as compared
with a side of the triangle.

111. Experiment XVII. — Concurrent lines of a triangle.
Draw acute, right and obtuse scalene triangles. In

each triangle, draw:

(a) The three altitudes.
(6) The three medians.

(d) The three perpendicular-bisectors of the sides.

(e) The bisectors of the three pairs of exterior angles.

TRIANGLES 59

•

Observe. — (a) The concurrence of the altitudes, the
orthocenter; (6) the concurrence of the medians, and where
on each median the point of concurrence is situated, the
centroid] (c) the concurrence of the angle bisectors, and the
special position of the point of concurrence with respect to
the sides of the triangle, the incenter; (d) the concurrence
of the perpendicular-bisectors of the sides, and the special
position of the point of concurrence with respect to the
vertices of the triangle, the circumcenter; (e) the concurrence
of each two exterior angle bisectors with the bisector of the
non-adjacent interior angle (three points), and the special
position of these points of concurrence with respect to the
sides and extended sides of the triangle, the excenters.

112. EXPERIMENT XVIII.— To find the center of gravity
of a triangle.

Help. — This point is found as in 111, part (b).

113. Experiment XIX.— Circles related to a triangle.

(a) The inscribed circle, tangent to the three sides.

(b) The circumscribed circle, containing the three vertices.

Help. — The centers of these circles are found as in 111,
parts (c), (d) and (e).

EXERCISES

1. Draw an isosceles triangle. Draw the altitudes, medians, angle-
bisectors, perpendicular-bisectors of the sides, and bisectors of
the exterior angles. Draw the inscribed, circumscribed and
escribed circles.

2. Draw as in Exercise 1, for an equilateral triangle.

114. REVIEW EXERCISES

Construct the following triangles from the given parts:

1. A right isosceles triangle whose equal sides are given; that is,
equal to a given sect or to a given measurement.

2. An isosceles triangle whose base and equal sides are given.

3. An isosceles triangle whose equal sides and base angles are
given.

4. An isosceles triangle whose base and vertex angle are given.

60 PLANE GEOMETRY

5. A triangle whose altitude and base angles are given.

6. A triangle whose altitude, base, and one base angle are given.

7. An equilateral triangle, whose altitude is given.

8. An isosceles triangle whose base and altitude are given.

9. A triangle of which one side, the opposite angle and the alti-
tude to another side are given.

10. A triangle of which a side, the altitude to the side and the
median to the side are given.

11. (a) A triangle of which two sides are 3.0 and 2.3. inches, and
the altitude to the shorter side is 1.5 in. (6) A triangle with
the same given sides and the altitude to the longer side 1.5 in.

12. A triangle of which a side, the altitude to that side, and the
radius of the circumscribed circle are given.

13. Are any of the constructions of examples 1 to 12 impossible
for certain relative values of the given parts?

14. Construct accurately without a protractor:

(a) an angle of 60°; (d) an angle of 15°;

(b) an angle of 30°; (e) an angle of 75°;

15. Draw a straight line through a given point, such that the per-
pendiculars to the line from two other given points shall be
equal.

16. Draw an angle ABC and a
point P within it. Find a
line DE passing through P,
such that the part of the
line included by the sides
of the angle is bisected at

point P. Help. — Apply a principle similar to that of 110.

17. Draw two sects a and b, and two unequal angles, C and C', of
which C>C". Construct a triangle having two sides and the
included angle equal respectively to a, b, C. Construct a second
triangle having two sides and the included angle equal respec-
tively to a, b, C'. Observe the relative lengths of the third
sides c and c' of the two triangles.

18. Draw four sects a, b, c, c'; of which c>c'. Construct a triangle
having three sides equal respectively to a, b, c. Construct a
second triangle having three sides equal respectively to a, b, c'.
Observe the relative size of the angles C and C'.

19. If a triangle is drawn from the measurements, a = 7, c = 12,
B = 75°, and a second triangle is drawn from the measurements
a' = 7, c' = 12, B' = 105°, how will the sides 6 and 6' of the
two triangles compare in length?

TRIANGLES 61

APPLICATIONS

115. SURVEYING AND ENGINEERING MEASUREMENTS. — In
each of the following problems make a sketch of the figure
with the proper lettering, and write all given values on
this sketch. Then select a suitable scale and construct the
figure accurately by the methods of 97-102. Find the
required values by measurement from the completed figure.

Accurate solutions of these problems may be made by
trigonometry, which is a development of geometry and
algebra for the purpose of calculating the required parts of
triangles from the given parts. A surveyor or engineer
would construct the figures to scale, as is required in these
exercises, as a check on the more accurate calculated values.

The line measurements given are made with a steel tape
and the angle measurements are made with a transit.

EXERCISES

1. A tree, C, is observed from two points, A and B, 1000 feet apart
on a straight road. The angles CAB and CBA are respectively
70° and 45°. How far is the tree from each point? Use a
linear scale either of 500 feet to the inch, or of 200 feet to the
inch.

2. A farm is bounded by three straight roads, the sides of the
farm being 850 feet, 1000 feet, 960 feet. What angles do the
sides of the farm make at the three vertices?

3. A vertical pole, 22 feet high, casts a shadow at mid-day 35 feet
long on a horizontal plane. What is the angle of elevation of
the sun?

4. An aeroplane is directly over a field that is known to be 3 \ miles
distant. The angle of elevation at the point of observation is
10°. How high above the ground is the aeroplane?

5. The distance, AB, across a pond is desired. A point C, is
selected from which CA and CB can be measured 1300 ft. and
890 ft. respectively, and at which the angle ACB is measured
42°. Find AB,

6. The distance of a lighthouse A, from a point B, on shore is
required. A line BC is laid off on shore 1500 ft. long. The
angles between this line and the direction of the lighthouse are
measured at points B and C, 73° and 81° respectivley. How
far is the lighthouse from B?

62 PLANE GEOMETRY

7. Two steamers leave the same port at the same time. One sails
northwest 15 miles an hour; the other east 19 miles an hour.
How far apart are they after 5 hours?

8. A bridge is to span a river. The positions of the abutments
on the banks of the river are marked by stakes A and B. A
stake C is placed on the same side of the river as A, 2460 feet
along the bank; the angles BAG and BCA are measured 80°
and 57° respectively. Find distance AB across the river.

9. The height of a flagpole is desired. At a distance of 130 feet
from the base the angle of elevation of its top is 50°. Find
the height of the pole.

10. The distance between two buildings is 1500 feet, and the dis-
tance of a point of observation from the nearer building is
2400 feet. The angle between lines from the point of obser-
vation to the buildings is 20°. How far is the point of observa-
tion from the other building?

The omission of what word
would make this construc-
tion ambiguous?

11. A surveyor often uses methods
involving several triangles, in
order to find some distance
that cannot be directly mea-
sured. A and B are two

f* n

buoys in a harbor, whose loca-
tions are to be marked on a chart. A line CD is laid out on
shore, and is also drawn to scale on the chart; C being nearer to
A, and D nearer to B. CD is measured 500 feet; angles ACD,
BCD, BDC, ADC, are measured respectively 115°, 45°, 110°,
60°. Plot accurately and measure AB.

CHAPTER V
TRIANGLES

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

116. The First Step in Classification. — The most elemen-
tary principles relating to triangles, based upon the experi-
mental results of Chapter IV, are:

Postulate 15. — Two triangles are congruent if the three
sides of one triangle, are equal respectively to the three
sides of the other triangle (97) .

Postulate 16. — Two triangles are congruent if two sides
and the included angle of one triangle, are equal respectively
to the two sides and included angle of the other triangle
(98).

Postulate 17. — Two triangles are congruent if two sides
and the angle opposite the greater of the two sides of one
triangle, are equal respectively to two sides and the angle
opposite the greater of the two sides of the other triangle
(99).

Postulate 18. — Two triangles are not necessarily congruent
if two sides and the angle opposite the less of the two sides
of one triangle, are equal respectively to two sides and the
angle opposite the less of the two sides of the other triangle
(100).

Postulate 19. — Two triangles are congruent if one side
and any two angles of one triangle, are equal respectively
to one side and the two homologous angles of the other
triangle (101, 102).

Postulate 20. — Two triangles are not necessarily congru-
ent if the three angles of one triangle are equal respectively
to the three angles of the other triangle (103).

These postulates mean that if two triangles are known to

64 PLANE GEOMETRY

have any three parts respectively equal (with two excep-
tions), they are alike in all respects; that is, the remaining
parts are also respectively equal.

117. The Second Step in Classification.— Most of the other
principles relating to triangles can be explained from the
elementary principles, or Postulates, of 116, and from the
previously analyzed and classified principles of Chapter III.

118. Theorem I (of triangles). — The sum of two sides
of a triangle is greater than the third side.

Help. — Apply Postulate 4.

Corollary. — The difference between any two sides of a
triangle is less than the third side.

Help. — Apply Axiom 3 to the conclusion of Theorem I.

119. Theorem II. — The sum of the angles of a triangle is
a straight angle (94).

State the hypothesis and conclusion.

Analysis.— (1) Extend AB to D and draw BE\\AC]
what postulate? — (2) x and y are equal to angles of the
triangle; what postulate and theorem of parallels? — (3) x+y
+ B = a straight Z; what axiom? — (4) substitute.

Write the preceding analysis in the form of 69, 72, etc.
This is one of the most famous theorems, and is believed
to have been analyzed by Pythagoras and Aristotle, as well
as by Euclid.

Corollaries. — 1. A triangle has at least two acute angles
(94).

2. When two angles of a triangle are known, the third
angle can be found.

3. The two acute angles of a right triangle are complemen-
tary.

4. When two angles of one triangle are equal respectively
to two angles of another triangle, the third angles are also
respectively equal.

TRIANGLES

120. Theorem III. — An exterior angle of a triangle equals
the sum of the two non-adjacent interior angles (94) .

Corollary. — An exterior angle of a triangle is greater than
either of the non-adjacent interior angles.

121. Theorem IV. — The base angles of an isosceles triangle
are equal (104).

Hypothesis. — Isosceles triangle ABC.

Conclusion. — B = C.

Analysis.

STATEMENT REASON

1. Draw AD bisecting /.A Postulate 12

2. Compare triangles ABD
and A CD

AB = AC

4. AD = AD

5. Zz = £y

6. AABD *

7. :.B = C

ACD

Hypothesis

Identical

Made so in drawing AD

Postulate 16

Homologous parts of ^ A

The method of analysis employed, is to draw an aux-
iliary line which will form two triangles with other lines of
the figure. Three homologous parts of these two triangles
are known to be equal from the method of drawing the
figure. Therefore the triangles are congruent, or alike in
all respects (with exceptions) ; and therefore any remaining
homologous parts of the two triangles are equal (116).
This method of analysis is much used. It will be found
an aid in understanding congruence of triangles, if equal
parts of the two triangles under consideration are overlaid
with colored crayons. Thus AB and AC may be overlaid
with a red crayon, AD for each triangle with a blue crayon,
and angles x and y with a yellow crayon.

Corollaries. — 1. An equilateral triangle is also equiangular
(107).

Help. — Such a triangle is isosceles in several ways.

2. Each angle of an equilateral triangle is two-thirds of a
right angle, or 60° (107).

PLANE GEOMETRY

122. Theorem V. — The bisector of the vertex angle of an
isosceles triangle bisects the base (104).

Helps. — Show that the two triangles formed in the figure
are congruent by Postulate 16; and therefore two homolo-
gous sides of these triangles are equal.

123. Theorem VI. — The bisector of the vertex angle of
an isosceles triangle is perpendicular to the base (104).

Helps. — Use the method of 121 and 122. Certain angles
of the two triangles are therefore equal. Then apply 84,
Theorem (9); or complete as that theorem is analyzed.

124. Theorem VII. — The median drawn to the base of an
isosceles triangle (a) bisects the vertex angle and (b) is per-
pendicular to the base (104).

125. Theorem VIII. — The altitude drawn to the base of an
isosceles triangle (a) bisects the vertex angle, and (b) bisects
the base (104). E

126. Theorem IX. — The perpendic-
ular-bisector of the base of an isosceles
triangle (a) contains the vertex of the
triangle, and (b) bisects the vertex angle
(104).

Hypothesis. — Isosceles A ABC, with
DE the perpendicular-bisector of BC.

Conclusion. — (a) DE contains point
A; (b) DE bisects ZA.

Analysis.
Part (a) :

STATEMENT

1. As DE may not contain
point A, draw AD
AD.LBC
But DEA.BC
.'. DE coincides with AD
5. .'.DE contains point A
127. A Postulate. Postulate 21. — If two straight lines
coincide, any property of one of the lines belongs to the
other line also.

2.
3.

REASON
Postulate 12

Theorem VII (124)
Hypothesis
Postulate 10
Postulate 21 (see 127)

TRIANGLES

128. A General Theorem of Isosceles Triangles.—
Theorems V to IX may be summarized thus : In an isosceles
triangle, the bisector of the vertex angle, median to the
base, altitude to the base, and perpendicular-bisector of
the base, all coincide in one line.

Corollary. — State a similar theorem for an equilateral
triangle.

129. Theorem X. — // two sides of a triangle are unequal,
the angles opposite the sides are un-
equal, the greater angle being opposite

the greater of the two sides (109).
Hypothesis. — A ABC with a>b.
Conclusion. — A >B.

A «*

Analysis.

STATEMENT

1. Measure CD = AC
and draw AD

2. x = y

3. A>y

4. A >x

5. x>B

6. A>B

REASON
What postulate?

What theorem?
Axiom 12
Axiom 13
What theorem?
What axiom?

Corollary. — // the three sides of a triangle are all unequal
the three angles are all unequal, the size of the angles being
in the same order as that of the sides to which they are opposite
(109).

130. Theorem XI. — // two angles of a triangle are equal,
the sides opposite these angles are equal and the triangle is
isosceles (108).

Helps. — Draw the triangle with two equal angles. Draw
an auxiliary line bisecting the third angle. Show that the
two triangles formed have three parts respectively equal
and are congruent; etc.

Corollary. — An equiangular triangle is also equilateral.

EXERCISE

Of what theorem is Theorem XI the converse?

68 PLANE GEOMETRY

131. Additional Axioms. — Just as the statement of a
new postulate was found necessary in 126 (stated in 127),
so some new axioms will be found necessary in Theorem
XII (133), as there analyzed.

EXERCISES

1. If there are two possible places of concealment of an object, and
one of these places has been examined without finding the object,
where must the object be?

2. If there are three possible places of concealment of an object, and
two of them have been examined without finding the object,
where must it be?

3. How many possible conditions may exist as to the relative size
of two objects, A and B? If, of these three possible conditions,
two have been shown to be impossible, what may we conclude?

This elementary reasoning leads to the statement of
the following axioms:

Axiom 14. — One of two quantities of the same kind may
be larger than, equal to, or smaller than the other.

Axiom 15. — If, of three possible relations of size between
two quantities of the same kind, two cannot exist, then the
third relation exists.

132. Indirect Analysis. — An analysis in which a principle
is established by disproving all other possible conditions,
is called indirect. This method is applied for the first
time in Theorem XII (133).

133. Theorem XII. — // two angles of a triangle are un-
equal, the sides opposite these angles are unequal, the larger
side being opposite the larger angle.

Hypothesis.— AABC with B>C.

Conclusion. — b > c.

Analysis.

STATEMENT REASON

1. Three relations of size are pos- Axiom 14

sible between b and c.
(a) b>c
(b)b = c
(c) b<c

TRIANGLES 69

2. Relation (b), b = c, is impos- Theorem IV
sible since then B would = C

3. Relation (c), b<c, is impossible

Since then B would be < C Theorem X

4. .'. Relation (a) is the true one, Axiom 15
or b>c

Corollary. — State a corollary for three unequal angles,
similar to the corollary of Theorem X.

EXERCISE

1. Of what theorem is Theorem XII the converse?

•

134. CONVERSE THEOREMS OF ISOSCELES TRIANGLES. — A
triangle is isosceles, in which:

(1) The bisector of an angle is perpendicular to the opposite
side.

(2) The bisector of an angle bisects the opposite side.

(3) A median bisects the angle from which it is drawn.

(4) A median is perpendicular to the side to which it is
drawn.

(5) The perpendicular-bisector of a side contains the opposite
vertex.

Help. — All of these theorems may be analyzed by showing
that the two triangles into which the given triangle is
divided, are congruent; therefore, etc.

135. The Position Assigned a Principle in the System
of Analysis and Classification. — All the principles of
Chapter IV which were discovered and proved experi-
mentally, have now been analyzed and classified in this
chapter, except the principles of 110-113. These princi-
ples cannot be analyzed until the logical system has been
carried to a more advanced stage. The system of classifica-
tion has, however, now been developed to the point where
certain principles of Chapter II, 28-32 inclusive, can be
analyzed.

PLANE GEOMETRY

136. Theorem XIII. — The rule and compass construction
for bisecting an angle is correct (28).

Hypothesis.— /.ABC, BD = BE, DF = EF, BF drawn.
Conclusion. — BF bisects Z.ABC; that is x = y.
Analysis.

STATEMENT

1. Compare &BEF and
BDF

2. BE = BD

3. EF = DF

4. BF = BF

5. A are ^

6. .*. x = y

REASON

Hypothesis

Identical

Postulate 15

Homologous parts of ^ A

137. Theorem XIV. — The rule and compass construction
for bisecting a sect is correct (29) .

Helps.— (I) Prove ACAD *
ACBD-, (2) .'. x = y; (3) ACAB
is isos.; (4) .'. CE bisects AB.

EXERCISES

1. Draw the two triangles separately
whose congruence is first proved.
Overlay the equal parts with colored
crayons. Draw the isosceles tri-
angle which is next considered, with
the bisector of its vertex angle.

2. Why cannot &CAE and CBE be proved congruent at once from
the three respectively equal parts, CA = CB, CE = CE, s = tt

3. After finding that x = y, because the A CAD and CBD are con-
gruent, can the &CAE and CBE then be proved also congruent?
Would this method of completing the analysis or that suggested
be better?

TRIANGLES 71

138. Theorem XV. — The rule and compass construction
for erecting a perpendicular to a straight line at a point in
the line is correct (30).

Helps. — Use congruent triangles and 84, Theorem (9.)

139. Theorem XVI. — State a theorem as indicated in
31.

Help. — Analyze by a method similar to that of 137.

140. Theorem XVII. — State the theorem indicated in 32.
Analyze it.

141. VARIATIONS IN THE ANALYSIS OF THEOREMS. — (1)
Analyze Theorem II (119) by drawing an auxiliary line
through C || AB. This was probably the method employed
by Pythagoras.

(2) Analyze Theorem II (119) by drawing lines parallel
to AC and BC from any point in AB.

(3) Analyze Theorem III (120) by employing the two
principles: the sum of the angles of a triangle is a straight
angle; and, an exterior angle and adjacent interior angle are
supplementary.

(4) Analyze Theorem VIII of Chapter III (81) by draw-
ing an auxiliary line through B and E. Two pairs of cor-
responding angles are equal, and therefore angles B and E,
which are the differences of these angles are equal. Simi-
larly for angles B and H.

(5) Analyze Theorem IX of Chapter III (82) by extending
lines BA, BC and EF until they intersect, thus forming two
right triangles containing respectively the angles B and E.
Relations between the angles of these triangles establishes
the equality of angles B and E. A

(6) Analyze Theorem XII (133),

by a direct method, as follows: c

Draw BD so that x = C; then

BD + DA>c: .'. b>c. T~

142. The Third Step in Classification.— A system of
classification in which all known experimental principles
have been assigned a place in a proper logical sequence,
may be extended into the realms of investigation and dis-

PLANE GEOMETRY

covery. Many principles are less evident than those
which form the framework of the system, and would
probably never have become known without the develop-
ment of the analytic, or logical method.

Some of the principles stated in 143 have probably been
thus discovered; as Theorems (8), (14), (15), (16), (17), (18),
(31), (42), and (47) to (53.)

ADDITIONAL THEOREMS

143. (1) The angle formed by two lines drawn from any
point within a triangle to two vertices
of the triangle, is greater than the
third angle of the triangle.

Helps.— Z1>Z2; Z2>Z3; /.
Z1>Z3.
Give reasons.

(2) The bisectors of any two angles
of a triangle intersect.

Helps. — Show that the angle formed by the bisectors is
(a) less than a straight angle, and (6) greater than the third
angle of the triangle. (The preceding theorem.)

(3) Two lines drawn perpendicular respectively to two
intersecting lines, also intersect.

Helps.— Zsl + 2>0°; .'• Zs3
+ 4<180°; .'. Z5>0°; .'. DE
and FG intersect.

(4) A straight line is terminated
by two parallel lines; through its
middle point another straight line
is drawn also terminated by the
parallels. The second line is
bisected by the first line.

(5) Two sects, AB and CD, are
drawn bisecting each other. Then

AC and BD are equal and parallel. State a similar property
of two other lines that may be drawn in the figure.

Helps. — (1) Use congruent triangles to show that AC =

TRIANGLES 73

BD; (2) a pair of homologous angles of these triangles are
equal, and therefore the lines are parallel by 72.

(6) // one of the equal sides of an isosceles triangle is pro-
duced through the vertex by a distance equal to the length of the
side, and the extremity of the produced sect is joined to the nearer
extremity of the base, this line is perpendicular to the base.

Helps. — (1) Two isosceles triangles are formed, having
equal angles (121) ; (2) two of these angles together form a
right angle.

Where has this principle been established experimentally?

(7) Two chords of a circle which are
drawn from the same point on the circum-
ference and which pass through the extrem-
ities of a diameter, are perpendicular.

Help. — Identify the figure with
Theorem (6).

(8) The shortest side of a right triangle whose acute angles
are 30° and 60°, is one-half the hypotenuse.

Helps. — Draw an auxiliary line dividing the right angle
into two parts of 30° and 60°. One of the triangles formed
is equilateral and the other is isosceles.

(9) The exterior angles at the base of an isosceles triangle
are equal.

(10) The hypotenuse of a right triangle is the longest side.

(11) AD±BC, AB = AC = AD. Then
ADBC is a right isosceles triangle.

(12) The bisector of the exterior angle at
the vertex of an isosceles triangle is parallel
to the base.

Where has this principle been established experimentally?

(13) State and analyze the converse of Theorem (12).

(14) The exterior angle at the vertex of an isosceles triangle
is double either (interior) base angle.

(15) State and analyze the converse of Theorem (14).

(16) A B is the diameter of a circle with its center at 0; AC
is a chord', OC is a radius. Then ZBOC = 2 /.BAC.

Help. — Apply Theorem (14).

74 PLANE GEOMETRY

(17) The bisectors of the base angles of an
isosceles triangle form an angle equal to an ex-
terior angle at the base of the triangle.

Help.— Apply 120.

(18) The angle formed by the altitudes drawn to
the equal sides of an isosceles triangle, is supple-
mentary to the vertex angle of the triangle.

(19) In a AABC, ZA is double LB. Then the
bisector of /.A meets BC in a point D such that
AD = BD.

(20) // equal distances are measured in succession on the
sides of a square, and these points are joined,
another square is formed.

Helps. — (1) The figure has equal sides; (2) its
angles are right angles.

(21) One of the equal sides of an isosceles triangle is greater
than one-half the base.

(22) The bisectors of the base angles of an isosceles triangle
form another isosceles triangle with the base.

(23) The bisectors of the exterior angles at the base of an
isosceles triangle form another isosceles triangle with the base.

(24) AABC is isosceles with AB = AC; BD = CE.
Then &BCD * ACBE. A

(25) The medians drawn to the
equal sides of an isosceles triangle are
equal.

Help. — Use congruent triangles (two ways).

(26) The altitudes drawn to the equal sides of an isosceles
triangle are equal.

(27) The bisectors of the base angles of an isosceles triangle
extended to meet the opposite sides, are equal.

(28) State a theorem concerning the lengths of
the medians, altitudes, and angle-bisectors of an
equilateral triangle.

(29) A ABC is isosceles with AB = AC; the
sides are produced so that AD = AE; BE, CD and
DE are drawn. Then BE = CD, and DE\\BC.

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(30) Homologous medians, altitudes and angle-bisectors of
two congruent triangles are equal.

Help. — The smaller triangles formed within the given
triangles, and containing the homologous lines as sides,
may be proved congruent.

(31) Triangle ABC has an
exterior angle C'} at C. The
bisectors of interior angle B and
exterior angle Cf, meet at E.
Angle E = % angle A.

Helps.— ZC" = ZA + Z5;
Zz = Z# + Z.y; ZC" = 2Zz; etc.

(32) Two isosceles triangles are congruent if their vertex
angles and equal sides are respectively equal.

(33) // two sides of one triangle are equal respectively to two
sides of another triangle, but the included angle of the first
triangle is less than the included angle of the other triangle,
then the third side of the first \/_
triangle is less than the third

side of the other.

E d\ F

Hypothesis— & ABC and DEF with a = d, c = f,

Conclusion. — A C < DF.

Analysis.— (1) Place AABC in the position A'EF;—(2)
draw EG bisecting £DEA';— (3) ADEG ^ AA'EG;—(4) . ' .
GD=GA';—(5) but FG + GA'>A'F;—($) .'. DF>AC.

(34) State the converse of Theorem (33.)

Helps. — Use the indirect method of analysis depending
upon Axioms 14 and 15.

(35) The symmetrical impression of a straight line with
respect to a center of symmetry is parallel to the given line.

PLANE GEOMETRY

Helps. — (1) Two triangles are congruent; (2) two angles
are equal; (3) .'. the lines are parallel.

(36) AB\\DE; any trans-
versal BC; BD and BE bisecting
Zs x and y. Then CD = CE.

Helps.— (1) Zl = Z2 and
Z3 = Z4;— (2) .'. &BCD and
BCE are both isosceles; —
(3) /. CD and CE are both equal to BC.

(37) Analyze Theorem X (129)
as follows: — (1) Draw CD bisect-
ing Z C, and make CE = CA;—
(2) A CAD * ACDE;— (3) .'.
Z# = /.A; — (4) .'. it can be shown A
that ZA

(38) The sum of two lines drawn from any point within a
triangle to the extremities of one

of the sides, is less than the sum
of the other two sides of the tri-
angle.

Helps.— (1) In A ACE, AC
+ CE>AD + DE;—(2) in A
DEB, DE+EB>DB;—(3) add these inequalities (Axiom 9) ;
—(4) eliminate DE (Axiom 7) ;— (5) the conclusion follows.

(39) Two oblique lines drawn from
the same point in a perpendicular and
cutting off equal distances from the foot, A
are equal (50).

(40) State the converse of Theorem
(39.)

(41) A perpendicular is the shortest distance from a point
to a straight line (50) .

Help.— Apply 133.

(42) The sum of three lines drawn from any point within a
triangle to the three vertices is (a) less than the sum of the sides,
and (6) greater than one-half the sum of the sides.

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Helps. — (a) Apply Theorem (38) for three positions of the
figure, and add the inequalities; (6) apply 118.

(43) In a quadrilateral A BCD, AB is the longest of the
four sides and CD is the shortest. Then ZC is greater than
/.A, and /.D is greater than /.B.

Help. — Draw a diagonal and apply 129.

(44) The sum of the three exterior
angles of a triangle (one at each vertex)
is a constant value.

Helps. — State the value of each

exterior angle in terms of interior

angles; add; substitute the known value of Zs a + b + c.

(45) // equal segments are laid off
on a straight line and parallels are
drawn through the points of division,
any other line cutting the parallels is
also divided into equal segments.

(46) Any side of a triangle is less
than one-half the perimeter.

(47) An altitude BD, and a me-
dian BE, are drawn to the hypotenuse
of the right triangle ABC. Then

/LDBE = ZA -- ZC.

(48) The bisectors of the exterior angles at
B and C of AABC, meet at D. ZD = 90°

(49) An altitude drawn to any side of a
triangle, divides the angle from whose vertex
it is drawn into two angles whose difference
equals the difference between the other angles
of the triangle.

(50) The perimeter of any quadrilateral is greater than the
sum of the diagonals.

Helps. — Apply 118 to four triangles in
such a way that the sides of the quadrilat-
eral are placed on the greater sides of the
inequalities; add and divide by 2.

78 PLANE GEOMETRY

(51) The sum of the diagonals of any quadrilateral is greater
than ike sum of either pair of two opposite sides.

Helps. — Apply 118 to two of the small triangles in such
a way that the segments of the diagonals are placed on the
greater side of the inequalities; add.

(52) In AABC, AOAB,
BD = CE. Then CD>BE.

Helps. — (1) ^ABC>
LACB; why?— (2) then apply
Theorem (33) of this section,
to the A's DBC and EBC.

(53) The median drawn to a side of a tri-
angle is less than one-half the sum of the other
two sides.

Helps.— (I) AE< what two lines?— (2)
AABD ^ AECD-, etc.

144. The Fourth Step in Classification.—
There is always an uncertainty as to the
correctness of principles whose discovery has depended
entirely upon analysis. For it must be remembered that
they are only an outgrowth of some of the postulates and
previously analyzed principles, or theorems. If, however,
any errors exist in the system of analysis and classification
up to this point, they can best be detected by putting such
deduced principles to an experimental test.
EXERCISES

1. Draw a right triangle whose acute angles are 30° and 60°.
Measure the shortest side and the hypotenuse and test the
relation of 143, Theorem (8).

2. Draw an isosceles triangle, with an exterior angle at the vertex.
Measure this exterior angle and the base angles of the triangle
with the protractor, and test the principle of 143, Theorem (14).

3. Draw a triangle by constructing (1) a base, (2) one of the
angles adjacent to the base, (3) a side any length, (4) an exterior
angle double the given base angle, (5) thus completing the
triangle. Measure the sides and test the^jproperty of 143,
Theorem (15).

4. Test the principle of 143, Theorem (16), by constructing the
figure and measuring the angles with a protractor.

TRIANGLES 79

5. Test the principle of 143, Theorem (17), by constructing the
figure and measuring the angles with a protractor.

6. Test the principle of 143, Theorem (18), by measurement.

7. Test the principle of 143, Theorem (31), by measurement.

8. Test the principle of 143, Theorem (42), by measuring the three
sides of the given triangle and the three lines drawn from the
point within it to the vertices. The sum of the three inner
lines should lie between two certain values, wherever their
point of concurrence is taken.

9. Test the principle of 143, Theorem (47).

10. Test the principle of 143, Theorem (48).

11. Test the principle of 143, Theorem (49).

12. Test the principle of 143, Theorem (50).

13. Test the principle of 143, Theorem (51).

14. Test the principle of 143, Theorem (53).

145. REVIEW EXERCISES

1. One acute angle of a right triangle is 25° 36'. What is the
other acute angle?

2. An exterior angle of a triangle is 98° and a non-adjacent interior
angle is 57°. Find the interior angle at the third vertex.

3. Calculate the third exterior angle of a triangle two of whose
exterior angles are 110° 32' and 75° 41'.

4. The exterior angle at the vertex of an isosceles triangle is 142°
30'. Calculate the base angles.

5. May two sides of one triangle be equal respectively to two
sides of another triangle, and still the two triangles be not
congruent?

6. May two sides of a right triangle be equal respectively to two
sides of another right triangle, and still the two triangles be
not congruent?

7. Does the median of a triangle bisect its angle?

8. What is the value of the angle formed by the bisectors of the
acute angles of a right isosceles triangle?

9. If three sticks are pinned together
with a single rivet at each intersec-
tion, can the triangle formed be
changed in shape without bending

or breaking the sticks? Upon what
geometric principle does the rigidity de-
pend?

10. If four sticks are pinned together in the
same way, is the frame rigid? How can
another stick be added to make it rigid?

80 PLANE GEOMETRY

11. If any point is taken within the triangle whose sides are 7, 11,
15 inches, and lines are drawn from the point to the extremities
of the 11 inch side, between what values will the sum of these
two lines be found?

12. If lines are drawn from the point (Exercise 11) to the three
vertices of the triangle, between what values will the sum of
the three lines be found? Will the sum of the three interior
lines vary with the position of the point?

13. Under which of the Postulates of 116 is each of the following
principles of right triangles included? Two right triangles are
congruent:

(a) If the two perpendicular sides of one triangle are equal re-
spectively to the two perpendicular sides of the other.

(6) If the hypotenuse and one of the perpendicular sides of one
triangle are equal respectively to the hypotenuse and a per-
pendicular side of the other.

(d) If one of the perpendicular sides and the adjacent acute
angle of one triangle are equal respectively, etc.

(e) If one of the perpendicular sides and the opposite acute angle
of one triangle are equal respectively, etc.

14. Under which of the Postulates of 116 is this statement included?
Two isosceles triangles are congruent if the two equal sides and
one of the base angles of the two triangles are respectively
equal.

15. Why is it sometimes necessary to separate the classification of
principles which evidently are closely associated and have been
so considered when they were investigated experimentally?
Give examples of such principles, some of which have been
analyzed in Chapter III and some in Chapter V.

APPLICATIONS

146. Surveying. — Land surveying is largely responsible
for the development of geometric science, which in its turn
has made scientific surveying possible.

EXERCISES

1. A surveyor measured the three angles of a triangular field,
37° 40.3', 81° 55.0', 60° 24.5'. What check exists as to the
accuracy of the measurements?

2. It is often necessary to find the value of an angle which cannot
be directly measured. In 115, Exercise 6, calculate the value

TRIANGLES

of the angle formed by the two lines converging at the light-
house.

3. If the sides of a triangular field have been measured with a
tape, may a map of the field be drawn to scale?

4. In what different ways may the least number of parts of a
triangular field be measured so that a map of the field may be
drawn?

5. AB is a known side of a field A BCD,
etc.; T is a tree whose position is
to be located on the map. What
different methods may be employed

to locate the tree, each giving suffi- A B

cient data for the construction of the triangle ABT to scale?

6. A farm is bounded by four straight lines. What measurements
of lines only may be made in order to construct a map to scale?

7. The distance AB across a pond may be found by laying out on
the ground a triangle congruent to

the triangle of which one side is the A
required distance. Point C is any
point from which AC and BC can
be measured. Explain how the re-
maining points are located, what line
equals the required line A B} and why the triangles are
congruent.

8. This method may be used when the required distance AB is
across a river. At A a right angle is turned

(see 143, Theorem (6), and any distance
AC is measured on the perpendicular. 'An
equal distance CD is measured. At D
another right angle is turned. The sur-
veyor finds a place E in the perpendic-
ular, from which he sights C and B in a
straight line, and drives stake E. What
line is equal to the required line AB?
What triangles are congruent and why?

9. A triangle is sometimes used in surveying,
formed by measuring A B along a given line
= 50 feet, and then holding a 100-foot tape
in the position ACB, with C at the 50-ft.
mark. What is the value of angle A ?

10. A surveyor may extend a line AB beyond

an obstacle, as a building or pond, and also determine the
unmeasured interval BD, by the method indicated in the figure.
The small triangles are constructed as in Exercise 9. What

PLANE GEOMETRY

measurements are made to locate points C and Z>? Why is
line DE an extension of AB1 How is BD found?

11. A line AB may also be extended beyond an obstacle by the
method illustrated in the figure. What measurements must
be taken in addition to the angular measurements indicated?
The distance BD cannot be found by this method except by
constructing the triangle BCD to scale on paper and measuring
the length BD to the scale of the drawing.

The methods described in Examples 7, 8, 9 and 10 are not
employed in high-grade work. More accurate methods are
described in the examples of 116, the required distances being
found by trigonometric calculation.
12. A perpendicular to a line
AB from a point C can
be laid out very accur-
ately as follows: Set any
point D on line AB, or
use the end point B if
not too far away and
visible from C. Set a
transit at D and measure
/.ADC. Set the transit
at C and lay off Z.DCE
= 90° - ZADCjandset
a stake at E on line AB.

TRIANGLES 83

The setting of a stake at E may be done accurately by using
two transits, one at B or D sighted on A, and the other at C sighted
along CE; or as follows: When

transit is set up at B or D set two H -H -H

or more stakes in line A B, as at F,

G, H. Stretch a string over the tops

of the stakes. Then stake E is set to this string, with the

transit set up at C.

147. LIGHT REFLECTED FROM MIRRORS. — The cases of a
single mirror, and of two parallel mirrors, were considered
in 86. The present problems involve principles of triangles.

EXERCISES

1. Show that if two mirrors, M and AT, are perpendicular, the
incident ray A is parallel to the finally reflected ray B.
Help.— Show that a + 0 = 180°.

2. Show that if two mirrors, M and M', are set at an angle of 60°
with each other, that the finally reflected ray B makes an angle
of 60° with ray A.

3. Show that if two mirrors, M and M' , are set at an angle of 45°
with each other, the reflected ray B makes a right angle with the
incident ray A.

PLANE GEOMETRY

Helps.— (1) 2 Z2 + y = 180°

(2) 2 Z3 + z = 180°

(3) Z2 + Z3 = 135°

(4) z + y +z = 180°

Eliminate Z2 and Z3 from Eq. (1), (2) and (3); substitute value of
y + z in Eq. (4).

4. Show that if two mirrors, M and M' ,
are set at an angle 0, the angle a be-
tween the directions of the incident and
finally reflected rays is equal to 2/3.

Helps.— (1) x = x' and y = y'; (2)
a = 2y - 2x (120); (3) 0 = y - x;
(4) /. a = 2/3.

Notice the application of 143, Theorem
(31), to the triangles aM'M and

&M'M>

5. Does the result of Exercise 4 agree
with those of the three preceding
exercises?

148. INSTRUMENTS WHICH MEASURE ANGLES BY THE USE
OF MIRRORS. The Optical Square. — This instrument (Fig .1)
is used for laying out a line perpendicular to a given line.

FIG. 1.

FIG. 2.

The instrument is held at a point A (Fig. 2 on line AB
so that object B can be seen through the windows Q. At

TRIANGLES

the same time the reflection of an object C lying on the
perpendicular AC is seen in the mirror Mf, apparently
in line with B. Upon which of the mirror formations
of 147 does the operation of this instrument depend?

The Sextant. — This instrument is used to measure angles
where rapidity is necessary rather than great accuracy, as
in army reconnoitering. It is also used on boats since a
transit cannot be used
on account of the
motion of the vessel.
It is the only accurate
angle-measuring in-
strument that can be
so used. The instru- M'

ment consists of a
graduated arc of a
circle of about sixty
degrees (hence its
name) , about the cen-
ter, 0, of which moves
an arm D. The arm
D is shown at the
zero reading of the
arc (Fig. 3). In this
position the mirrors
M and M' are parallel.
The mirror M' is fixed to the frame, and is so placed
that a reflection of an object may be seen in it by looking
through telescope T, at the same time that another object
may be seen directly through the unsilvered part of M'.
The mirror M is mounted on the movable arm D.

In order to read an angle BCA, or a, (Fig. 3), the sextant
is held, either horizontally or vertically as the case may be,
so that an object on one side of the angle is seen directly
ahead at B. The arm D is then moved to position D'
so that the object A on the other side of the angle is seen
through the telescope by double reflection from mirrors

FIG. 3.

86 PLANE GEOMETRY

M and M' ', as if it also were straight ahead and coincided in
position with B. Then by 147, Exercise 4, angle /3'=i a.
Since 0 = ft', the angle may be read on the scale ss', and
a = 2 arc sx. The scale ss' is divided into degrees of
one-half size and thus the angle a is read in its true value.

The sextant is used on ship-board to measure the alti-
tude angles of the sun, moon, planets or stars; that is, the
angle from the horizon to the star or to the lower or upper
edge of the sun or moon. From these angles the location
of the vessel in latitude and longitude are calculated.
Dark glasses are provided and must be turned into position
when the, sun is observed.

149. Navigation. — The location of a vessel at sea depends

entirely upon geometric
principles. The distance
travelled by a vessel is
measured directly by a log, which is dragged in the water
by a cord and which registers the distance by the revolution
of a propeller.

Angles are measured with a sextant (148).

EXERCISES

1. The distance of a ship from a lighthouse or other object is deter-
mined by the following method. At any place A, the angle x is
measured; the ship's course is kept in a straight line until a
position B is reached at which angle y is found to equal twice
/.x. The distance A B is measured meanwhile by a log. Prove
that BC = AB. This theorem is contained in 143.

\
\
\

\
\
\
\
\

B A

2. Another method of determining the distance at which a vessel
passes an object C, is to observe position A, at which /.x — 45°,
and position B at which /.y = 90°. What other measurement
is made? How is BC found?

TRIANGLES

These methods are known to navigators as " Doubling
the angle on the bow."

150. Engineering. — All engineering structures such as
cranes, roof and bridge trusses, scaffolding, etc., where
rigidity is necessary, are framed in a system of triangles.

AAAAA

EXERCISE

If a single cross brace as AB, were to be omitted in a framed
structure, would the structure be rigid? Could it then collapse
without breaking a member? (See 145, Exercises 9 and 10.)

151. BUILDING AND CARPENTRY. — Geometric principles
are employed in all construction work.

EXERCISES

1. A carpenter uses the following method to bisect an angle, ABC.
Measure BD = BE; place a carpenter's square so that the
edges GF and GH pass through points D and E, and so that the

•c "

scale readings from G on the two arms are equal. Show that
the method is correct. May the square be turned over so as to
bring the vertex G nearer to point B?

2. A construction similar to that of Exercise 1 may be made by
using your triangle, by laying off scales on the two perpendicular
sides. Bisect an angle by this method.

3. A carpenter may use the following method to bisect angle ABC,

PLANE GEOMETRY

Exercise 1. Measure BD = BE; draw DE; bisect Db meas-
urement at K; draw BK. State the princi-
pie involved.

4. The builder's plumb level is sometimes used
to set two points at the same level. Show
that B and C are at the same level when
the plumb line hangs at the middle point
Z), of the cross-piece.

and AB = AC.

5. A carpenter is sometimes required to cut two timbers x and y,
so that a third timber z shall make equal angles with both of
them. Explain how he may use the two squares and why AB
is the required line on which to cut x and y.

AAEF is isosceles,

6. A board may be divided into any number of strips of equal
width by the method shown. Explain why the perpendicular
line AB is divided equally (143, Theorem (45)).

152. PROBLEMS FOR FIELD WORK. — Suitable problems
depending upon principles previously determined and
classified are:

(1) To extend the curb of a road or path in the same
straight line on the other side of a building that stands on
the line of the road, using the method of 146, Exercise 10,
or of 146, Exercise 11.

(2) To find the distance from an accessible point to a
point on the other side of a stream, wall or street; (a) by
the method of 146, Exercise 8, or (b) by the method of
115, Exercise 8.

In method (a) the required distance is measured on the
ground, while in method (b) the plot is drawn to scale on
paper and the required distance is measured on the drawing.

(3) The plot of a piece of ground bounded by straight

TRIANGLES

lines may be made by the method of 146, Exercise 6.
Trees, etc., may be located by any of the methods of 146,
Exercise 5.

(4) To lay out a tennis court, basket-ball court, etc.

(5) Measure the angles of a triangular field, in order to
check the principle of 94 (d) and 119.

FIELD WORK
153. DEPARTMENT OF MATHEMATICS - —SCHOOL

To Lay out a Tennis Court

Geometrical Principles Employed. — (1) If one of the
equal sides of an isosceles triangle is extended through the
vertex its own length and the end of the extended line is
joined to the nearer end of the base, this line is perpendicu-
lar to the base (104, part (6); 143, Theorem (6)).
(2.) Elementary properties of a rectangle.

--^ /

3 I

//^"""

^-^

^-^^^ *"*•*•

^^^

^'"" A

r /

/2" - ^

METHOD. — (1) Set corner 1; set stake 2 so that 1-2 is
parallel to boundary line; measure 1-3 = 36 feet.

(2) Erect _L's 3-5 and 1-7; measure 3-6 = 1-8= 78
feet.

(3) Check by measuring 8-6 = 36 feet.

(4) Set stakes 9, 10, 11, 12, so that 1-9= 10 -3 =
etc. = 4' 6".

(5) Set stakes 13, 14, 15, 16, so that 9-13 = 18' and
9-14 = 60', etc.

(6) Set other stakes in order.

CHAPTER VI
POLYGONS

PKINCIPLES DETERMINED EXPERIMENTALLY

154. Polygons Defined. — A polygon is a plane figure
bounded by straight lines.

Almost all general descriptive words are used in the same
sense for polygons as for triangles, except that in a polygon
having an even number of sides, two angles or two sides are
opposite, while in a triangle and in polygons having an odd
number of sides, a side and an angle are opposite.

A diagonal is a straight line joining two non-consecutive
vertices.

A convex polygon is one in which no interior angle exceeds
a straight angle.

A concave polygon is one in which at least one interior
angle is reflex.

A regular polygon is both equilateral and equiangular.
Their properties are considered in Chapters XIV and XV.

155. Polygons Named According to the Nximber of
Sides.

A quadrilateral has four sides.
A pentagon has five sides.
A hexagon has six sides.
A heptagon has seven sides.
An octagon has eight sides.
A decagon has ten sides.
A dodecagon has twelve sides.
A pentadecagon has fifteen sides.

Other polygons may also be named from the Greek
numeral signifying the number of sides.

POLYGONS

156. Classification of Quadrilaterals.

quadrilateral

parallelogram

rectangle

square
oblong

oblique
parallelogram

rhombus
rhomboid

trapezoid
trapezium

Quadrilaterals are classified according to the parallelism
of opposite sides:

A parallelogram has both pairs of opposite sides

parallel.

A trapezoid has one pair of opposite sides parallel.
A trapezium has no sides parallel.
Parallelograms are classified according to the angles:
A rectangle has all right angles.
An oblique parallelogram has no right angles.
Rectangles are classified according to relative lengths of
sides :

A square is equilateral.

An oblong has one pair of opposite sides longer than

the other pair.

Oblique parallelograms are classified according to relative
lengths of sides:

A rhombus is equilateral.

A rhomboid has one pair of opposite sides longer

than the other pair.

157. Names of Special Parts of Quadrilaterals.
In a trapezoid :

The bases are the parallel sides.
The median connects the mid-points of the non-
parallel sides.
The altitude is a perpendicular to the bases,

included (comprehended) between them.
An isosceles trapezoid is one whose non-parallel sides are
equal.

In a parallelogram :

The bases are the horizontal sides, if the figure is so

92 PLANE GEOMETRY

drawn; but either pair of parallel sides may be
called bases.

The altitude is a perpendicular to the bases, com-
prehended between them.

168. EXERCISES

1. Draw a trapezium; place capital letters at the vertices. Name
the quadrilateral, reading the vertex letters consecutively.
Draw all possible exterior angles; all possible diagonals. What
parts are opposite?

2. Draw a pentagon, ABODE. What angle is opposite side ABf
What side is opposite /.Ef What angles are adjacent to BC?
What angle is included by BC and CD?

3. Draw a polygon of the general shape
of A BCD. Draw a polygon congruent
to it; a polygon similar to it. Name
one set of homologous sides of the
three polygons; one set of homologous
angles. Draw homologous diagonals
in each polygon.

4. Draw convex polygons of 3 sides ; 4 sides ; 5 sides ; 6 sides . Draw
(if possible) concave polygons of 3; 4; 5; 6 sides. Draw all
possible diagonals of these polygons. What special property
has at least one diagonal of a concave polygon?

5. Draw (approximately) regular polygons of 3; 4; 5; 6 sides.
Draw (if possible) equilateral polygons which are not equi-
angular, and also equiangular polygons which are not equi-
lateral, of 3; 4; 5; 6 sides.

6. What is the least number of sides a polygon may have? the
greatest number?

7. Draw the three kinds of quadrilaterals. Draw the four kinds
of parallelograms.

8. What other names are given to a regular triangle? What
other name has a regular quadrilateral?

9. Draw a trapezoid; the median; the altitude. Name the bases;
the non-parallel sides.

10. Draw an isosceles trapezoid.

11. Draw a trapezoid one of whose non-parallel sides is the altitude.

12. Draw a parallelogram whose sides form its altitudes.

13. Draw the two altitudes of a rhomboid.

159. Experiment I. — The properties of a rhomboid.
Draw a rhomboid by drawing two parallel lines cut by

POLYGONS 93

two other parallel lines at oblique angles and unequal
intercepts. Draw the two diagonals. Observe:

(a) The relation between opposite sides.

(6) The relation between opposite angles.

(d) The various pairs of congruent triangles formed.

(e) How the diagonals divide each other.
(/) The relative length of the diagonals.
State results.

160. Experiment II. — The properties of an oblong.
Observe, in addition to properties (a), (6), (c), (d), (e) of

Experiment I:

(a) That there are four pairs of overlapping congruent

triangles.

(6) The relative length of the diagonals.
State results.

161. Experiment III. — The properties of a rhombus.
Observe :

(a) The position of the diagonals with respect to each

other.
(6) How the diagonals divide the angles of the rhombus.

162. Experiment IV. — The properties of a square.
Observe the properties of the rhomboid, oblong, and

rhombus which belong to a square.

163. Experiment V. — Methods of constructing a rhom-
boid.

Construct a rhomboid :

(a) By drawing two pairs of parallel sides, as in drawing

the figure of Experiment I.

(b) By drawing two pairs of equal sides. Complete the

figure as the other two sides fall.

94 PLANE GEOMETRY

(d) By drawing the two diagonals, oblique, unequal, and
bisecting each other, and completing the figure as
the four sides fall.

164. Experiment VI.— Methods of
constructing.

From the diagonals:
(a) An oblong.
(6) A rhombus.
(c) A square.

165. Experiment VII. — Properties of a trapezoid.
Observe in any trapezoid :

(a) The position of the median.

(6) The length of the median compared with the lengths
of the bases.

(d) Lengths of diagonals.

(e) Angle properties of the figure.
(/) Pairs of congruent triangles.

166. EXERCISES

1. Do the diagonals of a rhomboid or oblong bisect the angles of
the figure?

2. Are the diagonals of all trapezoids equal?

3. Do the diagonals of any trapezoid bisect each other?

4. Two sides of a rhomboid are 21 and 36 feet respectively; what
are the other sides? One of the angles is 70°; what are the other
angles?

5. In what kind of quadrilaterals are the diagonals equal; equal
and perpendicular; unequal and perpendicular; equal without
bisecting the angles of the quadrilateral; perpendicular without
bisecting the angles of the quadrilateral; axes of symmetry of
the quadrilateral? In what kind may only one diagonal bisect
the angles of the quadrilateral? What quadrilaterals have a
center and axes of symmetry? A center of symmetry but no
axes of symmetry?

6. The bases of a trapezoid are 7 and 15 inches; calculate the
median.

POLYGONS 95

167. CONSTRUCTION OF POLYGONS FROM GIVEN PARTS.
EXERCISES

Construct the following figures:

1. A pentagon ABODE; AB = 1.5", Z£ = 130°, BC = 1.0",

ZC = 90°, CD = 2.0", ZD = 150°, DE = 1.6".

2. A pentagon ABODE; AB = 1.2", BC = 2.8", CD = 0.7",
DE = 3.0", EA = 2.2"; diagonals, 5# = 2.7", CE = 3.4".

3. A hexagon in which each side = 1.0", and each angle == 120°.

4. A parallelogram of which two sides and the included angle are
to be equal respectively to two given sects and to one given
angle.

5. A parallelogram of which two adjacent sides and a diagonal
are given.

6. A parallelogram of which one side, one diagonal, and the angle
included between the given side and diagonal are given.

7. A parallelogram of which two diagonals are respectively 3.0"
and 1.7", and their included angle is 50°.

8. A parallelogram of which one side, one diagonal, and the angle
included between the two diagonals are given.

9. A parallelogram of which one side = 1.6", a diagonal =4.0", the
angle included by the two diagnals = 45°.

10. A rectangle of which two adjacent sides are given.

11. A rectangle of which one side and one diagonal are given.

12. A rectangle of which one diagonal and the angle between the
diagonals are given.

13. A rectangle of which one diagonal and an angle between the
diagonal and a side are given.

14. A square of which one side = 2.5".

15. A square of which a diagonal = 1.7".

16. A rhombus of which the two diagonals are respectively 3.0"
and 2.4".

17. A rhombus of which one side and one diagonal are given.

18. A rhombus of which one side = 2.0", and one diagonal = 4.5".

19. A trapezoid of which the two bases and the altitude are given.

20. A trapezoid of which one base, the two adjacent angles, and
altitude are given.

21. An isosceles trapezoid of which one base, one of the non-parallel
sides, and an angle adjacent to the given base are given.

22. A trapezoid of which the median, lower base, and altitude are
given.

23. A trapezium of which the four sides and their order, and a
diagonal (and its position in the figure) are given

96 PLANE GEOMETRY

24. A trapezium of which the two diagonals and their included
angle are given.

25. A trapezium of which the four sides and their order and one
angle are given.

26. A trapezium of which the four sides in order are 2.0", 1.5",
2.4", 3.0", and the angle included by the 2.0" and 3.0" sides
is 80°.

27. Which of the preceding constructions are ambiguous (two
possible figures); which indefinite (any number of possible
figures); which impossible?

CHAPTER VII
POLYGONS

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

168. The Second Step in Classification. — No additional
postulates are required in the analyses of the principles of
Chapter VI, which depend upon the principles of parallels
(Chapter III) and of congruent and isosceles triangles (Chap-
ter V) . The polygons of most importance (other than tri-
angles) are parallelograms and trapezoids. In the Theorems
which follow, the analysis of which depends upon congruence
of triangles, draw the triangles separately and overlay the
equal parts with colored crayons.

169. Theorem I.— A diagonal of a parallelogram divides
it into two congruent triangles.

Helps. — Draw a rhomboid. Show that three parts of the
triangles are respectively equal (116).

170. Theorem II. — (a) Any two opposite angles of a
parallelogram are equal; (b) any two consecutive angles are
supplementary .

Helps.— (a) This follows from 169, or from 81. (b)
This follows from 79, Theorem (l).

171. Theorem III. — The opposite sides of a parallelogram
are equal. c D

Corollary. — Parallels are everywhere
equally distant.

Helps.— (1) AC\\BD, why? (2) AC _
= BD, why? A B

172. Theorem IV. — The diagonals of a parallelogram
bisect each other.

Help. — Be careful to select the triangles whose congru-
ence will prove that the parts of the diagonals are respec-
tively equal.

7 97

98 PLANE GEOMETRY

173. Theorem V. — The diagonals of a rectangle are equal.
Help. — The triangles selected for congruence must have

the entire diagonals for their respective sides.

174. Theorem VI. — The diagonals of a rhombus (a) are
perpendicular, and (b) bisect the angles of the figure.

Help.— Select two of the small adjacent triangles for
congruence; or apply properties of an isosceles triangle.

175. Theorem VII. — State and analyze all the properties
of a square.

CONVERSE PRINCIPLES

176. Theorem VIII. — // both pairs of opposite sides of a
quadrilateral are equal, the figure is a parallelogram (163,
part (6)). A ,B

Hypothesis. — Quadrilateral ABDC with AB = CD and
AC = BD.

Conclusion. — ABDC is a parallelogram; that is, AB\\CD
and AC\\BD.

Analysis.

STATEMENT REASON

1. Draw diagonal BC Postulate

2. Compare &ABC and BDC
3.

Find three respectively

_ equal parts
o.

Give reasons

6. .'. A are ^ What postulate?

7. .'. Zl = Z4 and Z3 = Z2 Horn, parts of ^ A

8. .'. AC\\BD and AB\\CD 72

177. Theorem IX. — // one pair of opposite sides of a
quadrilateral are parallel and equal, the figure is a paral-
lelogram.

Helps. — Construct the figure as in 163, part (c). The
figure may be proved to be a parallelogram by showing that

POLYGONS 99

both pairs of sides are parallel, each to each, as in 176; or
by showing that both pairs of sides are equal, each to each,
since we may apply the principle of 176.

178. Theorem X. — // the diagonals of a quadrilateral
bisect each other, the figure is a parallelogram.

Helps. — Both pairs of opposite sides may be proved
parallel, or the principles of either 176 or 177 may be
applied. Of these, the last mentioned makes the best
analysis.

179. Theorem XI. — // the diagonals of a quadrilateral
bisect each other and are equal, the figure is a rectangle.

Helps. — (1) The figure is a parallelogram (178); (2)
prove two overlapping triangles are congruent; (3) two
consecutive angles of the figure are equal; (4) these angles
are also supplementary; (5) and are therefore both right
angles.

180. Theorem XII. — A line which bisects two sides of a
triangle (a) is parallel to the third side; and (b) equals one-half
the third side.

Hypothesis. —
Conclusion . — (a) .
(6).
Analysis.

STATEMENT REASON

1. Extend DE] draw CF\\BA

2. AADE Q* AEFC

3. DB\\ and = CF

4. DFCB is a parallelogram

5. .*. DE\\BC

6. DE = EF

7. .'. DE = BC

100 PLANE GEOMETRY

181. Theorem XIII. — A line which bisects one side of a
triangle and is parallel to a second side, bisects the third side.

Helps.— (I) Extend DE and draw CF\\BA;~ (2) DFCB is
a parallelogram;— (3) BD = CF,—(4) AADE * AEFC:
—(5) /. AE= EC.

182. Theorem XIV. — A line which bisects one of the
non-parallel sides of a trapezoid and is parallel to the bases
bisects the other non-parallel side.
A B

Helps.— (1) Draw£G smdFH\\AD;—(2) ABGF * AFHC;
—(3) .% BF = FC.

183. Theorem XV. — A line which joins the middle points
of the two non-parallel sides of a trapezoid (i.e., the median)
is parallel to the bases.

A B

Helps.— (1) Draw EG\\AB and DC;—(2)G is middle point
of BC;— (3) F and G coincide (63, Postulate 5);— (4) .'. EF
and EG coincide (63, Postulate 1);— (5) /. EF\\AB and DC
(127, Postulate 21).

184. Theorem XVI.— The median of a trapezoid equals
one-half the sum of the bases.

Helps.— (1) Draw a diagonal; (2) the part of the median
intercepted within each triangle equals J a base; etc.

POLYGONS

185. Theorem XVII. — The medians of a triangle are
concurrent in a point which is two-thirds the length of each
median from its vertex.

Helps. — Part I. — (1) BE and CF intersect in point 0;
—(2) Draw EF', draw GH bisecting BO and CO respectively;
-(3) EF is || and = to GH;— (4) draw FG and EH;— (5)
FEHG is a parallelogram;— (6) .'.' OF = OH and OE = OG;
-(7) /. BO = f BE and CO = f CF.

Part II. (1) Similarly BO' = f BE and AO' = f AD;
—(2) .'. points 0 and 0' coincide, and the medians are
concurrent.

Notice that this is the first of the concurrent lines of a
triangle (111, part (&)) which has been analyzed.

186. The Third Step- in Classification. — The following
properties of polygons (187 and 188) might never have been
discovered if we were forced to rely upon experimental
methods of investigation alone (see 142).

187. Theorem XVIII. — The sum of the interior angles of a
polygon depends upon the number of sides of the polygon and
is independent of its shape.

Draw polygons of different numbers of sides and of differ-
ent shapes; let the number of sides of each polygon be
denoted by n.

Helps. — (1) Mark any point P, within each polygon,
and draw auxiliary lines from P to each vertex; (2) state
the number of triangles formed in each polygon in terms
of n; (3) state the sum of the angles of one triangle using
the straight angle as the unit; (4) state the sum of the
angles of all the triangles in terms of n; (5) deduct the sum
of the angles around point P; (6) the resulting expression
gives the sum of the interior angles of the polygon in

102

PLANE GEOMETRY

terms of n, the number of sides. Express the result also
in degrees.

188. Theorem XIX. — The sum of the exterior angles of a
polygon, one at each vertex, is a constant value and is inde-
pendent of the number of sides of the polygon.

Helps. — (1) State the value of the sum of an interior
angle and an exterior angle at each vertex of the polygon,
using the straight angle as the unit ; (2) state the sum of all
the interior and of all the exterior angles in terms of n;

(3) state the sum of all the interior angles alone (187);

(4) subtract; the difference is the value of all the exterior
angles. Express the result also in degrees.

189. — THEOREM. — Every triangle is isosceles.
This theorem is found in Mathematical Recreations and
Problems, by W. W. R. Ball ; published by Macmillan and Co.

cBS L

? /

\ *

\\
x \
\^x

/
;
/

/
/ t

/>£

// x

/ /

Hypothesis. — Any A ABC.
Conclusion. — A ABC is isosceles.
Analysis.

STATEMENT

1. Draw DG the J. -bisector of BC,
AO bisecting ZA, BO and CO,
GEL AC, OF _L AB.

2. There are four possible cases:
(a) DG and AO do not intersect,

REASON
Postulate

POLYGONS 103

(b) DG and AO intersect at point Z>,

(d) DG intersects AO outside AABC.
Consider Case (a) :

3. Then DG\\AO

4. .\AO±BC

5. .'. A ABC is isosceles 134. Theorem (1)
Consider Case (6) :

6. Then AO bisects BC

7. .*. AABC is isosceles 134. Theorem(2)
Consider Case (c) (Figure 1) :

8. Compare &AOF and AOE

9. AO = AO

10. x = y

11. s = t

12. .'. AAOF £ A40# Postulate 19

13. .'. AF = AE and OF = OE

14. Compare A50F and COE

15. OB = OC

16. OF = OE

17. u = w

18. .'. A£OF ^ C0# Postulate 17

19. .'. BF - C£7

20. /. AF + BF = AE + C# Axiom

21. Or AB = AC

22. .'. AABC is isosceles
Consider Case (d) (Figure 2) :

23. AAOF ± AAOE

24. .'. AF = AE and OF = 0£;

25. ABOF ^ C0£/

26. .'. BF = CE

27. .'. AF - BF = AE - CE

28. Or AB = AC

29. .'. AABC is isosceles

Thus in every case A ABC is isosceles. It has therefore
been proved that no triangle other than an isosceles triangle
can be drawn.

104

PLANE GEOMETRY

190. The Fourth Step in Classification. — The principles
of 187, 188 and 189, which were discovered by making use
of methods of analysis, and not by experiment, as were the
principles of Chapters II, IV and VI, must now be tested
experimentally, and verified or rejected as the result of such
tests shall indicate.

EXERCISES

1. Draw polygons of 4 sides, 5 sides, 6 sides, 7 sides and 8 sides,
respectively, of different shapes, both convex and concave.
Draw one exterior angle at each vertex. Note that the exterior
angle is negative at a vertex where the interior angle is reflex.
Measure with the protractor each interior and each exterior
angle of all the polygons. Find the sum of the interior and of
the exterior angles for each figure, and enter the results in such
a table as the following:

Number of sides
of polygon

Sum of interior
angles measured
in straight angles

Sum of exterior
angles measured
in straight angles

Does the result which may now be stated for the general
polygon of n sides, agree with the values deduced in 187 and 188?
2. May triangles other than isosceles triangles be drawn? Does
the conclusion of the theorem of 189 agree with experiment?
Do you find any flaw in the logical analysis? Either we must
discover wherein the logical error is concealed, or our entire
logical system may be invalidated.

The error will be found in having assumed incorrect, or im-
possible, positions for the point 0. Such an assumption is a Pos-
tulate; which having been found to be false must be rejected or
replaced by a correct Postulate. An accurate construction of
the figure will reveal the nature of the error. State as a
Postulate, thus discovered experimentally, the principle: If
perpendiculars are drawn from the point of intersection of an
angle-bisector and the perpendicular-bisector of the opposite side

POLYGONS 105

of a scalene triangle, to the other two sides of the triangle, these
perpendiculars fall, etc.

ADDITIONAL THEOREMS

191. (1) If the diagonals of a quadrilateral are perpendicular-
bisectors of each other, the figure is an equilateral parallelogram.

(2) // the diagonals of a quadrilateral are perpendicular-
bisectors of each other, and equal, the figure is a square.

(3) // two straight lines are equally distant from each other
at any two points (measured on perpendiculars) they are
parallel.

(4) Analyze Theorem XVII, 187, by dividing the polygons
into triangles by drawing diagonals, but so that no triangles
overlap.

(5) Each interior angle of an equiangular polygon can be
expressed in terms of the number of sides of the polygon.

(6) Each exterior angle of an equiangular polygon can be
expressed in terms of the number of sides of the polygon.

(7) The lines joining in order the middle points of the sides
of any quadrilateral form a parallelogram.

Helps. — Draw one or both diagonals of the given quadri-
lateral. Apply 180 (a) or (6). Note the three possible
variations in the analysis.

(8) The lines joining the middle points of the opposite sides
of any quadrilateral bisect each other.

(9) The lines joining the middle points of the opposite sides
of an oblong form a rhombus.

Help. — The new figure must be shown to be equilateral,
and its diagonals to be unequal.

(10) The diagonals of an oblique parallelogram are unequal.
Help.— Apply 143, Theorem (33).

(11) A parallelogram whose diagonals are unequal is
oblique. A E B

(12) In the parallelogram ABCD, E
and F are the mid-points of AB and
DC.j^Then DE is parallel and equal
toFB.

106

PLANE GEOMETRY

(13) In the figure of Theorem (12), the lines DE and FB
trisect the diagonal AC. A c R

Helps.— (1) In &ABH, GE bisects
AH; (2) similarly FH bisects GC; (3) / /G
/. AG = GH = HC. // >

(14) The bisectors of the angles of a D
quadrilateral form another quadrilateral whose opposite angles
are supplementary.

(15) Corollary of Theorem (14). When the given quadri-
lateral is a parallelogram, that formed by the angle bisectors
is a rectangle.

(16) A convex polygon cannot have more than three acute
angles.

Help.—li an interior angle is acute, what is the exterior
angle at that vertex?

(17) In a triangle ABC, AD is a median prolonged to E
so that DE = AD. Then CE is parallel to AB.

(18) The angles adjacent to either base of
an isosceles trapezoid are equal.

(19) The diagonals of an isosceles trape-
coid are equal.

(20) The sum of the angles at the points
of a five-pointed star is a straight angle.

(21) The perpendiculars drawn from
opposite vertices of a parallelogram to a
diagonal are equal.

(22) Any line drawn through the point of intersection of the
diagonals of a parallelogram, and terminated by the sides is
bisected.

(23) // a diagonal of a parallelogram bisects an angle, the
parallelogram is equilateral.

(24) Analyze Theorem
XVII, 185, as follows:
Draw AABC and medians
BE and CF intersecting at
0; draw AD through 0.
Prove that AD is a me-

POLYGONS

107

dian; i.e., BD = BC; and that 0 is a trisection point on
each median.

Helps.— (1) Prolong AD so that OG = AO and draw GB
and GC;— (2) in AAGC, OE\\GC;—(3) in AABG, OF\\BG;
-(4) .'. OCGB is a parallelogram;— (5) in AAGC, OE =
J GC,— (6) .'. OE = | BO = £ BE;— (7) BD = DC.

(25) // £/ie angles adjacent to either base of a trapezoid are
equal, the trapezoid is isosceles. ^

(26) // the diagonals of a trapezoid

are equal, the trapezoid is isosceles. .

Help. — Draw the altitudes from — \L
points A and B.

(27) Any obtuse angle equals a right angle.
Hypothesis. — A right /.CDA and

an obtuse /.FAD.

Conclusion.— /.CDA = /.FAD.

Helps. — (1) Construct rectangle
ABCD, extend CB to F, lay off AE
= AB, draw CE, draw HO perpen-
dicular-bisector of CB and KO per-
pendicular-bisector of CE, draw OD,
OC, OA, OE.

(2) HO intersects KO at 0; 143, Theorem (3.)

(3) ACDO * AEAO; 3 sides respectively equal; 116,
Postulate 15.

(4) .'. ZCDO = ^EAO.

(5) ZADO = ZDAO; base Zs of isosceles A; 121.

(6) /. ZCDA = /.FAD; 64, Axiom 3.

Find the error in the preceding theorem, which may be
found in Mathematical Recreations and Problems by W. W.
R. Ball; published by Macmillan and Co.

192. REVIEW EXERCISES

1. Why was it impossible to include the analyses of 180, 181 and
186 in the chapter on triangles (Chapter V) ?

2. Two consecutive sides of a parallelogram are 10 and 12 inches
respectively, and the included angle is 75°. Calculate the
remaining sides and angles.

108 PLANE GEOMETRY

3. The bases of a trapezoid are 7 inches and 10 inches. Calculate
the length of the median.

4. The median of a trapezoid is 13.5" and one base is 7.4". What
is the other base? Write a formula for the median m, in terms
of the bases &i and 62.

5. Calculate the sum of the interior angles of a triangle, quadri-
lateral, pentagon, hexagon, octagon, decagon, using the formula
of 187.

6. Calculate the value of each interior angle and of each exterior
angle of equiangular polygons of 3, 4, 5, 6, 8 and 10 sides.

7. If one of the interior angles of an equiangular polygon is 150°,
how many sides has the polygon?

8. Calculate the number of sides of an equiangular polygon if
one of the exterior angles is 18°.

9. How many sides has a polygon if the sum of the interior angles
is ten straight angles? If the sum of the interior angles is 630°?

10. How many sides has a polygon the sum of whose interior angles
equals the sum of the exterior angles?

11. How many sides has a polygon the sum of whose interior angles
is three times the sum of the exterior angles? a times the sum
of the exterior angles?

12. Five of the interior angles of a hexagon are 70°, 135°, 140°, 130°,
45° respectively. What is the other angle?

• 13. Under what circumstances should a postulate, which is sup-
posed to state a geometric principle of the most elementary
and most evident kind, be modified or rejected?

14. Can an argument based logically upon the accepted postulates
and theorems of a science prove a principle which is found to be
contrary to the fact as established by experimental investiga-
tion? Which possesses the higher degree of certainty, logically
deduced or experimental laws?

15. Find out, if possible, what was generally accepted by the scien-
tists of the world as the law of falling bodies before Galileo made
his famous experiments at the Tower of Pisa about 1590. Was
the logically deduced "law" confirmed or destroyed as the result
of these experiments?

16. Construct a regular hexagon each side of which is 1 inch. Cal-
culate the angles, and draw a side and measure an angle alter-
nately.

17. In the story of Sir Galahad will be found a description of the
"pen tangle" which adorned his shield. Design the "pent-
angle" from the description.

18. Make a list of all theorems of Chapters III, V and VII in which
the equality of two or more sects has been shown to be a neces-

POLYGONS 109

sary consequence of the constructed property of the figure.
Make a similar list concerning equal angles.

19. State theorems of congruent parallelograms and quadrilaterals,
in terms of the required number and arrangement of respec-
tively equal sides and angles.

20. Show experimentally that the relation between the number of
sides of a polygon and the total number of possible diagonals
is expressed by the formula, d = (n — l}n — [3 + (3 + 4 +
. . . n)]; where d = number of diagonals and n = number of
sides.

£.

21. Why is the edge AB of the parallel ruler shown in the figure,
parallel to the edge CD in all positions?

APPLICATIONS

193. Physics and Engineering. The Parallelogram of
Forces. — When several forces act simultaneously upon the
same body they may be replaced by a single force which will
produce an effect equivalent to the combined effect of the
acting forces. This single equivalent force is the resultant;
the acting forces are the components. The law of the rela-
tion of resultant and two component forces may be stated
geometrically as follows: If the sides of a parallelogram
are drawn in units representing the values of the component
forces, the angle between the sides being the angle between
the directions of the forces, then the length and direction
of the diagonal of the parallelogram represents the value
and direction of the resultant. This law is determined
experimentally.

110 PLANE GEOMETRY

EXERCISES

1. What lines in the figure represent the component forces?
What line represents the resultant?

2. Two forces of 10 and 20 Ib. respectively, act at an angle of 45°
to each other. Construct the parallelogram of forces to scale
and find the resultant and the angle which the resultant makes
with the direction of the larger force.

3. The resultant of two forces is 30 Ib., and makes an angle of 30°
with one of its components which has a value of 40 Ib. Find
the value of the other component and the angle between the
components.

4. Show how the resultant of three given forces may be obtained
by combining the resultant of two of them with the third. May
this combination be made in any order?

5. Let the resultant, r, of three forces, 4
and two of these forces, a and 6, be ^ /
given in value and direction. Find /

the value and direction of the third D L - w

component.

The parallelogram of velocities is
constructed in the same way as for forces.

6. A boat is travelling northward at a velocity of 20 miles per
hour, and a passenger is walking across the deck towards the
east at a velocity of 4 miles per hour. Find the actual direction
and velocity of the passenger relative to a fixed object.

7. A mail bag is thrown from a train at a right angle to the track,
at a velocity of 10 feet per second. The train is travelling
50 miles per hour. Find the direction and velocity with which
the mail bag travels until it hits the ground.

Help. — The velocity of the train is equivalent to 88 feet per
second.

8. An aeroplane is travelling 80 miles per hour in a northeasterly
direction, while the wind is blowing from the east at a velocity
of 40 miles per hour. Find the direction and velocity in relation
to the earth at which the aeroplane is travelling.

9. A vessel while moving at a rate of 30 miles per hour parallel
to the shore line, fires a shot at a target which is exactly abeam.
The projectile has a velocity of 200 feet per second. In what
direction must the gun be aimed if the shot is to hit the target?

The Polygon of Forces. — Several forces acting simultane-
ously upon the same body may balance each other so that
the resultant effect is zero. The forces are then said to be

POLYGONS

111

in equilibrium. The geometric condition for equilibrium
of a system of forces is that the lines representing in length
and direction the values and directions of the forces shall
form the sides of a polygon when placed end to end in the
same direction. The forces a, b, c in Fig. 1 are in equili-
brium if they form a triangle as in Fig. 2.

FIG. 1.

FIG. 2.

EXERCISES

10. It is evident that the force c balances the resultant r of forces
a and b. Show by construction that force a balances the result-
ant of forces b and c, and also that force b balances the resultant
of forces a and c.

11. Draw a point P and any four forces which are in equilibrium
acting at the point P, no two forces being in the same straight
line.

The principle of the polygon of forces is employed by
engineers in determining the breaking stress in framed
structures such as bridge and roof
trusses, buildings and cranes, (see
figures in 150). From the stress as 2,

thus found, the size of the piece, or
member, is calculated.

12. Let the bridge truss shown be designed to support a total uni-
form load of 40,000 Ib. A force of 10,000 Ib. may then be con-
sidered as supporting each end. The remaining 20,000 Ib.
rests upon the abutments and has no effect upon the truss
stresses. Draw" the polygon (triangle) of forces at the point
A, where the value and direction offeree x and also the direc-

112 PLANE GEOMETRY

tions of forces y and z are known. Determine the amount of
stress in both the upper and lower bridge members.

194. Surveying. — Many uses are found for the principles
of polygons.

EXERCISES

1. A surveyor has measured three angles of a four-sided field:
75° 30.5', 128° 51.2', 135° 3.7'. If these are correct, what is
the fourth angle?

2. A surveyor has measured the angles of a five-sided field; 110°
21.7', 141° 13.3', 148° 54.0', 47° 20.9', 92° 10.6'. What total
error has been made? Distribute the error equally among the
measured angles to find the most probable correct values.

3. If the four sides of a quadrilateral field are measured with a tape
may a map of the field be drawn to scale? What more must be
measured? If no transit is available what other line measure-
ment will determine the field?

'4. Explain the construction of a line through point C parallel to
line AB, by an adaptation of the principles of 163, part (d), and
178. In an out-of-doors problem line AB will be a street curb,
fence, etc., and a stake is to be driven at a point E so that CE
is parallel to AB.

C E

A B

5. A surveyor wishes to determine the line AF, point F being visible
from point A. Explain the method. What parallelogram prop-
erty is made use of in finding AF?

C D

6. Another method of laying out a line through point C parallel to
AB is shown by the figure. Explain how the measurements are
made and upon what principle it depends.

POLYGONS

113

195. Designing. — Triangles, parallelograms and regular
polygons appear in a great variety of artistic designs in
which symmetry is also an essential feature.

EXERCISE

The following designs are for tiling, parquetry flooring, lino-
leums and embroidery. Draw some of these designs to scale, or
copy or originate other similar designs. Suitable coloring may
be used.

XXXX

MMMX

A A A

UHt

I-I I

196. PROBLEMS FOR FIELD WORK.— In these problems the
principles of Chapters VI and VII are employed.

1. To lay out a line beginning at a given point, as a stake driven
in the ground, and parallel to a curb, fence or side of a building,
(a) Use the method described in 194, Exercise 4.
8

114 PLANE GEOMETRY

(6) Use the method described in 194, Exercise 6.
(c) The most accurate method is suggested in 68, Exercise 11.
This may be used as a check on the accuracy of methods (a)
and (6).

2. Measure the angles of a four-sided field (or at four stakes driven
in the ground), in order to verify the principle of 187. Distribute
the total error as in 194, Exercise 21.

197. STANDARD FORM SHEET FOR RECORDING FIELD
WORK.

FIELD WORK

DEPARTMENT OF MATHEMATICS SCHOOL

To lay out a line beginning at a given point, parallel to a
given line:

Geometric Principles Employed. — (1) If the diagonals of a
quadrilateral bisect each other, the figure is a parallelogram
(178).

(2) A line which bisects two sides of a triangle is parallel
to the third side (180).

(3) Two lines which are cut by a transversal making a
pair of alternate-interior angles equal, are parallel (72).

\K „---.''

/ -.-.-*"" V

AB is a curb, fence, side of a building, etc.; C is a point on the required

parallel line.

METHOD (a) :

(1) Drive stakes at D and E on line AB.

(2) Measure EC; divide by 2; measure to middle
point F.

(3) Measure DF; prolong until FG = DF.

POLYGONS 115

METHOD (6) :

(1) Measure DC; prolong until CH = DC.

(2) Measure HE; bisect at K.
METHOD (c).

(1) Measure ADEC.

(2) Lay out Z.ECL = ^DEC.

Points C, K, G, and L should lie in the same straight line.

CHAPTER VIII
LOCI

PRINCIPLES DETERMINED EXPERIMENTALLY

198. A Locus Defined. — A line may be considered as
being made up of an indefinite number of points. When
all the points which compose a line possess some common
property of position, the line is called a locus.

A locus is (in general) a line which is made up of all the
points which satisfy a given condition. The line must
contain all possible points which satisfy the condition and
no other points. A point which moves according to a
fixed law generates a locus or part of a locus.

A locus may also be a group of lines, a line and a point, a
series of detached points, a surface, etc.

199. Experiment I. — To find the locus of points which
are equidistant from two given points.

Draw two points A and B. To find a point of the
required locus, draw two arcs with centers at A and B
respectively, and with any equal radii, which intersect
each other. Obtain about 8 or 10 points in this way.
Draw the line which contains all the determined points.
This line is the required locus, since if all possible points
equidistant from points A and B were found, the points
themselves would form the line.

RESULT. — The locus of points which are equidistant
from two given points is the perpendicular-bisector of the
sect joining the given points.

200. Experiment II. — To find the locus of points which
are at a fixed (or constant) distance from a given point.

Mark the given point A and draw the given fixed distance
BC. Construct the locus by finding a number of detached

116

LOCI 117

points and then drawing the line which an indefinite number
of such points would form. B c

RESULT. — The locus of points which H — H

are at a fixed distance from a given point
is a circle whose center is the given point
and whose radius is the fixed distance.

201. Experiment III.— To find the
locus of points which are at a fixed
distance from a given straight line.

Help. — The locus is composed of two ^
lines.

202. Experiment IV. — To find the locus of points within
an angle which are equidistant from the sides of the'angle.

The perpendicular dis-
tances from each point of
the locus to the sides of
the angle are equal. Such
a point may be found by
trial by using the com-
passes to measure the ~£
equal distances, and plac- A
ing them so that they measure a perpendicular to the side
of the angle, the perpendicular being carefully judged, or
a right triangle being used for accuracy.

203. EXPERIMENT V. — To FIND THE Locus OF POINTS
WHICH ARE TWICE AS FAR FROM ONE OF Two GIVEN POINTS

AS FROM THE OTHER.

Mark two points A and B.

To find a point of the locus, draw any straight line M N;
mark a point DI, on MN, and measure DiDz = MD\.
Draw an arc with center at A and radius = MD\; and a
second arc with center at B and radius = MDz; the points
of intersection of these arcs, D and D', are points of the
required locus.

Repeat with radii MEi, and ME2, which determine the
points E and Ef of the locus. Find enough points to
completely determine the locus, which will be found to be a

118 PLANE GEOMETRY

circle. A second circle may be found if the relative dis-
tances of the points of the locus from A and B are reversed.

M F,D, E, F2 D2 Ez N

The result may be stated if desired, by stating (1) the
form of the locus, (2) the position of the center of the circle,
(3) the radius of the circle. (2) and (3) are stated in
terms of the sect AB.

204. REVIEW EXERCISES

Construct the following loci by finding only sufficient points in
each to determine the required locus.

1. The locus of points which are equidistant from the ends of a sect.

2. The locus of points which are equidistant from two parallel
straight lines.

3. The locus of points which are equidistant from two intersecting
straight lines.

4. The locus of the centers of circles which contain two given points.

5. The locus of the centers of circles of fixed radius tangent to a
given straight line.

6. The locus of the centers of circles of fixed radius which contain
a given point.

7. The locus of points which are equidistant from two concentric
circles.

205. Algebraic Equations Considered as Loci. — Pupils
who have studied graphs and plotting points from their
coordinates, will understand the following geometric inter-
pretation of algebraic equations in two variables,

LOCI

119

-r

EXERCISES

1. Plot a number of points for which x = y. If all possible points
were plotted, they would make up a line bisecting one of the
pairs of vertical angles formed

by the coordinate axes, x —
y = 0 is therefore the equa-
tion of this line, or locus.

2. What is the equation of the *"'
locus of points equidistant «
from the sides of the other -10 -5

pair of vertical angles formed / +5 */0

by the coordinate axes? /^ — 1-5"

3. Combine these equations into >
a single equation with a ^/
double (±) sign. Of what

may this equation be con-
sidered the locus?

4. Write the equation of the locus of points which are at a distance
+5 or —5 from the axis of abscissas.

5. Plot the locus x — 2y = 12; that is, the locus of points of which
the abscissa exceeds twice the ordinate by 12. It is preferable
to use squared paper to facilitate plotting.

6. Plot the locus z2 + y2 = 100. To what locus of this chapter
does the graph correspond?

206. Constructions Depending upon Loci. — The location
of a point which fulfils two independent given conditions
may be determined by drawing the two loci which contain
the required point. The intersection of these loci determines
the point.

It often happens that more than one point fulfils the
given conditions. It is understood that all such points are
required.

Certain peculiarities or special cases of the problem which
may occur should also be noted.

EXERCISES

1. Find a point equidistant from two given points and also equi-
distant from two given parallel lines.

Let A and B be the given points, and CD and EF be the
given parallels.

120

PLANE GEOMETRY

The required point, in order to be equidistant from points A and
B, lies on the perpendicular-bisector of the sect AB; and in order
to be equidistant from the parallels CD and EF, it lies on the
line PR, parallel to them and
equidistant from them. The
point 0, which is the intersec-
tion of the two loci is therefore
the required point.
Special Cases.— (a) If PR is
parallel to MN, there are two
points at infinity.
(6) If PR coincides with MN,
every point of MN is a required
point.

2. Find a point equidistant from

two given points and also equi-distant from the sides of a
given angle.

Help. — The sides of the angle may, if necessary, be extended
through the vertex and the bisector of the angle may therefore
also be extended through the vertex.

3. Find a point equidistant from two given points and situated
upon a given straight line.

4. Find a point equidistant from two given points and also at a
given (fixed) distance from a given straight line.

Help. — There are in general two points which fulfil these condi-
tions, and several special cases.

5. Find a point equidistant from two given points and situated
on a given circumference.

Help. — There are in general two points. Special cases occur
(a) where there is one point, and (6) where there is no point.

6. Find a point equidistant from the sides of an angle and situated
upon a given straight line.

7. Find a point equidistant from two given points and at a given
distance from a third given point.

8. Find a point equidistant from three given points.

Help. — Find a point equidistant from points (1) and (2), and
also equidistant from points (2) and (3).

9. Find a point within a triangle and three points outside the
triangle, which are equidistant from the three sides of the tri-
angle (or sides extended).

10. Find a point equidistant from two intersecting straight lines
and situated upon a third given straight line.

LOCI

121

11. Find a point equidistant from two intersecting straight lines
and situated upon a given circumference.

Help. — Besides the general case, there are five special cases.

12. A pirate buried his treasure 250 feet from an oak tree O, and
70 feet from the line marked by two

poplars P and P'. A sailor who discovered
the pirate's chart and directions failed to
locate the treasure. What may have been
his error?

13. A tree is 20 feet from a straight fence. A
gardener directs his assistant to plant a
bush 30 feet from both the tree and the

fence. Are the directions sufficient? For what case would
they be sufficient? Draw a sketch.

14. Find a point equidistant from two perpendicular axes, that is,
on the locus x ± y = 0; and also at a distance of 5 units from
the vertical axis, that is, on the locus x = ± 5.

Help. — There are four points.

APPLICATIONS

207. SURVEYING. — A point equidistant from three given
points may be located by the method of 206, Exercise 8.

The lines are bisected by measurement, and the perpen-
diculars are erected either by measuring a right angle with
a transit or by some method in which a tape is used.

208. Higher Mathematics. — Coordinate geometry, de-
scribed briefly in 205, forms the basis of calculus and of a
great deal of applied mathematics, such as engineering,
military science and navigation. An understanding of
loci is therefore a requisite for such professional studies.

209. THE ELLIPSE. — This curve is the locus of points the
sum of whose distances from two given

points is a constant.

To construct an ellipse: (1) draw the
axes AB and CD bisecting each other
at right angles; (2) draw two arcs with
center at C and radii equal to AG,
intersecting AB at points E and F, which are called the
foci; (3) tie two knots in a string at a distance apart
equal to AB; fix the knots with pins at points E and F,

122 PLANE GEOMETRY

and slip a pencil around the line AHKCB keeping the
string taut.

The locus is correctly described since EA + AF = EH
+ HF = EC + CF = etc. = a constant, which = AB.

EXERCISES

1. Construct an ellipse whose axes are 5 and 3 inches respectively.

2. Lay out an elliptical flower bed 24 feet by 12 feet.

3. Draw an elliptical window with symmetrical colored glass design.

4. Draw an elliptical picture-frame, platter, or embroidered doily
or table-cover.

5. Observe occasional elliptical outlines in architecture as arches,
ceilings, column bases, windows. These are, in general, more
pleasing than circular arcs.

CHAPTER IX
LOCI

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

210. The Method Employed. — After it has been deter-
mined experimentally that a certain line (or lines) is made
up of points having some given or assigned property; it is
necessary to show, in order to explain this principle: (a)
that every point in the determined line possesses the
assigned property; and (6) that every possible point which
possesses the assigned property is in the determined line.

If only property (a) were known about a determined line,
the line still might be only a part of the locus. If only
property (6) were known about the determined line, the
entire line might be more than the required locus. If both
properties (a) and (6) belong to the line, then the determined
line is the complete locus.

Only two of the loci determined experimentally in
Chapter VIII are important in advanced analysis. The
four concurrent line properties of a triangle (111) which
have not as yet been analyzed, are explained from the
properties of loci (213-216).

211. Theorem I. — The locus of points equidistant from
the ends of a sect is the perpendicular-bisector of the sect.

Part (a). — Any point on the perpendicular-bisector of a
sect is equidistant from the ends of the sect.

c
123

124

PLANE GEOMETRY

Hypothesis. — Sect AB with perpendicular-bisector CD,
and a point E on CD.

Conclusion. — Point E is equidistant from points A and B;
i.e., AE = BE.

Analysis.— (1) AECA ^ ECB; (2) .'. AE = BE. Write
out the complete analysis in standard form.

Part (6). — Any point which is equidistant from the ends
of a sect is on the perpendicular-bisector of the sect.

Hypothesis. — Sect AB with perpendicular-bisector CD,
and a point F equidistant from A and B; i.e., AF = BF.

Conclusion. — Point F is on CD.

Analysis.— (1) Draw CF;—(2) AFC A * AFCB;—(3) .'.

CF _L AB;— (4) .'. CF coincides with CD and F is on CD.

Write out complete analysis in standard form.

Corollaries. 1. — Two points equidistant from the ends
of a sect determine the perpendicular-bisector of the sect.

Helps. — (1) Each point is on the perpendicular-bisector;
why? (2) .'. the line joining the two points coincides with
the perpendicular-bisector.

2. // a perpendicular to a sect contains a point that is
equidistant from the ends of the sect, it bisects the sect.

Helps. — This may be analyzed by principles of congruent
triangles, or of isosceles triangles, or as follows: — (1) the
perpendicular-bisector of the sect contains the given point;
why? — (2) .'. the two perpendiculars coincide, etc.

LOCI

125

212. Theorem II. — The locus of points within an angle
equidistant from the sides is the bisector of the angle.
State Part (a).

State the hypothesis, and the conclusion (EF = EG)
State Part (6).

State hypothesis (HK = HL); and the conclusion.
' Helps.— (I) Consider &HLB and HKB;—(2) if x = y,
BH bisects /.ABC;— (3) /. BH coincides with BD, and
point H is on BD.

Corollary. — One point equidistant from the sides of an
angle determines with the vertex the bisector of the angle.

213. Theorem III. — The three perpendicular-bisectors of
the sides of a triangle are concurrent.

Hypothesis. — A ABC with perpendicular-bisectors of
sides, DE, FG and HK.

Conclusion. — DE, FG and HK are concurrent.

126

PLANE GEOMETRY

Analysis.

STATEMENT

1. DE and FG intersect at a point 0

2. Pt. 0 is equidistant from pts. A

and B; that is, AO = OB

3. Pt. 0 is equidistant from pts. B

and C; that is, OB = OC

4. .'. Pt. 0 is equidistant from pts. A

and C; that is, OA = OC

5. .'. HK contains pt. 0

REASON

143, Theorem (3)
Locus Theorem I,

Part (a)
Locus Theorem I,

Part (a)
Axiom 1

Locus Theorem I,
Part (6)

, FG and HK are concurrent
Corollary. — Point 0 is equidistant from the three vertices
of the triangle.

214. Theorem IV. — The three angle-bisectors of a triangle
are concurrent.

Helps. — (1) Two of the angle-bisectors intersect at a
point 0; what theorem? (2) point 0 is in the third angle-
bisector.

Write out the complete theorem using statements
identical with those of Theorem III, except that point 0 is
here equidistant from the sides of the angles.

Corollary. — Point 0 is equidistant from the three sides of
the triangle.

215. Theorem V. — The three altitudes of a triangle are
concurrent.

Helps.— (1) Draw GH, HK, KG\\AB, BC, CA respec-
tively;—(2) AD.LKH; why?— (3) KACB is a parallelogram;
.'. KA = BC;—(4) also AH = BC; why?— (5) /. KA =

LOCI 127

AH;— (6) .'. AD bisects KH.—(7) Similarly BE and CF
are perpendicular-bisectors of the sides KG and GH of
AGHK. — (8) Consider now that A ABC is removed, leaving
AGHK and the three perpendicular-bisectors of its sides;
they are concurrent; why?

Write out the complete analysis.

216. THEOREM VI. — The bisectors of each two exterior
angles of a triangle and of the non-adjacent interior angle are
concurrent.

State the hypothesis and the conclusion.

Helps. — (1) AD and BE intersect in a point 0; — (2) which
is equidistant from CA and CB; — (3) .'. CF contains point 0.
Show that there are two other similar points.

State a corollary similar to the corollaries of Theorems
III and IV.

ADDITIONAL THEOREMS

217. (1) If any point of the median to the base of an
isosceles triangle is joined to the ends of the base, another
isosceles triangle is formed.

Help. — The median is the perpendicular-bisector of the
base.

(2) The middle point of the hypotenuse of a right triangle
is equidistant from the three vertices.

Help. — Show that the perpendicular-bisectors of two of
the sides of the triangle are concurrent at this point.

(3) The circumcenter, incenter, orthocenter and centroid
of an equilateral triangle coincide.

(4) The diagonals of a quadrilateral divide it into four

128 PLANE GEOMETRY

triangles whose circumcenters are at the vertices of a
parallelogram.

(5) The perpendicular-bisector of a chord of a circle
contains the center of the circle.

(6) The line of centers of two intersecting circles is a
perpendicular-bisector of the chord joining their points of
intersection.

218. REVIEW EXERCISES

1. Sketch the locus of points which are twice as far from one of
the sides of an angle as from the other: (a) when the points lie
within the angle; (6) when they lie outside of the angle.

2. Construct by points, the locus of the
vertex of an angle of constant size,
whose sides pass through two given
points.

Help. — Use an angle of the drawing -
triangle. Place the triangle in differ-
ent positions such that the sides form-
ing the angle contain the two marked points.

3. If a locus is known to be a straight line, how many points are
required to determine it?

4. Construct the locus of the centers of circles which are tangent
to the two sides of an angle.

5. Find a point (or points) which is equidistant from two given
points and also at a given distance from a given circumference.

6. Find a point which is equidistant from two intersecting straight
lines and at a given distance from their point of intersection.

7. Plot the locus xy = 12, using both plus and minus values of the
coordinates.

8. A stone wall across a farm runs north and south; a second wall
crosses it east to west; 200 ft. east of the point of intersection a
third wall crosses the second from S.E. to N.W. at an angle of
50°. The owner of the farm finds in an old paper: "trap-door
to concealed passage leading to treasure vault, is at point equally
distant from three walls." Draw map and locate the trap-door.
In how many possible places may it be located?

9. The first mathematician to use the coordinate method of plotting
algebraic equations was Rene Descartes (1596-1650). Coor-
dinate geometry is also called Cartesian geometry in his honor.
Find in an encyclopedia some facts concerning Descartes.

CHAPTER X

THE MEASUREMENT OF SECTS, AND SIMILAR
TRIANGLES

PKINCIPLES DETERMINED EXPERIMENTALLY

219. In this chapter we propose to investigate more
complex relationships of sects than those merely of equality
or inequality, these relationships forming an entirely new
group of geometric principles. Some of the more evident
of these principles have certainly been known as early as
2000 B.C., and researches in this field were made by Pytha-
goras about 500 B. C.

220. Measurement of Sects. — A ratio is a quotient arising
from the division of one quantity by another of the same

kind; as r, a -r- 6, a : b.

a c
A proportion is an equality of ratios; as T = ^, a -r- b =

c -T- d, a : b = c : d.

The measurement of a quantity is a ratio of the quantity
measured, to the unit of the same kind. See also Chapter I.

The ratio of two quantities of the same kind is the ratio
of their measures in terms of a common unit.

EXERCISES

1. What units are generally used to measure sects of about the
length of this page? very short sects? the side of a room? geo-
graphical distances? astronomical distances?

2. Measure sects a and b with the unit c. What is the ratio of £ ?

3. Measure the edges of this page in inches. Draw two sects less
than two inches whose ratio equals the ratio of the two edges.
9 129

130 PLANE GEOMETRY

221. Properties of a Proportion. — , = ^, or a : b = c : d,

represents any proportion.

The terms are the numbers a, b, c and d.

The means are the second and third terms, 6 and c.

The extremes are the first and fourth terms, a and d.

The four numbers (quantities, values) are proportional.

The fourth term of a proportion is a fourth proportional
of the three preceding terms.

When the two means are the same, either is a mean
proportional of (between) the first and fourth terms.

A reciprocal of a number is unity divided by the number.

The numbers are reciprocals, as x and -.

EXERCISES

7 14

1. Name the means and extremes of the proportions, g = -g-;

5 : x = 15 : 2; x : x + 1 = 6 : 8.

2. What is the fourth proportional of 5, 8 and 25?

3. What is the mean proportional of 2 and 8; of 1 and 25? Express
the proportions.

4. Write four proportional numbers.

5. Write a series of three equal ratios of which the first ratio is 7,-

6. Write as many simple proportions (two ratios in each) as possi-

a c e
ble from the ratio series, r = -; = j.

7. Combine these proportions into a series, v = - and r = -..

What axiom is used?

1 2

8. Write the reciprocal of 5; of gjof ^; of .25.

222. Division of a Sect. — A sect is divided by a point, or
intersecting line, into two parts or segments.

When the point is between the ends of the sect it divides
the sect internally; when the point is in the sect extended it
divides the sect externally. The segments extend from the
point of division to each end of the sect.

A sect is divided in a given ratio when the ratio of the
segments equals the given ratio.

Two sects are divided proportionally when the four seg-
ments are proportional.

MEASUREMENT OF SECTS, AND SIMILAR TRIANGLES 131

EXERCISES

1. The sect AB is divided internally at C and externally at D.

Read the segments of AB formed by each point of division.

2. Draw a sect 3 inches long. Mark a point which divides it in-
ternally in the ratio of 1 : 2. Mark a point which divides it

internally hi the ratio of „•

3. Divide the sect of Exercise 2 externally in the same ratios.
Help. — Let x = segment DB; then x + 3 = segment DA; and

x _ 1
x + 3 ~ 2*

4. Draw sects of 3 and 4 inches. Divide them proportionally so

that a segment of the 3-inch sect = „• inch.

5. A mile is divided into segments of 2000 and 3280 feet. Divide
an inch proportionally, the segments being calculated to two
decimals.

6. A sect is said to be divided harmonically when it is divided by
two points, one internal and the other external, so that the ratio
of the segments formed by one point of division equals the ratio
of the segments formed by the other point.

Divide a sect AB = 2 inches internally at C so that EC = .5",
and harmonically at D. Calculate DB.
„. DA CA
Help'~~DB = OB'

223. Necessary Accuracy of Experimental Work. — In the
experimental work of this chapter, measure sects to the
nearest hundredth of an inch and angles to the nearest
quarter degree. Drawings must be made very accurate in
order to obtain reliable results.

224. Experiment I. — The ratio of two sects.

Draw two sects of different lengths, each of several
inches.

(a) Find their measures in inches. Calculate the ratio
of their lengths as a decimal, to three places.

(6) Find their measures in centimeters or in any arbi-
trary unit as H H Calculate their ratio.

Observe the relation between the ratios.

132 PLANE GEOMETRY

RESULT. — The ratio of two sects is independent of the
unit of measurement.

225. Experiment II. — A line parallel to a side of a
triangle.

x H ff/

Draw any oblique scalene triangle ABC. Draw a line
parallel to a side of the triangle, intersecting the other two
sides. Measure the segments into which the two sides are

AD
divided by the parallel line. Calculate the ratios ^5 and

7^77. each to three decimals. Compare these ratios. Re-

JtLL

peat with other parallel lines, as FG, HK, MN

RESULT. — A line parallel to a side of a triangle divides
the other two sides proportionally.

EXERCISES

1. The sides of AABC are AB = 2", AC = 3", BC = 2.4". If
DE cuts AB so that AD = 1.5", calculate AE.

2. In the triangle of Exercise 1, MN cuts AC so that NC = 2.2";
calculate MB.

3. In the same triangle, FG cuts AB prolonged so that BF = 0.5;
calculate CG.

4. In the same triangle AH = | AC; calculate AK.

5. If point D is the middle of AB, to what previous simpler prin-
ciple does this principle reduce?

6. The sides of a triangle are 100 ft., 250 ft., 280 ft. A line parallel
to the 280 ft. side cuts the 250 ft, side 60 ft. from its intersection
with the 100 ft. side. At what point does this line cut the 100 ft.
side? From the same point in the 250 ft. side, a line runs par-

MEASUREMENT OF SECTS, AND SIMILAR TRIANGLES 133

allel to the 100 ft, side. At what point does it cut the 280 ft.
side?

7. Two sides of a parallelogram are 10" and 15", and a diagonal is
18". At a point in this diagonal, 3 inches from one end, lines
are drawn parallel respectively to the four sides of the parallelo-
gram. Calculate the points at which these lines intersect the
sides.

226. Experiment III. — The position of a line which
divides two sides of a triangle proportionally.

Draw any triangle ABC; mark a point D on AB. Meas-
ure AD, AB and AC. Calculate the position of a point E

on AC such that T7*=T7?' Measure AE and mark

point E; draw DE. Observe the position of line DE.

RESULT. — A line which divides two sides of a triangle
proportionally, etc.

EXERCISES

1. The sides of a triangle are 4", 6", 8". A line cuts the 4" and 6"
sides at points 1" and 1.5" respectively from their point of inter-
section. Is this line parallel to the 8" side?

2. Two adjacent sides of a rectangular room are 10 ft. and 15 ft.
A line joins two points on the sides of the room, measured
respectively 3 ft. 6 in. and 5 ft. 3 in. from a corner of the room.
Is this line exactly or approximately parallel to a diagonal of
the room.

227. Similar triangles are defined in 92.

228. Experiment IV. — The relation of the sides of similar
triangles.

134 PLANE GEOMETRY

Draw any oblique scalene triangle; measure the three
angles with a protractor. Draw a second triangle of a
different size whose angles are equal to the angles of the
first triangle. Measure all the sides of both triangles.

Calculate the ratio of each pair of homologous sides, — , ^, —f,

to three decimals. Observe the relation between these
ratios. State result.

EXERCISES

1. The sides of a triangle are 3", 5", 5.4". The side of a similar
triangle, homologous to the 3" side, is 6". Calculate the other
sides of the second triangle.

2. The sides of a AXYZ are x = 27", y = 31", z = 22". The
side y' of a similar AX'Y'Z' is 42". Calculate sides xf and z'.

229. Experiment V. — Triangles whose sides are pro-
portional.

Draw any triangle; measure the three sides. Multiply
each side by the same number, as .0.73, or 1.5, etc. Con-
struct a second triangle from a set of three sides thus
obtained. Measure the three angles of both triangles.
Observe the kind of triangles and state result.

EXERCISES

1. The sides of two triangles are respectively 7", 5", 3.4", and
21", 15", 10.2". Are the triangles similar?

2. The sides of two triangles are respectively 4.5 ft., 5.8 ft., 7.1 ft.,
and 5.4 ft., 6.96 ft., 9.94 ft. Are the triangles similar?

3. If a triangle is drawn with sides a = 6", b = 8", c = 10", the
angles will be found to measure approximately, A = 37°, B =
53°, C = 90°. Obtain the values of the angles of a triangle
DEF, of which the sides are d = 15", e = 25", / = 20", without
constructing the triangle.

230. Experiment VI. — Triangles with two sides pro-
portional and the included angles equal.

Draw any triangle; measure two sides and the included
angle. Construct the second triangle. Measure the other
two angles and the third sides of both triangles. Observe
(a) the kind of triangles obtained, and (b) the ratio of the
third sides.

MEASUREMENT OF SECTS, AND SIMILAR TRIANGLES 135

EXERCISES

1. If two triangles are constructed from the parts, respectively,
a = 5", b = 7", C = 40°, and a' = 10", V = 14", C" = 40°;
are the triangles similar?

2. If A ABC is constructed from the values a = 10", 6 = 30",
C = 45°, the remaining parts will be found to be approximately,

c = 24", A = 17°, B = 118°. In a ADEF, d = 15", / = 45",
E = 45°. Determine without constructing the triangle, the
values of parts e, D and F.

231. Projection. — The projection of a point on a line is the
foot of the perpendicular from the point to the line. The
given line is the line of projection.

The projection of a sect on a given line is the sect included
by the projections of its end points.

EXERCISES

1. Draw a point A and a line BC; pro-
ject A upon BC. Draw a sect DE;
project DE upon BC. Name the
line of projection.

2. When does the projection of a sect
become a point? When is it equal

to the sect? Is it ever greater than B ~
the sect?

3. Draw a sect AB intersecting a line CD at A. Construct the
projection of AB on CD.

4. Draw two intersecting sects; con-
struct the projection of each on
the other.

5. Draw a triangle: project each two
sides on the third side.

6. In what kind of triangle do the projections of any two sides upon
the other side fall within the side upon which the projection is
made? In what kind of triangle do the projections of some
sides fall on a side prolonged? In what kind of triangle is the
projection of one or more sides a point?

7. In every triangle, how do the projections of any two sides upon
the third side compare with the length of the third side?

8. Construct a sect AB = 3", at such an angle to line CD that its
projection on CD = 2.2".

136 PLANE GEOMETRY

232. EXPERIMENT VII. — THE RELATION BETWEEN THE
SEGMENTS OF A SECT AND THE PROJECTIONS OF THE SEGMENTS.

Draw a sect; divide it into two or more unequal segments.
Draw any other indefinite line, not parallel to the sect.
Project the segments of the sect upon the other line.
Measure the segments and their projections. Observe
how certain ratios compare. State result.

EXERCISES

1. A sect, AB, 3" long is divided at point C into segments AC =
1.2" and CB = 1.8". The segments are projected on a line
DE such that the projection of AC = 0.8". Calculate the
projection of segment CB.

2. The projections of the segments AC and CB of a sect AB are
3.5" and 6.2". Sect AB = 12". Calculate the segments of
AB.

233. Experiment VIII.— The relation between the sides
of a right triangle.

Draw one or more right triangles; measure the sides.
Square the value of each side. Observe the relation be-
tween the squares of the three sides of each triangle.
State result.

EXERCISES

1. The perpendicular sides of a right triangle are 3 and 4 inches.
Calculate the hypotenuse.

2. The hypotenuse and base of a right triangle are respectively
13 and 5 inches. Calculate the altitude.

3. May the lengths 8, 15, 17 inches be taken as the sides of a right
triangle? Why?

234. Experiment IX. — The condition for similar polygons.
Draw a trapezium (a polygon of four sides).

(a) Draw a polygon having its angles equal, each to
each, to those of the given polygon, but its sides in no
constant ratio to the sides of the given polygon.

(b) Draw a polygon having its sides proportional to the
sides of the given polygon, but its angles in no relation to
'those of the given polygon.

MEASUREMENT OF SECTS, AND SIMILAR TRIANGLES 137

(c) Draw a polygon having its angles equal, each to
each, to the angles, and its sides proportional to the sides,
of the given polygon.

Observe the following principles :

(a) May two polygons be mutually equiangular without
being similar?

(6) May two polygons have their sides proportional
without being similar?

EXERCISES

1. A rectangle has adjacent sides of 4" and 2". A second rectangle
has homologous sides of 6" and 3", Are the rectangles similar?

2. The sides of two quadrilaterals are respectively, 3", 5", 4.8", 2";
and (in the same order) 6", 10", 9.6", 4". Are they similar?

3. The sides of one of two similar pentagons are 7", 10", 5", 8", 6".
A side of the second pentagon, homologous to the 10" side of
the first, is 12". Calculate the other sides of the second pentagon.

235. Abbreviation. — Similar, is similar to, ~.

APPLICATIONS

236. Forestry. — The following methods of determining
the heights of trees with sufficient accuracy, are described
in a bulletin issued by the U. S. Department of Agriculture.

(1) The Shadow Method. — Thales employed this method
to find the height of the great pyramids of Egypt, to the
admiration of native scholars.

EXERCISES

1. Are the sun's rays, AC and DF, substantially parallel?

2. Prove that the triangles, formed by tree, shadow and sun's ray

138

PLANE GEOMETRY

is similar to that formed by the vertical pole, shadow of pole
and sun's ray.

3. The shadow of a tree is measured 130 feet at the same time
that the shadow of a 12-foot pole is 15 feet. Calculate the
height of the tree.

4. May this method be used to find the height of a flagpole, build-
ing, stack, etc.

5. Will the two triangles be similar if the ground is sloping? If
the tree is leaning?

6. Find a few facts concerning Thales.

(2) The Graduated Rod Method.

(a) The observer standing.
(6) The observer lying down.

EXERCISES

7. Explain both methods as illustrated.

8. Place letters on the figures and read the similar triangles.

9. What measurements are taken in each method?

10. Assume reasonable measurements and calculate the heights of
the trees by both methods.

11. Is this method applicable if the ground is sloping?

(3) The Mirror Method.

EXERCISES

12. Explain the method illustrated.

13. Where are the similar triangles?

14. Assume reasonable measurements and calculate the height of
the tree.

15. Is this method applicable on sloping ground if the mirror is
leveled?

MEASUREMENT OF SECTS, AND SIMILAR TRIANGLES 139

(4) Faustmann's Hypsometer. — A simplified form of this
instrument is shown. The instrument is sighted to the top

0 '

of the tree and the reading is taken at F on the scale EF.
It is necessary to measure AB, which may be done by
pacing, and to measure or estimate BG. This may be done
by sighting through a hand level and noting the point B
on the level line AB, and then measuring from B to the
ground. After calculating CB, the distance BG is added
to find the total height of the tree.

In the instrument as usually made, the point of attach-
ment D of the plumb-line is on a movable scale, so that the
distance AB in feet may be set off on the scale equal to ED.
The graduations on scale EEf are made in the same units as
employed on scale ED, and the height of the tree in feet
is therefore equal to the reading of the scale distance EF.
A narrow adjustable mirror hinged at point E is provided,
and the observer may thus read the scale at F while sighting
the instrument on C, the figures of the scale EE' being
reversed.

EXERCISES

16. Name the two similar triangles. Prove that they are similar.

17. If DE = 5", EF = 6", AB = 40', BG = 7'; calculate CG.

18. If BG = 8.5 feet, AB = 120 feet; and if ED is set at 120 units
and EF reads 95 units; what is the value of BC in feet, and
what is the height of the tree?

237. Surveying. — The principles of similar triangles afford
some useful methods of determining distances which cannot
be measured directly, and of laying out perpendiculars.

140

PLANE GEOMETRY

The Distance Across a River. — The distance AB is
required. Turn a right angle at A; measure any distance
AC; turn a right angle at C;
measure any distance CD. Set
a stake at point E on both lines
DJSandAC.

EXERCISES

1. Name the similar triangles of
the figure. Prove that they
are similar.

2. What proportion can be stated whose solution will give the
distance AB?

3. What additional measurement, not mentioned above, must be
made, in order to give sufficierit data for the calculation of AB?

4. If AC = 500', CD = 320', CE = 180' 6", calculate AB.

5. Upon what principle of this chapter does the calculation of A B
depend?

6. How does this method differ from that of 146, Exercise 8?

The Distance Between Two Points on Opposite Sides of a
Building on Pond. — The distance AB is required. Set a
stake at any point C from
which both A and B can
be seen; measure CA and
CB. Mark any point E on
CB such that a line DE
will pass the building.
Calculate the position of
point D on CA such that

DE is parallel to A B, and set a stake at D. Measure DE
and calculate A B.

EXERCISES

7. How is the distance CD found, such that DE will be parallel
to AB?

8. How is the distance AB calculated, after all measurements are
completed?

9. Calculate AB if CA = 240', CB = 300', CE = 200', DE =
238'.

10. Upon what principles of this chapter do the calculations of CD
and A B depend?

MEASUREMENT OF SECTS, AND SIMILAR TRIANGLES 141

11. How does this method differ from that of 146, Exercise 7?

12. Is it possible to modify the method, by setting point E (a) on
CB extended through B; (6) on CB extended through Cf
Sketch the figures in these cases.

A Perpendicular to a Given Line at a Point in the Line.—
This is a method quickly used and
often sufficiently accurate. Mea- pN^

sure CD = 40'; hold a 100-foot 8j ^?N

tape with one end at C and the A" c " ~~0X-^ — £
80-foot mark at D; take hold of

the 30-foot mark and hold the tape so that both parts EC
and ED are taut. Mark point E.

EXERCISES

13. Upon what principle of this chapter does the method depend?

14. What is the largest triangle that can be laid out in this way
with a 100-foot tape, with exact foot-lengths for its sides?

15. Can the shorter side of the triangle be measured on CD?

A Perpendicular to a Given Line from a Point Without the
Line. — The perpendicular is to be laid out from C which is
more than 100 feet from line AB.
Place any stake D on line AB;
measure CD; place a stake E on AB
at about the foot of the required
perpendicular, and measure DE;
erect a perpendicular EF; place a
stake at G on both.EF and CD;
measure GC.

EXERCISES

16. How is EH calculated such that CH Q
will be parallel to EG?

17. After calculating EH, how is the required perpendicular
marked on the ground?

18. Let CD = 250.4', DE = 175.8', GC = 36.5'; calculate EH.

19. May the point E be placed beyond H and point G in line CD
extended? Draw a figure illustrating this and describe the
method of calculating EH.

20. Upon what principle of this chapter does the method depend?

142

PLANE GEOMETRY

Locating a Curved Line. — A straight line AE is staked out
near the curved line; and divided into 40-foot intervals.
Perpendiculars, or offsets, CI, DK, etc., are measured.

EXERCISE

21. Plot the bank of a creek from the given field notes, to a scale
of 100 feet to the inch. The points lie on a straight line at
intervals of 40 feet. The offsets are measured to the center of
the creek.

Pts.

Offsets
at 40'

Pts.

Offsets
at 40'

Pts.

Offsets
at 40'

Pts.

Offsets
at 40'

+ 60

+60

-90

+ 30

+100

+45

-35

+ 50

+ 95

-70

CHAPTER XI
SIMILAR TRIANGLES AND SIMILAR POLYGONS

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

238. The First Step in Classification.

POSTULATES

1. Mutually equiangular triangles are similar.

2. The homologous sides of similar triangles are pro-
portional.

3. Two triangles are similar if two sides of one triangle
are proportional to two sides of the other and if the included
angles are equal.

4. Two triangles are similar if all the sides are propor-
tional.

5. Two polygons are similar if they are mutually equi-
angular, and also if their homologous sides are proportional;
and conversely.

239. The Second Step in Classification. — All other
principles of similar triangles and similar polygons can be
analyzed in terms of the postulates of 238 and other previ-
ously analyzed principles.

240. Theorem I. — A proportion can be changed according
to certain laws.

(a) Law of Products. — The product of the means equals
the product of the extremes.

a c
Hypothesis. — Any proportion, , = ,•

Conclusion. — ad = be.

Help. — Clear of fractions. What axiom is used?

Corollary. — If the product of two numbers equals the
product of two other numbers, the numbers of one product may
be made the means and those of the other product the extremes,
of a proportion.

143

144 PLANE GEOMETRY

Hypothesis. — xy = st.

Conclusion.— (1) * - \ (2) ~ = *-, (3)* = I etc.
by i y jc i

Help. — (1) Divide by sy; etc.

(6) Law of Inversion. — The ratios of a proportion may be

inverted.

a c
Hypothesis. — r = -v

b d

Conclusion.— - = -•
u c

Help. — Apply law (a) ; divide by cd, etc.

(d) Law of Composition. — The sum of the numerator
and denominator of each ratio may be made the numerators

of the ratios.

a c
Hypothesis. — 5 = ^'

a + 6 c +d
Conclusion. — — ^— — ^—

Help. — Add 1 to each ratio.

(e) Law of Division.

a — b c — d
Conclusion. — — ^ — = —^ —

(/) Law of Ratio Series.

ace
Hypothesis. — ^ = ^ = ->= etc.

a + c + eH a c

Conclusion.— ^_ d + f + = b =d =

a c

Analysis.— (1) ^ = r;^ = r; etc.

(2) a = br; c = dr; etc.

(3) a + c + e + - - =br + dr + fr + •

a + c + e

(4) r = t^+j^r.

What axioms are used?

SIMILAR TRIANGLES AND SIMILAR POLYGONS 145

EXERCISES

3 4

1. Solve these proportions for the unknown term: (1) ^ = -; (2)

? _? ,^*-4 m-5-31
2 - 3, W g - 3> ^x - 3

Help. — Apply law (a).

2. Solve for the unknown term: (1) - = =; (2) ~ = — •

X / £ ju

Help. — Apply law (a).

7 21

3. Change the proportion 5 = ~=; according to laws (a) to (/).

;J Z /

Is each new proportion obtained a true equation?

4. Change the ratio series ~ = « = Tc = i~ci according to law (/).

5. Solve x : re + 1 =5:7 for x, and check.

6. Show that the proportion T = -, is equivalent to: (1) T = -;(2)

q c , a c a+c «

c +d d

241. Theorem II. — A line parallel to a side of a triangle
divides the other two sides proportionally.

Hypothesis.— AABC with DE\\BC.
„ , . AD AE

Conclusion.-

Helps. — Show that there are two similar triangles; use
Postulate 2, 238. Draw the triangles separately.
DE AD AE

AD _AI3 AD _ AE AD DB
(2) AE ~ AC' DB ~ EC' AE ~ EC' e

rfAD AE AB AC AB - AD

.-li- = ', then

AC-AE AD AE AD _ AE

AE ' AB-AD~ AC-AE' DB~EC

146 PLANE GEOMETRY

242. Theorem III. — A line which divides two sides of a
triangle proportionally is parallel to the third side.

Hypothesis. — AABC with DE drawn so that -,-g = -j-~-

Conclusion.— DE\\BC.

Helps.— (1) AADE ~ AABC; Postulates; (2) .'. x =
y; (3) .'. DE\\BC; why? Draw the triangles separately.

243. Theorem IV. — Two triangles similar to the same
triangle are similar to each other.

Hypothesis.— &DEF and GUI ~ AABC.

Conclusion.— ADEF ~ AGHI.

Helps.— LA = £D; ZA = Z.G; .'. ZD= Z.G; etc.

244. The Third Step in Classification.— The principles
which follow, 245 to 256, are probably the result of the
application of deductive reasoning, and, with the exception
of 250 in an elementary form, were not discovered experi-
mentally.

In making use of similar triangles in analysis it is ad-
visable to mark the mutually equal angles of the two
triangles under consideration, x and x', y and y', z and z'.
The homologous sides are then readily identified as being
opposite the homologous angles. In some figures it is also
helpful to draw the triangles under consideration apart
from the general figure, and to letter all angles and sides as
they are lettered in the general figure. The use of colored

SIMILAR TRIANGLES AND SIMILAR POLYGONS

147

crayons, overlaying homologous sides of similar triangles
with the same color, will be found a great help.

Notice that, if two angles of one triangle are equal, each
to each, to two angles of another triangle, the triangles are
similar. Why?

245. Theorem V. — The bisector of an angle of a triangle
divides the opposite side into segments proportional to the
adjacent sides.

State the hypothesis and the conclusion.

Helps. — (1) How are the auxilliary lines drawn? (2)

CA CD

-rjj, = jyF>', why? (3) A ABE is isosceles; how shown?

C A CD

(4) .'. AB = AE; (5) .

Draw the elementary figures which make up the figure of
the theorem.

246. Theorem VI. — The bisector of an exterior angle of a
triangle divides the opposite side externally into segments, etc.
(complete the theorem after analysis).

CA CD
Helps.— (1} Draw BE; (2) -^ = =^; (3) AABE is

isosceles; (4) AB = AE; etc.

Draw the elementary figures as in 245.

148

PLANE GEOMETRY

247. Theorem VII. — The altitude drawn to the hypotenuse
of a right triangle divides the triangle into two triangles which
are similar to the given triangle and to each other.

•c D

•c 0

Helps. — The three triangles to be considered are ABC,
BDC and ADB.

248. Theorem VIII. — The altitude drawn to the hypotenuse
of a right triangle is a mean proportional between the segments
into which it divides the hypotenuse.

Helps. — Each pair of triangles of 247 affords three pro-
portions. If these are stated in full, it will be found that a
proportion with a repeated mean (or extreme) is found in
each set of three. One of these mean proportionals is the

a c
required relation. A repeated extreme, T = -, may also be

regarded as a mean proportional, since an equivalent pro-

b a

portion is - = —
a c

249. Theorem IX. — Each perpendicular side of a right
triangle is a mean proportional between the hypotenuse and
the projection of the side upon the hypotenuse.

Help. — The required relations are found among the
proportions suggested in 248.

Corollary. — The square of a perpendicular side of a right
triangle equals the product of the hypotenuse and the projection
of the side upon the hypotenuse.

250. Theorem X. — The sum of the squares of the two

SIMILAR TRIANGLES AND SIMILAR POLYGONS

149

perpendicular sides of a right triangle equals the square of the
hypotenuse.

Hypotenuse. — Right AABC.
Conclusion. — b2 — a2 + c2.
Analysis.

STATEMENT

1. a2 = bn and c2 = bm

2. a2 + c2 = bn -f 6m

3. a2 + c2 = 6(n + w)

4. a2 + c2 = 6.6

5. .*. a2 + c2 = 62
Corollaries.— (1) a2 = 6
(2) a =

- c

REASON

Theorem IX, Corollary
Axiom

Algebraic Axiom
Axiom

Algebraic Axiom
;2 = 62 - a2.

c)(&- c); etc.

The relation of Corollary 2 is convenient in numerical
calculation.

251. Historical Interest of Theorem X. (250).— This
relation is supposed to have been known for simple number
values (at least for the numbers 3, 4, 5) by the Egyptians
as early as 2000 B. C., and also by the ancient Hindoos
and Chinese. The general principle was discovered by
Pythagoras (582-500 B. C.), whose method of analysis
was probably somewhat like the method here presented.
The theorem is known as the Pythagorean Theorem. An
entirely different method was given by Euclid (Chapter
XVII). Numbers which possess the relation a2 + 62 = c2
are called Pythagorean numbers; many such series exist
in higher values.

252. Theorem XI. — In any triangle, the square of a side
opposite an acute angle equals the sum of the squares of the
other two sides minus twice the product of one of those sides
by the projection of the other side upon it.

150

PLANE GEOMETRY

FIG. 1. FIG. 2.

Hypothesis. — A ABC with an acute /.A, and BD±AC.

Conclusion. — a2 = 62 + c2 — 26m.

Analysis.

STATEMENT

1. m + n = 6,or 6 -f- n = m

2. n = b — m or n = m — b

3. n2 = 52 _ 26m + m2

REASON
Axiom 11
Axiom 3
Algebraic axiom

4. n2 + p2 = 62 - 26m + m2 + p2 Axiom 2

5. a2 = 52 _ 26m + c2 250 and Axiom 13
253. Theorem XII. — In any obtuse triangle, the square

of the side opposite the obtuse angle equals, etc. (Complete the
theorem after analysis.)

Write the hypothesis and conclusion.

Helps. — (1) n — m = 6; (2) .*. n = b + m. Complete
as in 252.

254. Theorem XIII. — The projection
of a sect can be calculated when the angle
which the sect makes with the line of pro-
jection is (a) 60°, (6) 30°, (c) 45°.

Part (a).

Hypothesis. — A sect s making an
angle of 60° with the line of projection
AD, and its projection m.

SIMILAR TRIANGLES AND SIMILAR POLYGONS 151

Conclusion. — m = f s.
Analysis.

STATEMENT

1. ZC = 90°; Z5 = 30°

2. m = |s = .5s

Part (6).

REASON
143, Theorem (8)

A CO

State hypothesis and conclusion.

3s2

Helps.— (1) p = is; (2) m2 = s2 - p2; (3) m2 = ^~ ;

(4) m = = .866s.

Part (c).

C D

s\/2

Helps.— (1) m = p; (2) m2 + p2 = s2; (3) m =
= .707s.

255. Theorem XIV. — An altitude of a triangle can be
expressed in terms of the three sides.

AH -6 M

« 1\

152 PLANE GEOMETRY

Analysis.

STATEMENT

1. In ABDA, Fig. 1, pb2 = c2 - m2.

2. In AABC, a2 = 62 + c2 - 2bm

A2 I r2 _ «2

3. From 2, m = D ±^-

4. Substituting from 3 in 1,

=,.-(»-
.(.+i

c2 -

REASON

What theorem?
What theorem?

Axioms
Axiom

X
ck —

/ 62 + c2 - a2\ 41 - u .

f c — sr— — J Algebraic axiom

(a - 6 + c)(a + 6 - c)
— ^~- —

A1 , .
Algebraic axiom

7. Let a + 6 + c = 2s
then

— a + 6 + c = 2s — 2a = 2(s — a) Algebraic axiom
a — 6 + c = 2s — 26 = 2(s — 6), Algebraic axiom
etc.

8. Substituting from 7 in 6

2s-2(s - a)-2(s - 6)-2(s - c)
b = ~ Axiom

. _
9. pb = T \/s(s — a)(s — b)(s — c) Algebraic axioms

Show that the same result is obtained in Fig. 2.
Helps.— (1) pb2 = c2 -- m2; (2) a2 = 62 + c2 + 26m;

a2-62-c2
(3) m = -- 25 -- >' etc-

Write the values of pa and pc by interchanging letters in
the formula for pi.

256. Theorem XV. — A median of a triangle can be ex-
pressed in terms of the three sides.

Analysis.

1. In ABED, pb2 = mb* - t*

2. In ABEC, a2 = pb2 + + t = b2 + + bt

SIMILAR TRIANGLES AND SIMILAR POLYGONS 153

3. In ABEA, c2 = pb2

4. Adding 2 and 3; a2 + c2 = 2p62 + +

5. Substituting value of pb2 from 1 in 4;

62
a2 + c2 = 2(w62 - t2) + -2 + 2*2 =

2(a2 _|_ C2) ._ 52

6. Solving 5 for m^\m^ • ^

Show that the same result is obtained if Z.A is obtuse.
Write the values of ma and mc by interchanging letters in
the formula for mb.

257. Theorem XVI. — Two similar polygons can be divided
into the same number of triangles similar each to each and
similarly placed.

0 0

Helps.— (I) AABC~AA'B'C'; 238, Postulate 3;— (2) .'.
Z 1 = Z2 and Z3 = Z4; and -p^r, = -^-,-g, = ^7^>; — (3) /5
= Z6, why? and /. &ACD~AAC'D'; etc.

154 PLANE GEOMETRY

258. Theorem XVII. — Two polygons that are composed of
the same number of triangles similar each to each and simi-
larly placed, are similar.

Helps.— (I) ZA = ZA'; Zl = Z2; Z3 = Z4;— (2) Z5 = Z6;
Z7 = Z8; etc., — (3) .'. polygons are mutually equiangular.

... AB BE BC EC .

~(4) A7B' = WE' = PC7 = WC' = etc--® ' • Polygons
have their homologous sides proportional.- — (6) .'. the poly-
gons are similar; 238, Postulate 5.

259. The Fourth Step in Classification. — The principles
of] 245 to 256 cannot be considered as established beyond
all doubt until they have been tested by actual measure-
ments. This we will now proceed to do.

EXERCISES

1. Draw a triangle (full size) whose sides are 4, 5.6 and 6.2 inches.
Draw the bisector of each interior angle. Calculate by 245
the segments into which each bisector divides the opposite side.
Measure these segments and note the agreement of the meas-
urements with the calculated values.

Thus to calculate the segments of the largest side, calling the seg-
ments x and 6.2 — x, we have 7 = ' - „ — , whence x = 2.58",

and 6.2 — re = 3.62". Calculate in the same way the segments
of the other sides.

2. Draw the bisectors of each pair of exterior angles of the triangle
of Exercise 1, and prolong each bisector until it meets the oppo-
site side prolonged. Calculate by 246 the segments into which
each bisector divides the opposite side externally. Measure
these segments (from the external point of division to each end
of the side), and note the agreement with the calculated values.

Thus to calculate the segment of the largest side, calling the
segments x (to the nearer end of the side) and 6.2 + x (to the

farther end of the side), we have ^ = ' , fi — » whence derive

SIMILAR TRIANGLES AND SIMILAR POLYGONS 155

values of x and 6.2 + x. Calculate in the same way the seg-
ments of the other sides.

3. Draw a right triangle (full size) whose sides are 3, 4 and 5 inches.
Draw the altitude to the hypotenuse. Calculate by 249 the
projection of each perpendicular side upon the hypotenuse, i.e.,
the segments into which the altitude divides the hypotenuse.
Measure these projections (or segments) and note the agree-
ment with the calculated values.

Thus to calculate the smaller segment x, of the hypotenuse, we

have I = „ whence x = 1.8". Calculate in the same way the

other segment y.

4. In the triangle of Exercise 3, calculate by 248 the length of the
altitude drawn to the hypotenuse. Measure the altitude and
compare with the calculated value.

5. Draw a sect 3 inches long and through one end of it a line mak-
ing with the sect an angle of 60°. Project the sect on the
second line. Calculate by 254(a) the value of the projection.
Measure the projection and compare with the calculated value.

6. Draw a sect of 4 inches, and a line making an angle of 30° with
it. Project the sect on the other line. Calculate by 264(6)
the value of the projection. Measure and compare results.

7. Draw any sect and a line making an angle of 45° with it. Pro-
ceed as in Exercises 5 and 6.

8. Draw a triangle (full size) of which two sides are 5 and 3 inches
and the included angle 60°. Calculate the third side by 252.

Thus: the projection of the 3" side on the 5" side is (by 254(a))
1.6".

Then a2 = 32 + 52 - 2(5) (1.5). Complete the calculation.
Measure the third side and compare with the calculated value.

9. Draw two triangles in each of which two sides are 5 and 3 inches
respectively; in one triangle the included angle being 30°, and
in the other 45°. Calculate the projections of the 3" side on
the 5" side by 254(6) and 254 (c), and complete calculations for
the third side of each triangle as in Exercise 8.

10. Draw a triangle of which two sides are 5 and 3 inches and the
included angle 120°. Calculate the third side by 253.

Thus: the projection of the 3" side on the 5" side is 1.5".
Then a2 = 32 + 52 + 2(5) (1.5); complete the calculation.
Measure the third side and compare with the calculated value.

11. Draw triangles with the sides as in Exercise 9, the included
angle in one triangle being 150° and in the other 135°. Calcu-
late the projections of the 3" side on the 5" side as hi Exercise 9,
and the values of the third side of each triangle by 253. Meas-
ure and compare values.

156 PLANE GEOMETRY

12. Draw a triangle (full size) whose sides are 4, 5, and 6 inches.
Draw the three altitudes. Calculate the lengths of the alti-
tudes by 265.

Thus, to calculate the altitude to the longest side:

4+5+6

• — Q — I .O

s - 4 = 3.5, s - 5 = 2.5, s - 6 = 1.5

P = I V(7.5)(3.5)(2.5)(L5)

Complete the calculation. Measure the altitude and note the
agreement with the calculated value. Calculate and measure
the other altitudes in the same way.

13. Draw the medians in the triangle of Exercise 12. Calculate
their lengths by 266. Measure these lengths in the triangle
and compare with the calculated values.

14. In every case in the preceding exercises do the calculated values
show a sufficiently close agreement with the deduced principles
as to confirm the accuracy of these principles?

260. ADDITIONAL THEOREMS

(1) Two isosceles triangles with equal vertex angles are
similar.

(2) All equilateral triangles are similar.

(3) Homologous altitudes of similar triangles are propor-
tional to any two homologous sides.

Help. — Show that the triangles containing the altitudes
as homologous sides are mutully equiangular.

(4) Homologous medians of similar triangles are propor-
tional to any two homologous sides.

Help. — Use 238, Postulate 3, to show that the triangles
under consideration are similar.

(5) Homologous angle-bisectors of similar triangles are
proportional to any two homologous sides.

(6) In a triangle ABC, altitudes AD, BE and CF are drawn
Show that AADC is similar to ABEC, and AADB is
similar to ACFB, and that there is another pair of similar
triangles formed.

(7) Two right triangles are similar if they have a pair of
equal acute angles.

(8) Two isosceles triangles are similar if they have a base
angle of each respectively equal.

SIMILAR TRIANGLES AND SIMILAR POLYGONS

157

(9) The projections of the segments of a sect are propor-
tional to the segments (232).

(10) A system of parallel lines divides any two transversals
proportionally.

(11) The diagonals of a trapezoid divide each other propor-
tionally.

(12) State another theorem relating to the proportion-
ality of the segments into which the diagonals of a trapezoid
divide each other.

(13) A pencil of three concurrent lines
divides two parallel lines which are cut by all
the lines of the pencil, proportionally.

Helps. — Use two pairs of similar tri-
angles. The two proportions thus occur-
ring have a common ratio.

(14) The diagonal of a square equals the side multiplied by
\/2; or d = s\/2.

(15) The squares of the perpendicular sides of a right
triangle are proportional to their projections on the hypotenuse.

Help. — Apply 249, corollary.

(16) The altitude of an equilateral triangle may be expressed

s\/3
in terms of the side; or p - — - — .

(17) The side of a square may be expressed in terms of the
diagonal.

(18) The side of an equilateral triangle may be expressed in

terms of the altitude; or s = ^ .

(19) Find a method of drawing
a square HGKL inscribed in the
triangle ABC. Prove that the
method found is correct.

Helps.— (I) AF = FE;—(2)

°^~ - — - m ?P - ?. (4\ •
FE BF~(6) AF ~

GL _HG
FE ~AF; etc'

158

PLANE GEOMETBY

(20) A relation exists between a perpendicular on one side
of a diameter of a circle and the segments into

which it divides the diameter.

Helps. — Draw DA and DB; see 143, Theorem
(7). Then apply one of the preceding theorems
of this chapter relating to right triangles.

(21) A relation exists between a chord of a circle, the diam-
eter drawn through the end of the chord, and the projection of
the chord on that diameter.

(22) A relation exists between the product of the perpendicu-
lar sides of a right triangle, the hypotenuse, and the altitude to
the hypotenuse.

(23) Any angle of a triangle may be evaluated, whether acute,
right or obtuse, by the relation of the squares of the three sides.

(24) Two triangles whose sides are parallel, each to each,
are similar.

Helps. — From 81, we may have; (a) x -\- xf = 180°,
y + y' = 180°, z + z' = 180°; or (b) x + x' = 180°, y +

y' = 180°, z =

or (c) x + X* = 180°, y = y

= z

or (d) x = x' , y = yf, z — zr. Show that (d) is the only
possible numerical relation of the angles of the two triangles.

(25) A relation exists between two triangles whose sides are
perpendicular, each to each.

(26) The segments into which the bisector of angle A of

A „ ~ , . . , . , ab _ ac

A ABC divides side a, are =— - and r— ; —

b + c b + c

x b

Help. — If x is a segment of a,

(27) Point D lies within
LA.BC. Find a method of
drawing a sect through point
Determinated by the sides of
the angle and bisected by
point D. Prove that the
method found is correct.

(28) In the figure of
Theorem (27) , find a method

SIMILAR TRIANGLES AND SIMILAR POLYGONS 159

of drawing a sect through point D so that it is divided by point
D in any given ratio. Prove that the method found is correct.

(29) The product of a side of a triangle multiplied by the
altitude drawn to that side, equals the product of any other
side of the triangle multiplied by the altitude drawn to it.

(30) Two polygons similar to the same polygon are similar
to each other.

Help. — It is necessary to prove both that the polygons
are mutually equiangular and that their homologous sides
are proportional.

(31) Homologous diagonals of similar polygons are pro-
portional to any two homologous sides.

(32) The perimeters of two similar polygons are propor-
tional to any two homologous sides.

Help.—(l) The sides -, = ^ = -t = etc.; (2) apply 240
a o c

co-

(33) A sect is divided harmonically into segments of which
x is one internal segment and y is the corresponding external

segment, when y = _ „ •

(34) The bisectors of adjacent interior and exterior angles of
a triangle divide the opposite side harmonically.

(35) The sum of the squares of the sides of a parallelogram
equals the sum of the squares of the diagonals.

Help. — Express the value of each diagonal from 252 and
253, and add.

(36) Four times the sum of the
squares of the medians drawn to
the perpendicular sides of a right
triangle, equals five times the square

of the hypotenuse. &

Helps. — (1) Express x and y from 256 ; (2) square these
values and multiply out the coefficients 2 and J ; (3) add
and reduce.

(37) The difference of the squares of two sides of any triangle
equals the difference of the squares of the projections of these
sides upon the third side.

160 PLANE GEOMETRY

Help. — Draw an altitude, thus forming two right tri-
angles.

261. REVIEW EXERCISES

1. The perpendicular sides of a right triangle are 12 and 16. Cal-
culate the hypotenuse.

2. Calculate the diagonal of a square whose side is 10.

3. Calculate the side of a square whose diagonal is 12.

4. Two sides and the included angle of a triangle are 10, 15, 135°.
Calculate the third side.

5. Two sides and the included angle of a triangle are 20, 45, 60°.
Calculate the third side.

6. The perpendicular sides of a right triangle are 6 and 10. Calcu-
late their projections on the hypotenuse.

7. To check the layout of a tennis court, 36 feet by 78 feet, calcu-
late and measure the diagonals.

8. A derrick mast 50 feet high is guyed by a rope running from
the top of the mast to a tree trunk 100 feet distant. Allow
5 feet for fastening, and calculate the length of rope required.

9. Calculate the resultant of two forces of 75 Ib. and 32 lb., acting
at an angle of 90°.

10. The resultant of two forces acting at an angle of 90° is 12 lb.;
one of the forces is 3 lb. Calculate the other force.

11. Calculate the resultant of two forces of 10 lb. and 15 lb. acting
at an angle of 45°; at an angle of 135°.

12. Calculate the resultant of three forces of 10, 12 and 15 lb.
respectively, the first two forces acting at a right angle to each
other, and the third force acting at a right angle to the resultant
of these two.

13. Show that the values a2 + 62, a2 — 62, 2ab may represent the
sides of a right triangle. Where is the right angle?

14. Pythagoras stated that the values 2n + 1, 2n(n + 1), 2n2 -f
2n + 1 are the sides of a right triangle. Prove that this is so.

15. Let n = 1 in the general values of Exercise 14; show that the
sides of the triangle are in this case, 3, 4, and 5. Let n = 2,
and find the sides of the right triangle from the formulas of
Exercise 14. Similarly, let n = 3, 4, 5, etc.

16. One of the perpendicular sides of a right triangle is 2 more than
the other; the hypotenuse is 10. Calculate the sides. Solve
by forming an algebraic equation. Check.

17. One of the perpendicular sides of a right triangle is 20 inches
and the hypotenuse is 10 more than the other side. Calculate
side and hypotenuse.

SIMILAR TRIANGLES AND SIMILAR POLYGONS 161

18. An ancient Chinese problem which appears in many texts: In
the middle of a circular well 10 feet in diameter, grows a reed
which projects 1 foot above the surface. When the reed is
bent over, remaining straight from its root, the top just reaches
the edge of the well. How deep is the well?

19. A builder plans a barn with the ridge of
the roof 10 feet above the loft floor, and
the floor 24 feet wide. How long must
he order the rafters, allowing 2 feet
overhang at the eaves and 6 inches for
trimming the ends?

20. Two chords of a circle, 8 and 12 inches respectively, are drawn
from the same point in a circumference to the extremities of a
diameter. Calculate the diameter.

21. A perpendicular to the diameter of a circle is 10 inches and one
of the segments into which it divides the diameter is 5 inches.
Calculate the diameter.

22. The altitude to the hypotenuse of a right triangle is 3 inches
and the projection of one of the perpendicular sides upon the
hypotenuse is 8 in. Calculate the three sides of the triangle.

23. The rise CB in 100 feet of slope AB
is 17 feet. Calculate the horizontal
distance in 100 ft. of slope. __^

24. A 40-foot ladder is placed against a

building with its foot 10 feet from the wall. How high does
it reach on the building?

25. Can a rod 40 inches long be placed within a trunk 24 inches by
30 inches, by 16 inches high; from an upper corner to a lower
opposite corner?

26. The sides of a triangle are 10, 25, 30. Calculate the segments
into which each angle-bisector divides the opposite side.

27. Calculate the position of the point where the bisector of the
exterior angle opposite the 30 side of Exercise 26 intersects this
side produced.

28. Calculate the length of the three altitudes of the triangle of
Exercise 26.

29. Calculate the lengths of the three medians of the triangle of
Exercise 26.

30. Two men are walking on opposite sides of a street. A tree
near the curb is between them. The man on the farther side
walks uniformly; must the man on the near side walk uniformly
or otherwise, in order to keep the tree always between them?

31. Show how the methods of these figures can be used to divide a
given sect 4J3. into any number of equal parts. Explain the
11

162

PLANE GEOMETRY

principles on which the result depends. Draw similar figures
showing the division of the sect into 5 equal parts.

Line DE is parallel to
A

32. Construct a right triangle when the sum a, of the two perpen-
dicular sides, and an acute angle a

or, are given. H — — i—

33. Construct a right triangle when
the perimeter and a similar tri-
angle are given.

34. Calculate the value of \/352 — 242
by the method of 250, Corollary 2.

35. In AABC, AB = 10", AC = 15".
BC and AE is 3" more than

AD. Find the lengths AD and

AE. Check the result.

36. In the triangle of Exercise 35,
AE is 10" more than AD.
Calculate AD and AE and check result.

37. Test the principle of 260, Theorem (13), by accurate measure-
ment.

38. Test the principles of 260, Theorem (14), by measurement.

39. Test the principle of 260, Theorem (15), by measurement.

40. Test the principle of 260, Theorem (16), by measurement.

41. Calculate the diagonal of a square if the side = 50'.

42. Calculate the altitude of an equilateral triangle if a side is 120'.

43. Calculate the side of a square if the diagonal is 24".

44. Calculate the side of an equilateral triangle if the altitude is 24".

45. Test the principle of 260, Theorem (20), by accurate measure-
ment.

46. The diameter of a circle is 24". Calculate the length of the
perpendicular on one side of the diameter erected at a point 6"
from one end.

47. The sides of a triangle are 10", 12", 16". Calculate the seg-
ments into which the bisector of the largest angle divides the
opposite side, using the formulas of 260, Theorem (26).

48. Determine which angles of the following triangles are acute,
right or obtuse. . (1) a = 5, 6 = 12, c = 13; (2) a = 15, b = 10,
c = 18; (3) a = 68, b = 51, c = 100; (4) a = 7, 6 = 25, c = 24.

SIMILAR TRIANGLES AND SIMILAR POLYGONS

163

51.

52.

49. The sides of a right triangle are 3, 4, 5. Calculate (a) the alti-
tude to the hypotenuse; (6) the segments into which the altitude
divides the hypotenuse; (c) the length of the median to the
hypotenuse; (d) the length of the bisector of the right angle;
(e) the segments into which this angle-bisector divides the
hypotenuse.

50. To what simpler formula do the formulas of Theorems XI and
XII of similar triangles reduce if A is a right angle? What
does the projection of c on 6 then become?

Calculate the median to the base of an isosceles triangle whose
sides are 4", 4", 3"; (a) using the formula of 250 ; (6) using the
formula of 255 ; (c) using the formula of 256.
A carpenter lays out a perpendicular by measuring a string
24 feet long, with knots at 8 feet and 14 feet from one end.
The ends are tied together and the string is stretched around
three nails so that the knots are at the nails. Make a sketch
and explain the principle upon which the method depends.

53. A man 5 feet 6 inches tall stands 8 feet from a lamp-post and
casts a shadow on the ground 10 feet long. How high is the
lamp-post?

54. Calculate the altitudes of a triangle, of which the sides are 102,
104 and 106 feet.

55. What is the diameter required, of a bar of round tool steel in
order that a square bar f inch on a side may be cut from it?

56. What is the largest square bar that can be cut from a round
bar of steel 1$ inches in diameter?

57. Calculate the chord of a circle whose radius is 12", which sub-
tends an arc of 90°.

58. Calculate the length of a tangent drawn to the circle of Exercise
57, from a point which is 30" from the center of the circle.

59. Calculate the length of the shortest chord which can be drawn
in the circle of Exercise 57, through a point 10" from the center
of the circle.

The radii of two concentric circles are 10" and 12". Calculate

the length of a chord of the larger circle which is tangent to the

smaller.

Show that the side of a regular octagon con-

structed by cutting the corners from a

square is s' = s(\/ 2 — 1).

60.

.— s =s' + 2x = V +

; solve for s'.

62. If an octagon of side s' is required, show
that it may be drawn by first drawing a
square of side s, such that s = s'(\/2 + 1).

164

PLANE GEOMETRY

63. Find the side of a regular octagon formed by cutting the corners
from a square whose side is 20.

64. What is the side of a square which must be first drawn, in order
that the regular octagon formed by cutting off the corners shall
have sides equal to 20?

APPLICATIONS

262. Drawing Maps and Designs to Scale. — All reduc-
tions and enlargements of figures depend upon the following
general property of similar figures: If all angular measure-
ments of two figures are equal, all linear measurements are
proportional; and conversely. Enlargements of stage
scenery, wall decorations, architectural designs, etc.,
are made as illustrated in the following exercises.

EXERCISES

1. Measure the bearing of A from B on the map of Fig. 1. Measure
the distance AB. Measure the approximate coast line of the
island.

SCALE OF MILES
05 10 15

FIG. 1-

FIG. 2.

FIG. 3.

2. Enlarge Fig. 2 on a scale of 1 to 3. Draw a similar rectangle,
divided into the same number of squares, and follow proportion-
ally located points in the enlarged figure.

3. Enlarge the design of Fig. 3 on a scale of 1 to 2 5.

SIMILAR TRIANGLES AND SIMILAR POLYGONS

165

The pantograph is an instrument for enlarging or reducing
drawings. A simplified form of the instrument is shown.

EXERCISES

4. Sketch the pantograph adjusted so as to enlarge in a ratio of
1 to 5.

5. How is the instrument used if a design is to he reduced to one-
fifth the original size?

Proportional Dividers. — This instrument is used by
draughtsmen to obtain a sect having a given ratio to a
given sect. It can thus be used to transfer a drawing, one
measurement after another, to a different scale.

EXERCISE

6. Explain the operation of the instrument.

263. THE PLANE TABLE. — This is an instrument used in
surveying to locate points within or adjacent to a plot of
ground, as corners of buildings, trees, fence corners, etc.,

166

PLANE GEOMETRY

upon a map. It consists of a drawing board mounted on a
tripod. One line, as AB, of a plot is measured and drawn to
scale on a sheet of paper which is fastened to the Plane
Table. The table is set up in the field over each end in turn
of the line AB. The lines are drawn on the paper in the
directions of the points to be located. The angular values
are thus preserved on the paper, and the resulting figure is
similar to the true figure lying on the ground.

Objects in a lake or bay may be located with the Plane
Table.

264. THE GEOMETRIC SOLUTION OF NUMERICAL PROB-
LEMS.— These methods are chiefly interesting because of
their antiquity, since they were used by Greek mathemati-
cians many centuries before arithmetic was sufficiently per-
fected and before algebra had been invented.

Any unit may be used in measuring the sects, the same
unit being used throughout a problem.

A Fourth Proportional of Three Given Numbers. — Let the
three given numbers be represented by sects a, 6 and c.
Draw two intersecting lines MN and MP; lay off the sects
a, 6 and c, as shown; draw QR and ST parallel to QR.

Then x is the fourth proportional required, since -r = -•

In order to find the numerical value of the result, measure x
with the same unit used in drawing the given sects, a, 6, c.

SIMILAR TRIANGLES AND SIMILAR POLYGONS

167

b K

C H

The Product of Two Numbers. — Let db

, , 1 b

ab = a?-l;and- = --

a x

The Quotient of Two Numbers. — Let T = x; then r = ~'
Draw the figure as in the preceding problems.

168 PLANE GEOMETRY

A Number Divided into Parts Proportional to Two Given
Numbers. — Let it be required to divide line a into parts pro-
portional to m and n.

n i-

EXERCISES

Solve these problems by the geometric method and check the
result arithmetically:

1. Find the fourth proportional of 2, 5, 8.

2. Find the third proportional of 2 and 3.

3. Find the product of 2.5 and 1.8.

4. Find the square of 2.6.

5. Find the quotient of 4.3 by 1.7.

6. Divide 7 into parts proportional to 2 and 3.

The Mean Proportional Between Two Numbers. — Let
a and b be the given numbers. It is required to find the

value of x in the proportion = v.

Show that this value is determined in the constructed
figure (260, Theorem (20)).

The, Square Root of a Number. — Let x = \/a; then x2 —

a; and x.x = a.l; - = . Construct the figure as in the
x a

preceding problem, assuming a unit of measurement u.

SIMILAR TRIANGLES AND SIMILAR POLYGONS 169

EXERCISES

7. Find the mean proportional of 2 and 8.

8. Find the square root of 4.

9. Find the mean proportional of 1.5 and 2.4.

10. Find the square root of 3.

11. Solve geometrically, x = \/2.7 X 4.5.

A Table of Square Roots. — The accompanying figure may
be regarded as a geometric table of square roots. The
diagonal lines give the square roots of consecutive integers.

EXERCISES

12. Explain why x = \/2; y = \/3; etc.

13. Construct the figure as far as the square root of 6. Make a
numerical table and compare with the known arithmetical
values.

265. FIRST PRINCIPLES OF TRIGONOMETRY. — This great
science has been referred to in several places. It has been
developed from the principles of similar triangles. If a
sufficient number of lines and angles of a figure are known to
permit the definite construction of the figure, then the
remaining, or unknown parts, may be determined by
measuring the completed figure. The object of trigonometry

170

PLANE GEOMETRY

is to furnish methods of calculating the unknown parts with
any desired degree of accuracy.

The Functions of an Acute Angle. — Draw any acute angle
x; draw a line 0 perpendicular to side A. In the triangle
HO A, the ratio of any two sides is called a function of
angle x.

Four of the six functions are called:

side opposite
hypotenuse

sine x

0
H

side adjacent A

cosine x = -r— ~ = 7?

hypotenuse H

0
A

_ side opposite
side adjacent

side adjacent A

cotangent x = . , — r— = 7:

side opposite 0

EXERCISES

1. Show that the value of a function is independent of the position

0 0'

of the perpendicular; i.e., ^ = ^7, etc.

2. Draw angles of 10°, 20°, 30°, etc., to 80°. In each figure draw a
perpendicular and measure the three sides of the right triangle
formed. Calculate the values of the sine, cosine, tangent and
cotangent of each angle, taking the results to two decimals.
Thus, in the figure, x = 20°, 0 = ?, A = ?, H = ?, sin 20° =

^ = = ?, cos 20° = jr = = ?, tan 20° = -v = = ?,

fj[ fj. A\.

cot 20° =£ = =?

SIMILAR TRIANGLES AND SIMILAR POYLGONS
3. Make a table of the results as follows :

171

Angle

Sine

Cosine

Tangent

Cotangent

10°

.17

20°

.34

30°

.50 ,

40°

50°

60°

70°

80°

Problems Solved by Trigonometry. — A triangle whose sides
are many feet or miles in length, or so large that its sides
reach from the earth to the sun or from earth to a star,
may be similar to one of the triangles used in calculating
a table of functions. The function values of the acute
angles of these larger triangles are therefore equal to the
values found from the constructed triangles.

EXERCISES

4. The distance AB across a river is required. A surveyor meas-
ures a distance EC at a right angle to the required line; and at
point C he measures Z.C = 30°. To which of the triangles of
Exercise 3 is this triangle (full size) similar?

2000 FT.

A R

To what function of 30° is TV equal?

> taking the proper value from

Form the equation, on?)?)

the table. Solve for AB.

In solving the problems which follow, use the more exact func-

tion values given in the table at the end of this section.

172 PLANE GEOMETRY

5. The horizontal distance from a point A on the ground, to the
base of a flagpole is 130 feet. At A the angle of elevation of
the top of the pole is 50°. Find the height

of the pole above the transit. Add 4 feet
for the height of transit above the ground
at the base of the pole (Compare 115,
Exercise 9).

6. A hillside slopes at an angle of 20° with the
horizontal. What is the rise in 500 feet
measured along the slope?

7. An aeroplane is observed over an aviation field that is known
to be 3.5 miles distant, at an angle of elevation of 10°. What
is the altitude of the aeroplane (Compare 115, Exercise 4)?

8. Find the base of an isosceles triangle whose vertex angle is 40°
and whose equal sides are 20 in.

Help. — Draw the bisector of the vertex angle.

9. A pendulum 10 feet long swings from end to end of a straight
line 3.4 feet long. Through what angle does the pendulum
swing?

Help. — Draw the bisector of the total angle.

10. The radius of a circle is 10 inches; radii drawn to the extremi-
ties of a chord form a central angle of 80°. Find the length of
the chord.

11. Calculate the projection of a sect 5 inches long on a line which
forms an angle of 40° with the sect.

12. Calculate the angles of the isosceles triangle of 97, Exercise 4.
Help. — Cosine of one of the base Zs = % = .6667.

13. Calculate the base of the isosceles triangle of 98, Exercise 3.

Help.— ~ = sine \ vertex angle /. base = 6 sine 35°.
o

14. A vertical pole 12 feet high casts a shadow on a horizontal
plane 13.33 feet long. Find the angle of elevation of the sun.

15. The projection of a sect 20 inches long is 17.5 inches. Find the
angle which the sect makes with the line of projection.

16. Calculate the functions of 30°, 45° and 60° from the values
obtained in 254.

Oblique Triangles can also be solved by dividing them into
two adjacent or overlapping right triangles.

EXERCISES

17. The distance AB, across a pond is required. A point C is
selected from which the lines CA = 280.5 feet, and CB =

SIMILAR TRIANGLES AND SIMILAR POLYGONS

173

620.8 feet are measured. Angle C
is measured 37°.

Helps.— (1) In AACD, calculate p
and w; — (2) n = a — m; — (3) in
AABD, calculate Z.B and then side C
c; or, calculate c = \/p2 + n2.

18. An island C is located on a map by selecting
two known points A and B on the shore,
measuring AB = 1250 feet, /.CAB = 67°;
^CBA = 48°. Find AC and EC.

Helps. — (1) In AABD, calculate p and m;
—(2) ZC = 180°- (A +5);— (3) in ABCD,
calculate n and a; — (4) 6 = m + n.

19. A line is to be extended past a
building. A surveyor measures
/.DAB = 153°. AB = 325 feet,
/.ABC = 105°. Calculate BC

so that point C shall be on the D

line DA extended, and Z.BCE -"

so that CE shall be the prolongation of line DAC.

20. The height of a stack is found as follows : A transit is set up at
point A and the angle of

elevation of point C is mea-

sured, /.DAC = 32°; the

height of the transit above

the ground is measured on

the wall of the building;

DE =4.7 feet. The tele-

scope of the transit is turned *-

over and a point B is located E

in the plane of AACD. We will suppose that the transit

stands at the same height at B as at A. Angle ABC is measured

24°; AB is measured on the ground = 125.4 feet.

Helps.— Calculate (1) AF; (2) ZACF; (3) AC; (4) CD; (5) CE.

Solve the triangle of 99, Exercise 1.

21.

Helps.— (1) = sine 57°;-(2)

sine A; — (3) -r-^

cosine 57°; — (4) ^-= = cosine A; — (5) c = m -f- n; — (6) C = 180° —
(B + A).

Geometrical methods of calculation which are limited to
special cases in geometry, may be generalized by trigonome-
try.

174 PLANE GEOMETRY

EXERCISES

22. Find a formula for calculating the projection m, of a sect s, on
a given line for any angle x, between the sect and the line of
projection (see 254).

TT T Wl

Helps. ——= cosine x.
.'. m = s. cosine x.

23. Calculate the projection of a sect of 24" which makes an angle
of 30° with the line of projection.

m = 24 X cosine 30° = 24 X .8660 = etc.

24. Calculate the projection of a sect of 38" which makes an angle
of 51° with the line of projection.

25. Find a general formula for the third side of a triangle when
two sides and their included angle are given (see 252).

Helps. — a2 = 62 + c2 — 26m; — = cosine A
.'. a2 = 62 + c2 — 26rcosine A

26. Calculate the third side of a triangle of which two sides and
their included angle are respectively, 6 = 1.5", c = 2.9", and
A = 57° (see 98, Exercise 1).

Helps.— a* = (1.5)2 + (2.9)2 - 2(1.5)(2.9) cos 57°.

27. Calculate the angles B and C of the triangle of Exercise 26.
Helps. — (1) ™ = cosine A; — (2)n = 6 - m; — (3) cosine C = - ;
(4) B = 180° - (A + C)

Some geometrical methods of calculation may be simplified
by trigonometry.

EXERCISES

28. Find a formula for the diagonal of a square in terms of the
side s (260, Theroem (14)).

Helps.-(l) = sine 45°; (2) d =

29. Calculate the diagonal of a square whose side is 50 (261,
Exercise 41).

30. Calculate the distance from home plate to second base in a
base ball diamond.

31. Find a formula for the altitude of an equilateral triangle in
terms of the side of the triangle (260, Theorem (16)).

Help.— a = sine 60°.
s

32. Calculate the altitude of an equilateral triangle of which the
side is 120 (261, Exercise 42).

SIMILAR TRIANGLES AND SIMILAR POLYGONS

175

TABLE OF FUNCTIONS

Angle

Sine

Cosine

Tangent

Cotangent

0°

0000

1 0000

0000

1°
2°
3°
4°

.0175
.0349
.0523
.0698

.9998
.9994
.9986
.9976

.0175
.0349
.0524
.0699

57.2900
28.6363
19.0811
14.3007

88
87
86

5°
6°
7°
8°
9°

.0872
. 1045
.1219
.1392
.1564

.9962
.9945
.9925
.9903
.9877

.0875
.1051
.1228
.1405

.1584

11.4301
9.5144
8.1443
7.1154
6.3138

85
84
83
82
81

10°
11°
12°
13°
14°

.1736
.1908
.2079
.2250
.2419

.9848
.9816
.9781
.9744
.9703

.1763
.1944
.2126
.2309
.2493

5.6713
5.1446
4.7046
4.3315
4.0108

80
79

78
77
76

15°
16°
17°
18°
19°

.2588
.2756
.2924
.3090
.3256

.9659
.9613
.9563
.9511
.9455

.2679
.2867
.3057
.3249
.3443

3.7321
3.4874
3.2709
3.0777
2 . 9042

75
74
73
72
71

20°
21°
22°
23°
24°

.3420
.3584
.3746
.3907
.4067

.9397
.9336
.9272
.9205
.9135

.3640
.3839
.4040
.4245
.4452

2.7475
2.6051
2.4751
2.3559
2.2460

70
69
68
67
66

25°

26°
27°
28°
29°

.4226
.4384
.4540
.4695
.4848

.9063
.8988
.8910
.8829
.8746

.4663
.4877
.5095
.5317
.5543

2.1445
2.0503
.9626
.8807
.8040

65
64
63
62
61

30°
31°
32°
33°
34°

.5000
.5150
.5299
.5446
.5592

.8660
.8572
.8480
.8387
.8290

.5774
.6009
.6249
.6494
.6745

.7321
.6643
.6003
.5399
.4826

60
59
58
57
56

35°
36°
37°
38°
39°

.5736
.5878
.6018
.6157
.6293

.8192
.8090
.7986
.7880
.7771

.7002
.7265
.7538
.7813
-8098

.4281
.3764
.3270
.2799
.2349

55
54
53
52
51

40°
41°
42°
43°
44°
45°

.6428
.6561
.6691
.6820
.6947
.7071

.7660
.7547
.7431
.7314
.7193
.7071

.8391
.8693
.9004
.9325
.9657
1.0000

.1918
.1504
.1106
.0724
.0355
.0000

50
49
48
47
46
45

Cosine

Sine

Cotangent

Tangent

Angle

176

PLANE GEOMETRY

Trigonometry has been developed so that it meets the
needs of the most exact scientific work, such as surveying,
engineering, navigation, military science, and astronomy.
Seven place tables of the logarithms of the functions are
published, and ten place tables are used in some calculations.
There are also special formulas for solving oblique tri-
angles, finding areas, and for many other uses. A text on
trigonometry should be used if further study is desired.

In using the table on page 175, angles from 0° to 45° are
read in the left hand column and the names of the functions
are found at the top of the table; angles from 45° to 90° are
read in the right hand column and the names of the functions
are found at the bottom of the table.

266. PROBLEMS FOR FIELD WORK.

1. The distance across a stream, or inaccessible street,
etc., by the method of 237.

2. The distance between two points on opposite sides
of a building or pond, by the method of 237.

3. Lay out a baseball diamond when the position of
home plate and the direction of second base are given.

cp/ ^
V

//«-

Let H be the home plate, and HC the direction of center
field.

(1) Find the distance to second base, H2 = \/902 + 902.

(2) Divide by 2 and measure to point A.

SIMILAR TRIANGLES AND SIMILAR POLYGONS

177

(3) Erect perpendiculars A B, AD.

(4) Measure Al = A3 = HA = ?

(5) Check by measuring HI = 12 = 23 = 3H = 90 feet.

4. Make a map of some curved path or road or border of a
stream, by the method of 237.

5. Find the distance from a position A to a place B as
follows: First observer proceeds on a line perpendicular
to line AB, to C; second observer proceeds

along line BA to D, until observer at C signals
him that the angle BCD measured by an optical
square (148) is a right angle. The distances AC
and AD are paced or measured with a tape.
A B may then be calculated (248).

This method of estimating the distance to an
enemy station, etc., is used in warfare.

6. Find the distance from a point inside

the school grounds to an inaccessible point outside the
grounds by the method of 265, Exercise 4, or Exercise 18.

CHAPTER XII
CIRCLES

PRINCIPLES DETERMINED EXPERIMENTALLY

267. Circles Defined. — Refer to 14 for general definitions.

Congruent, or equal, circles are circles which have the
same radii.

Two points on the circumference of a circle divide the
circumference into a minor and a major arc.

A central angle is formed by two radii. -

An inscribed angle is formed by two chords meeting in the
circumference.

. A chord is said to subtend the two arcs opposite the chord,
and conversely. A central or an inscribed angle subtends
or intercepts the included arc. To subtend means to be oppo-
site to.

A segment is a part of a circle bounded by a chord and the
subtended arc.

A sector is a part of a circle bounded by two radii and the
intercepted arc.

An inscribed angle whose sides pass through the extremi-
ties of a chord is inscribed in the segment cut off by the chord.

Two circles may in some positions have a common chord
or common tangents.

A secant is a straight line intersecting a circle in two points
and extending outside the circle in one or both directions.

A polygon is inscribed in a circle when all its vertices
lie in the circumference. The circle is then circumscribed
about the polygon.

A polygon is circumscribed about a circle when all its sides
are tangents of the circle. The circle is inscribed in the
polygon.

178

CIRCLES 179

EXERCISES

1. What part of a circumference is an arc of 60°; 45°; 25°; 72° 10'?

2. Define complementary, supplementary and explementary arcs
(see 17).

3. Draw a circle with a chord subtending an arc of (approximately)
75°. What major arc does the chord subtend?

4. What chord subtends equal arcs of a circumference?

5. What must be given in order to determine the position and size
of a circle?

6. Draw a circle with a central angle subtending an arc of 90°.
How many degrees are subtended by the explementary central
angle?

7. Draw an inscribed angle subtending an arc of 90°.

8. Draw an angle inscribed in a segment of (approximately) 150°.

9. Define a semicircle as a segment.

10. Define a semicircle as a sector of a certain number of degrees.

268. Experiment I. — Relations between central angles
and the subtended arcs.

(a) Equal central angles in the same or in two equal circles.

Draw a circle; draw several pairs of equal central angles.
Observe the relative lengths of the subtended arcs of each
pair of equal angles. State the result.

(6) Unequal central angles in the same or in two equal circles.

Draw several unequal central angles. Observe the relative
lengths of the subtended arcs. Result.

parts (by trial) into three equal angles; divide one of these
parts again into three equal angles; divide one of these parts
into five equal angles. This will give a central angle of one
degree (of angle). Observe the part of the circumference
which is subtended by the central angle of one degree; the
number of degrees of arc subtended by each central angle;
the number of degrees of arc which will be subtended by
any central angle. State result.

(d) The measures of the arcs subtended by a central angle
of two concentric circles.

Draw two concentric circles; draw a central angle.

180 PLANE GEOMETRY

Observe the relative lengths of the two subtended arcs
when measured in inches on the circumference; and their
relative lengths when measured as fractional parts of
the two circumferences, i.e., in degrees of arc. Result.

EXERCISES

1. What arc measured in degrees of arc will a central angle of 75°
subtend in a circle of 2" radius? In a circle of 5" radius?

2. If an arc of 30° measures 5" in a given circle, what will an arc
of 90° measure in a circle of double the radius ?

269. The Protractor.— The principle of 268(d) explains
the use of a protractor, which can be used to (a) measure a
given angle; (b) construct an /

angle of a given measure; (c)
measure a given arc; (d) con-
struct an arc of a given mea-
sure on a given circumference.

EXERCISES

1. Draw an arc AB of about
one inch in radius. Measure
it in degrees.

2. Construct an arc of 35° with a radius of 6 inches.

270. Experiment II. — Properties of chords.

(a) The relative lengths of arcs subtended by equal chords,
in the same or in two equal circles.

(b) The relative lengths of arcs subtended by unequal chords.

(d) The relative distances of unequal chords from the
center of the circle.

(e) The relation of a diameter perpendicular to a chord, to
the chord and to the two subtended arcs.

(/) The property of a chord equal in length to the radius of
the circle.

In considering Part (6), observe whether the minor or
major subtended arcs vary in the same way as the chords;
also whether there is a simple numerical relation between

CIRCLES 181

the lengths of chords in inches and the subtended arcs in
degrees.

In considering Part (d), observe whether there is a
simple numerical relation between the length of a chord
and its distance from the center of the circle.

In considering Part (/), draw a number of chords equal
in length to the radius of the circle, end to end around the
circle.

State all results.

EXERCISES

1. Construct an equilateral triangle with its three vertices in the
circumference of a circle.

2. Construct a symmetrical six-pointed star by drawing a circle
and using Part (/) above.

3. Draw a circle and lay off an arc of 120° by using Part (/) above

271. Experiment III. — A method of
bisecting an arc.

(a) By rule and compass.

(6) By measurement.

Part (a) : Use the principle of 270 (e) .

272. Experiment IV. — Properties of tangents.

(a) The relative position of a tangent and the radius
drawn to the point of contact (or point of tangency).

(b) The relative lengths of two tangents drawn to a circle
from the same external point.

In considering Part (c), draw two circles of different
radii, that are (1) externally out of contact, (2) externally
tangent, (3) intersecting, (4) internally tangent, (5) internally
out of contact. Observe the number of common tangents
that can be drawn in each case. State all results.
EXERCISES

1. What position have two tangents drawn at the ends of a
diameter?

2. In what position are two circles if three common tangents can
be drawn to them?

182 PLANE GEOMETRY

3. How many common tangents have two concentric circles?

4. Invent a rule and compass construction for drawing a line
through a given point parallel to a given line, based on a com-
mon tangent to two equal circles.

5. Draw a circle and mark a point on the circumference. Show
how the direction of the tangent at this point can be accurately
determined from a radius drawn to the point of tangency.

273. Experiment V. — Properties of the line of centers
of two circles.

(a) The relation between the line of centers of two inter-
secting circles and of their common chord.

(6) The position of the point of contact of two tangent
circles with respect to the line of centers.

(d) The position of the points of intersection of a pair of
internal or external common tangents with respect
to the line of centers.

In considering Part (a), draw (1) two equal circles, and
(2) two unequal circles.

In considering parts (6) and (c), draw (1) two externally
tangent circles, and (2) two internally tangent circles.

In considering part (d), draw two circles that are (1)
externally out of contact, (2) externally tangent, (3)
intersecting. State results.

EXERCISES

1. Draw a circle of 1^" radius; mark a point on the circumference.
Draw a second circle of 2" radius tangent externally to the first
circle at this point.

Helps. — Draw the radius of the first circle to the point of tangency
and extend it outside the circle. Use Part (6) above.

2. Draw an arc of 3" radius and mark a point on it. Draw a
second arc (or circle) of 1" radius tangent internally to the first
arc at the given point.

274. Experiment VI. — Properties of inscribed angles.

(a) The size of an angle inscribed in a semicircle.

(b) The size of an angle inscribed in a minor segment.

CIRCLES 183

(d) The relative size of angles inscribed in the same segment.

(e) The measure of an inscribed angle in terms of the
subtended arc.

State results.

275. EXPERIMENT VII. — A method of drawing a line
perpendicular to a given line.

(a) When the point is in the given line.

(b) When the point is without the given line.
Adapt Part (a) of 274 to both constructions.

276. EXPERIMENT VIII. — The number of circles that
can be drawn through given points.

(a) One given point.

(b) Two given points.

(d) Four or more given points.

(e) The conditions that four or more given points shall
lie on the same circle.

In stating the results of Parts (a), (6), (c), state where the
centers of the circles which contain the given points are
found, as well as the number of such circles ; that is, describe
the locus of their centers.

Part (d) : If the points are placed at random, will any
circle contain them all? RESULT. — In general, etc.

Part (e): A certain set of lines must be concurrent in
order that the points shall be on a circle (i.e., concyclic).

EXERCISES

1. What is the locus of the centers of circles which contain the ends
of a sect?

2. Can a circle be drawn containing the vertices of a square? Of
an oblong? Of a rhombus? Of an isosceles trapezoid?

3. Are the vertices of an oblique scalene triangle concyclic?

4. How many points may be concyclic? How many points must
be concyclic?

5. If three points chance to lie in the same straight line, where is
the center of the circle which may be said to contain them?

184 PLANE GEOMETRY

277. EXPERIMENT IX. — THE NUMBER OF CIRCLES THAT
CAN BE DRAWN TANGENT TO GIVEN STRAIGHT LINES.

(a) One given line.

(b) Two given lines.

(d) Four or more given lines.

(e) The condition that four or more given straight lines
shall be tangent to the same circle.

Part (6): Consider two cases, (1) the given lines inter-
secting, (2) the lines parallel.

Part (c): Consider the lines, (1) two lines parallel and
the third line intersecting them, (2) the three lines parallel,

(3) the lines inclosing a triangle, (4) the lines concurrent.
In (3), note that four circles can be drawn tangent to the
three given lines.

Part (e): Find the set of lines that must be concurrent
as the necessary condition.

EXERCISES

1. What is the locus of the centers of the circles in Part (b), (1)
and (2)?

2. What are the tangent circles called in respect to the triangle
enclosed by the three lines in Part (c), (3)?

3. Can a circle be drawn tangent to the sides of a square? Of an
oblong? Of a rhombus? Of an isosceles trapezoid?

278. EXPERIMENT X. — ANY OTHER PROPERTIES OF
CIRCLES WHICH CAN BE DISCOVERED EXPERIMENTALLY.

279. Construction of Circles from Given Data. — Find
all possible solutions of each problem. Discuss special and
impossible cases, which should be illustrated by sketches.

CIRCLES

185

Use rule and compass or measurement methods, and avoid
trial and error solutions.

Construct a circle (or circles) of which there are given:
1. The length and position of a chord, and the radius.

Helps. — AB is the given chord; r is the given radius.

In general two circles can be drawn.

Special cases: (1) When r<% AB, no circle is possible. (2)

When r = ^ AB, one circle, as C, can be drawn.

2. The length and position of a
chord and the position of a tangent-^"

through an end of the chord.

3. (The position of) a tangent,
the point of tangency, and a
point on the circumference.

4. Two intersecting tangents
and the radius.

5. A chord and a straight line containing the center of the circle.

6. A tangent circle and the center of the
required circle.

7. Two non-intersecting circles, both of

which are tangent to the required circle, \ ^ / x

and the radius of the required* circle. \ / B

8. A tangent circle and a point in the cir-
cumference of the required circle and the radius of the re-
quired circle.

9. The length and position of a chord and a circumference con-
taining the center of the required circle.

10. A tangent circle and the point of tangency, and a point in the
circumference of the required circle.

186

PLANE GEOMETRY

11. State other conditions for the construction of a required circle
(or circles). Make the conditions definite but not conflicting.

280. Constructions Relating to

rule and compass methods:

Circles. — Construct by

1. A tangent to a given circle parallel to a
given straight line.

2. A tangent to a given circle perpendicular
to a given straight line.

3. A chord in a given circle of given length
and parallel to a given straight line.

4. A chord of a given circle passing through a given point within
the circle and bisected at that point.

5. A right triangle when the ^ H

lengths of the hypotenuse and p {

of one of the perpendicular
sides are given. Construct (1)
by 274 (a); (2) by any other
method.

6. A right triangle when the
hypotenuse and the altitude to
the hypotenuse are given.

7. A chord of given length in a

given circle, which shall^pass^through a given point within
the circle.

8. Four circles within a given square, each tangent to the sides of
the square and to the two adjacent circles.

9. A chord of given length in a given circle, which shall pass (if
produced) through a given point outside the circle.

281. REVIEW EXERCISES

1. Draw two tangent arcs of radii 2" and 3.5", 45° and 60° respec-
tively, in two positions of contact.

2. Construct the locus of
the centers of circles
tangent to a given circle
at a given point on the
circle.

3. Construct an arc of 60°,

6" radius, without using a protractor.

4. Construct a circle (or circles) tangent to two given intersecting
straight lines and having its center in a third given straight line.

5. Construct a circle tangent to a given circle and to a given
straight line and having a given radius.

CIRCLES 187

6. Two chords intersect on the circumference of a circle and have
their other ends at the ends of a diameter of the circle. The
chords are 5" and 6" long. Calculate the diameter.

7. A chord of a circle is 10" and its distance from the center is 6".
Calculate the radius of the circle.

8. The perpendicular from the center of a circle to a chord sub-
tending an arc of 120° is 3". Calculate the radius and the
length of the chord.

9. A chord subtending an arc of 60° is 5" from the center. Calcu-
late the radius and the chord.

10. The radius of a circle is 10". Calculate the length of a chord
subtending an inscribed angle of 45°.

11. The center sect of two circles is 10"; the radii are 4" and 2".
Calculate the length of an internal common tangent.

12. Draw two externally tangent circles. Prove experimentally
that the internal common tangent bisects the two external
common tangents.

13. Draw a number of chords of a circle concurrent in a point of
the circumference. What is the locus of their middle points?

14. Draw a number of secants of a circle concurrent in a point out-
side the circle. What is the locus of the bisection points of
both the major and minor segments of the secants?

15. Make an experiment similar to that of Exercise 14 for a pencil
of chords intersecting within the circle.

16. Draw a circle or an arc of more than 180° without marking
the center. Find the center by using the right angle of the
draughtsman's triangle (274 (a)).

17. Construct a circle tangent 6
to the two circles A and

B and containing the
point C. Draw (a) the
locus of points equidis-
tant from circle A and
point C; (6) a similar
locus for circle B and
point C; (c) the required circle with center at O.

18. Exercise 17 is the geometric solution of the problem of locating
the German " super-gun" which shelled Paris in 1917-18. See
Mathematics Teacher for Dec., 1918.

Three stations, A, B, C, were located and plotted. A listening
apparatus was installed at C, and observers were stationed at
A and B. Each observer timed with accurate stop watches
the interval between hearing the discharge of the gun at C and
hearing it at his station, A or B. Let this interval be 1.7

188 PLANE GEOMETRY

seconds for the A observer, and 2.5 seconds for the B observer.
Sound travels about 1100 feet per second. Draw the circle A
with a radius 1.7 X 1100 feet, and circle B with radius 2.5 X
1100 feet. Circle O is found as in Exercise 17, or by trial; at
the center, O, of which the gun is located.

APPLICATIONS

282. Railroad and Road Curves. — A preliminary survey
of a proposed railroad consists in staking out a series of
straight lines, locating the points of intersection, measuring
the lengths, AB, BC, etc., (Fig. 1) and the inter section angles,
x, y, etc.; also in estimating the radii of the connecting
circular arcs which will best conform to the surface of the
ground.

The lines AB, BC, etc., are known as tangents.

FIG. 1.
EXERCISES

1. Show that the central angle at O equals the intersection angle x
(Fig. 2).

FIG. 2.

CIRCLES

189

2. How do the tangent lengths BH and BK compare (Fig. 2)?

3. Plot the tangents on a scale of 1000 feet to the inch, from the
following notes (Fig. 1):

Tangents

Lengths,
feet

Intersection
IS.

Values

Direction
of turn

2000

40°

To right

BC
CD
DE

2500
1500
3000

C
D
E

90°
50°

35°

To left
To right
To right

To Lay Out a Circular Arc by Chords. — The necessary
calculation are made by trigonometry. Thus if, in Fig. 2,
x = 40°, £HOK = 40°, and the arc HK may be divided
into 10 equal arcs of 4° each. The radius HO = 1433 feet,
the tangents, BH = BK = 522 feet; and the chords HL =
LM = MN = 100 feet. Set up a transit at H, sight on
B, turn 2° and measure 100 feet to L; turn another 2° and
measure LM = 100 feet; etc.

EXERCISES

4. Plot the curve described above on the drawing of Exercise 3.

5. Why are the angles BHL = LHM = MHN = etc. = 2° (Fig.

2)?

The Two -transit Method of Laying Out a Circular Arc.—

The arc HK may also be laid out with two transits, without
making any measurements with a tape.

After calculating BH and BK (Fig. 2) and locating points
H and K, set a transit at each of these points. In order to
locate points L, M, N, etc.; turn ZBHL = 2° with the H-
transit, and an angle BKL = 18° with the X-transit, and set
a stake on both lines of sight; then turn /.BUM = 4° and
£BKM = 16°, and set stake M as before; etc.

190

PLANE GEOMETRY

EXERCISE

6. Arrange the angles of each transit in a table:

Point on arc

Z turned by //-transit

Z turned by /^-transit

2°

18°

Etc.

Compound Curves. — Such curves are composed of two
or more arcs successively tangent to each other.

A reverse curve is a compound curve composed of two
arcs of equal radii, joining two parallel tangents.

FIG. 1.

FIG. 2.

EXERCISES

7. Design, by trial, a compound curve in which arc AB (Fig.l)
has a radius of 300 feet and arc BC a radius of 500 feet, and in
which the intersection angle x = 90°.

8. Design a reverse curve joining two parallel tangents the distance
between which is 30 feet, the radius of both arcs being 30 feet.

£)2 I |^2

9. Show that in Fig. 2, R = — ~ — .

Help.—OiE = 2R -W't (hO~22 = Ch

10. Calculate R when D = W; also when D = W = 30 ft.; also
when D = 30 ft. and W = 20 ft.

CIRCLES

191

Y-curves. — Such curves connect two railroad tracks as
shown, and are sometimes used to reverse the direction of
an entire train.

EXERCISES

11. Show that EC = EF = ED.

12. Show that EC = <\/frr, where R and r are the radii of the curves.
Helps. — AAEB is a right A. Apply 248.

13. Calculate EC when R = r = 200 ft. Also when R = 200 ft.
and r = 150 ft.

283. THE CIRCLE IN ARCHITECTURE. — Arches, columns,
domes, doorways, windows, and other architectural features
exhibit many beautiful and interesting uses of the circle.

The Gothic Arch. — A B is the span; CD is the altitude.

EXERCISES

1. Design a Gothic arch with as pan of 40 ft., and an altitude of
30ft.

2. Design a Gothic arch with a span of 30 ft., and an altitude of
40ft.

The Breccia Arch. — A ABC is equilateral.
3. Design a Breccia arch with a span of 40 ft. and an altitude DE
of 30 ft.

192

PLANE GEOMETRY

The Persian (ogee) Arch. — Trisect AC at E and F; draw
FG±AC; prolong GE to H; G and H are the arc centers.
The arch may also be drawn with E at the middle of AC.

4. Design Persian arches with 30 ft. span and 40 ft. altitude; (a)
with E at the trisection point; (6) with E at the bisection point.

The Basket-handle Arch.

5. Design a basket-handle arch whose span = 60 ft., altitude =
25ft.

The Tudor Arch.

6. Design a Tudor arch whose span = 60 ft., altitude = 30 ft.

Circular Arches.

7. Design circular arches: (a) span = 120 ft., altitude
(6) span = 120 ft., altitude = 40 ft.

60 ft.;

CIRCLES

193

284. A PROBLEM FOR FIELD WORK. — To lay out a circu-
lar arc tangent to two straight lines, by the two-transit

method (282). Select two straight curbs or center lines of
paths, and mark such points as A, B, C, D, by driving stakes
or with a chisel. Set transits over points A and D and sight
on points B and C respectively, and set stake E on the two
lines of sight. Set the transit at E and measure angle
AED. Measure any equal distances EF and EG, say 200
feet. The central angle 0 of the arc connecting points
F and G = 180° - LAED = Z.FOG. Divide Z.FOG by
10 = /.FOH. Set the transits over points F and G. Sight
the /^-transit on point G and turn an angle GFH = 9
times | /.FOH. Sight the (7-transit on point F and turn
an angle FGH = \£FOH. Set stake H on the line of
sight of both transits. Proceed in this way until the 9
stakes H, 7, J, K, etc., are located.

This problem is greatly simplified by selecting two perpen-
dicular lines, or by laying out two perpendicular lines from
any selected point E.

CHAPTER XIII
CIRCLES

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

285. The first step in classification is the selection of
the most elementary experimental properties of circles as a
basis of analysis.

CIRCLE POSTULATES

1. Any point within a circle is at a less distance from the
center than the radius; and conversely.

2. Any point without a circle is at a greater distance from
the center than the radius; and conversely.

3. If two or more points are equally distant from another
point, a circle may be drawn with its center at that
other point which will contain all of the equidistant
points.

4. A circle is determined by (a) the center and radius;

(6) by a diameter.

5. A diameter bisects a circle.

6. Equal chords of the same circle or of equal circles
subtend equal arcs; and conversely.

7. Equal chords of the same circle or of equal circles are
equally distant from the center; and conversely.

8. Unequal chords of the same circle or of equal circles
subtend unequal arcs; the greater chord subtending
the greater minor arc.

9. Unequal chords of the same circle or of equal circles
are unequally distant from the center, the greater
chord being at the less distance from the center.

10. A diameter is the greatest chord.

194

CIRCLES 195

11. (a) A tangent is perpendicular to the radius drawn
to the point of tangency. (6) A line perpendicular
to a radius at its extremity is tangent to the circle.

12. (a) Every point of a tangent lies outside the circle
except the point of contact. (6) If every point of
a straight line lies outside a circle except one point,
the line is a tangent.

13. Equal central angles of the same circle or of equal
circles subtend equal arcs; and conversely.

14. A central angle has the same measure in degrees
of angle as the subtended arc has in degrees of arc.
Or, a central angle equals its subtended arc.

15. The line of centers of two tangent circles contains the
point of tangency.

286. The second step in classification is the analysis of
other experimental properties of circles in terms of more
elementary properties.

287. Theorem I. — In the same circle or in equal circles,
equal chords subtend equal central angles;

and conversely.

Parti.

Hypothesis. — Circle 0 with equal
chords AB and CD, subtending central
angles x and y.

Conclusion. — x = y.

Analysis.

1. Arc AB = arc CD What postulate?

2. .'. /.x = /.y What postulate ?
Part II.

Hypothesis. — Circle 0 with equal central angles x and y,
subtending chords AB and CD.

Conclusion. — Chord AB = chord CD.
Analysis.

1. Arc AB = arc CD What postulate?

2. .*. chord AB = chord CD What postulate?
Help. — The postulates used are 6 and converse, and 13

and converse.

196

PLANE GEOMETRY

288. Theorem II. — The perpendicular bisector of a chord
contains the center of the circle.

Help. — Use a locus theorem or a theorem of isosceles
triangles. (See also 217, Theorem (5) .)

Corollaries. — (1). The perpendicular bisectors of two
non-parallel chords of a circle determine the center of
the circle.

2. Three points on the circumference of a circle determine
the center of the circle.

289. Theorem III. — A diameter perpendicular to a chord
bisects it and both major and minor subtended arcs.

Corollary. — A radius perpendicular to a chord bisects
it and the subtended arc.

290. Theorem IV.— Two tangent
circles have a common tangent at the
point of tangency.

Helps. — (1) Draw AB, which
contains point C; Postulate? (2)
Draw CD tangent to circle A; (3)
CD.LAC; why? (4) /. CD is tangent to circle B; why?
Analyze also for two internally tangent circles.

291. Theorem V. — A perpendicular to a A B
tangent at the point of tangency contains the

center of the circle.

Helps.— (1) Draw OA; (2) OA coincides
with AC; etc.

292. Theorem VI. — Two tangents drawn
to a circle from an exterior point are equal.

Helps. — (a) Draw a line from the point to the center of
the circle, and radii to the points of tangency. Prove
that the triangles are congruent. Or, (6) connect the
points of tangency and draw radii as before. Use principles
of isosceles triangles. Or, (c) draw a circle concentric to
the given circle and containing the given exterior point.
Prolong the tangents so that they shall be chords of this
larger circle. Use circle Postulate 11 (a) and 7 converse,
and Theorem III or Corollary.

CIRCLES

197

ADDITIONAL THEOREMS

293. (1) The line of centers of two intersecting circles is
a perpendicular bisector of the common chord (See also
217, Theorem (6).)

(2) The line which joins the center of a circle with the center
of a chord is perpendicular to the chord and bisects the arc
subtended by the chord.

(3) A radius which bisects an arc bisects the chord sub-
tended by the arc and is perpendicular to

the chord.

(4) // a straight line intersects two con-
centric circles the two sects intercepted between
the circles are equal.

(5) Two chords which are both perpen-
dicular to the same chord are equal.

Help.— Show that OA = OB.

(6) Two equal chords which intersect
within a circle divide each other so that the
segments of one chord equal the segments of
the other.

(7) A line drawn from the center of the
circle to the point of intersection of two
equal intersecting chords, bisects the angle
formed by the chords.

Analyze also when the chords intersect
the circle.

(8) // two secants which intersect outside a circle make
equal angles with a line drawn from the center of the circle to
the point of intersection of the secants, the whole secants are
equal and the segments into which they are divided by the circle
are respectively equal.

(9) A side of an equilateral inscribed hexagon is equal to
the radius of the circle.

Helps. — Draw radii; the central angles are equal; each
central angle is 60°; each triangle formed is equiangular;
.'. each triangle is equilateral.

(10) Two equal chords which intersect within a circle inter-
cept one pair of equal opposite arcs.

without

198

PLANE GEOMETRY

(11) Two chords which make equal angles with a diameter
drawn through their point of intersection are equal.

(12) A chord which is drawn through a
given point within a circle perpendicular
to the diameter through the given point,
is shorter than any other chord drawn
through the given point.

Helps. — Draw OB; show that
OA>OB;etc.

(13) A chord of the larger of two con-
centric circles which is tangent to the
smaller circle is bisected at the point of
tangency.

(14) The line joining the point of
intersection of two tangents with the center

of the circle: (a) bisects the angle formed by the tangents;
(b) bisects the central angle formed by the radii to the
points of tangency; (c) bisects the arcs subtended by the
tangents; (d) is a perpendicular bisector of the chord of contact
of the tangents.

(15) The side of a circumscribed square equals the diameter
of the circle.

(16) The tangents drawn at the extremities of a diameter are
parallel.

(17) The tangents drawn at the extremities
of a chord make equal angles with the chord.

(18) Chords of the larger of two concentric
circles which are tangent to the smaller of the
two circles, are equal.

(19) The sum of two opposite sides of a
circumscribed quadrilateral equals the sum of the other two
opposite sides.

Help. — Apply 292.

(20) The central angle formed by radii which are drawn to
the points of tangency of two tangents which intersect outside
the circle, is supplementary to the angle formed by the tangents.

(21) The line of centers of two circles externally out of con-

CIRCLES 199

tact contains the points of intersection (a) of the internal
common tangents, and (6) of the external common tangents.
Helps. — Draw lines from the point of intersection of the
tangents to the centers of the circles. Show (a) that these
lines lie in the same straight line; (b) that they coincide.

294. The Third Step in Classification. — The more impor-
tant experimental results of Chapter XII have been classi-
fied, but there are many measurement properties of angles
and sects associated with circles, which are not evident
from the inspection of a constructed figure and can only
become known by extending the methods of analysis into
the realms of discovery.

295. Angle Properties of Lines Drawn to a Circle. — Any
two intersecting lines which intersect or meet or touch a
circle, form an angle whose measure is related to the measure
of the subtended or intercepted arc or arcs. The simplest
of these angle measures are stated in Postulates 13 and 14.
Other similar properties will now be deduced.

296. Theorem VII. — An inscribed angle has one-half
the measure of the intercepted arc; or, it is measured by one-
half the intercepted arc; or, it equals one-half, etc.

This principle is best considered in three cases.
Case 1. — When one side of the inscribed angle is a diameter.

Hypothesis. — Inscribed /.x, AB being a diameter.
Conclusion. — Zz = ^ arc AC.
Analysis. — (1) Draw OC.

(2) y = arc AC

(3) x = %y.

(4) .'. x = etc.

200

PLANE GEOMETRY

Case 2. — When the center of the circle is within the inscribed
angle.

Analysis. — (1) Draw diameter BD.

(2) y = \ arc AD.

(3) z = \ arc DC.

(4) .*. x = etc.

Case 3. — State and analyze this case.
Corollaries. — (1) Angles inscribed in the same segment
are equal.

2. Inscribed angles subtending equal arcs are equal.

3. An angle inscribed in a semicircle is a right angle.
(See also 143, Theorem (7).)

4. A principle relating to an angle inscribed in a major
or a minor segment.

EXERCISES

1. What inscribed angle intercepts an arc of 95°?

2. What arc does an inscribed angle of 95° intercept?

297. Theorem VIII. — The angle formed by a tangent and
a chord drawn from the point of tangency has a measure in
terms of subtended arcs.

Helps. — There are two of these angles. Draw a diameter
through the point of tangency; one of the angles to be con-
sidered is the sum, and the other is the difference, of a right
angle and an inscribed angle. The measures of these angles
in terms of arcs can be expressed from known principles.
State the theorem discovered.

CIRCLES 201

EXERCISES

1. A chord subtends an arc of 80°, and a tangent is drawn at one
extremity of the chord. Calculate the value of the angle formed.

2. A chord meets a tangent at the point of tangency at an angle of
75°. What is the value of the arc subtended by the chord (or
intercepted by the angle) ?

3. A chord subtends an arc of 100°, and tangents are drawn at each
end of the chord. What angle is formed by the tangents at
their point of intersection?

298. Theorem IX. — The angle formed by two chords which
intersect within a circle has a measure in terms of subtended
arcs.

Helps. — Draw a chord connecting extremities of the given
chords, forming a triangle; the angle to be considered is
related to angles of this triangle, which are measured by
certain arcs as previously analyzed. Draw the elementary
figures. State the theorem.

EXERCISES

1. Two chords which intersect within a circle intercept opposite
arcs of 70° and 100°. What angle do the chords form?

2. Two chords intersect within a circle forming an angle of 30°;
one of the two intercepted opposite arcs is 20°. What is the
value of the other of the opposite arcs?

299. Theorem X. — The angle formed by two secants which
intersect outside the circle has a measure in terms of the two
intercepted arcs.

Helps. — Draw a chord as shown. The angle to be con-
sidered has a relation in terms of two inscribed jangles.
Draw the elementary figures. Analyze also by drawing a
different chord from that shown in the figure.

202

PLANE GEOMETRY

300. Theorem XI. — The angle formed by a secant and
a tangent which intersect outside the circle has a measure in
terms of the two intercepted arcs.

301. Theorem XII. — The angle formed by two tangents
which intersect outside a circle has a measure in terms of the
two intercepted arcs.

Corollary. — The angle formed by two tangents which inter-
sect outside a circle has a measure in terms of the smaller
intercepted arc alone.

Helps.— (1) A = J(* - 2/); (2) x + y = 360°; (3)
eliminate x.

EXERCISES

1. Two secants which intersect without a circle intercept arcs of
80° and 40°. What angle do the secants form?

2. A secant and a tangent which intersect without a circle intercept
arcs of 100° and 40°. What angle do the secant and tangent
form?

3. Two tangents which are drawn from the same exterior point
intercept arcs of 200° and 160°. What angle do the tangents
form?

4. Two secants meet at an angle of 35°; the nearer of the inter-
cepted arcs is 50°. What is the value of the farther arc ?
Help.— 35° = \(x - 50°).

ADDITIONAL THEOREMS

302. (1) The arcs intercepted on a circle
by two parallel chords are equal.

(2) The arcs intercepted on a circle by
a tangent and a chord parallel to it are
equal.

(3) The bisector of an inscribed angle
bisects the subtended (or intercepted) arc.

(4) The bisectors of all the angles
inscribed in the same segment are con-
current.

State where the bisectors are concurrent.

(5) Opposite angles of an inscribed quadrilateral are
supplementary.

CIKCLES

203

Help. — Consider the arcs which measure these angles.

(6) Corollary of Theorem (5). The sum of one pair of
opposite angles of an inscribed quadrilateral equals the sum
of the other pair of opposite angles.

(7) The value of angle x can be expressed in terms of arcs
b and c. Investigate, and state the theorem discovered.

(8) The value of angle x can be expressed in terms of arcs
b} c and d.

Help.— Show that x = b + }(c + d).
(9) An exterior angle of an inscribed quadrilateral equals
the interior angle at the opposite vertex.

(10) An angle is inscribed in each of the
four outside segments of a circle, which
are formed by the sides of an inscribed
quadrilateral. Investigate the measure of
the sum of these four angles.

(11) State a general theorem concerning

the measure of an angle in terms of subtended arcs, accord-
ing as the vertex is within, on, or without the circle.

(12) An angle formed by two lines passing through the
extremities of a diameter is acute, right, or obtuse, accord-

204 PLANE GEOMETRY

ing to the position of the vertex. Investigate and state the
theorem.

(13) The center of the circumscribed circle of a triangle
lies within the triangle, on a side, or without the triangle,
according to the angle-shape of the triangle. Investigate
and state the theorem.

303. The Fourth Step in Classification. — Make an experi-
mental test of the principles of circles which have been
deduced from the postulates and other experimental
principles.

1. The angle formed by a tangent and a chord drawn from
the point of tangency (297).

Draw the figure of 297 large enough, or extend the angle
lines of the figure as shown, so that angles BAC and AOC
can be measured with a protractor. Compare these measure-
ments and check the principle of 297.

2. The angle formed by two chords which intersect within
a circle (298).

<K
-0

Measure angle AEC, and angles FOG and HOK. Note
that the latter angles measure the subtended arcs LM and
NP. Compare with the principle of 298.

CIRCLES

205

3. The angle formed by two secants which intersect outside
a circle (299).

Measure /.ABC, and ^sDOE and FOG; etc.

4. The angle formed by a secant and a tangent which
intersect outside a circle (300).

5. The angle formed by two tangents which intersect outside
a circle (301).

6. The principle of 302, Theorem (5).

7. The principle of 302, Theorem (7).

8. The principle of 302, Theorem (8).

9. The principle of 302, Theorem (9).
10. The principle of 302, Theorem (10).

304. The Third Step in Classification. An Investigation
of Sect Properties of Lines Drawn to a Circle. — The sects
intercepted by a circumference on any two intersecting
lines which intersect or meet a circle in two points or which
are tangent to the circle, possess some relative or propor-
tional values. The simplest of these sect properties are
stated in 63, Postulate 11, and in 292. Other sect relations
which are not evident from a constructed figure, will now
be deduced from the principles of similar triangles.

305. Theorem XV. — A relation exists between the lengths
of the segments of two chords which intersect

within a circle.

Helps.— (1) Draw AC and BD. (2) Ai
AAEC ~ ABED; because /.A = /.D and
Z.C = Z£; why? (3) State the proportion
between sides AE} CE} BE and DE. (4)
Change the proportion into products.

206 PLANE GEOMETRY

Draw the two triangles under consideration separately,
and overlay the equal angles with colored crayons. Then
overlay the homologous sides of the triangles which enter
into the required proportions, with colored crayons.

Corollaries. — (1) The segments of all chords of a pencil
of chords intersecting within a circle are
related.

2. A line drawn perpendicular to the
diameter of a circle is a mean proportional
between the segments of the diameter. (See
also 260, Theorem (20).)

EXERCISES

1. Two chords 10" and 12" respectively, intersect at a point which
is 4" from the end of the 10" chord. How far from the end of
the 12" chord is the point of intersection?

2. A third chord of the circle is concurrent with the two chords of
Exercise 1, and intersects them at a point which is 6" from its
nearer end. What is the length of the chord?

3. The diameter of a circle is 400 feet and a perpendicular is erected
to the diameter at a point which is 40 feet from one end of the
diameter. Calculate the length of the perpendicular from the
diameter to its point of intersection with the circle.

306. Theorem XVI. — A relation exists between the lengths
of two secants which intersect outside a circle and the external
segments of the secants. ^ ^

Helps. — The overlapping triangles
are similar. State the proportions
in the form of equal products. Fol-
low the suggestions given in 305.

Corollary. — The lengths of all secants
of a pencil of secants intersecting outside of a circle, and of
their external segments, are related.

EXERCISES

1. Two secants are 20" and 24" measured from their point of inter-
section to the farther point of intersection with the circle. The
20" secant intersects the circle 12" from its point of intersection
with the other secant. At what distance from their point of
intersection does the 24" secant cut the circle?

CIRCLES

207

2. The secants of Exercise 1 meet a third secant at the same point.
A length of 1" is intercepted on this third secant between the
two points of intersection with the circle. What is the total
length of this secant?
Help.—x(x + 1) = 240.

307. Theorem XVII. — A relation exists between the lengths

of a secant and a tangent drawn to a /- \

circle from the same point outside a A B / \ C

circle, and the external segment of

the secant.

Helps. — AABD ~ AACD.
State the required relation as a
proportion. Follow the suggestions given in 305.

Corollary. — The square of the tangent equals, etc.

EXERCISES

1. In the figure of Theorem XVII, the segment AB = 10", and
segment BC — 12". Calculate the length of the tangent AD.

2. The length of a tangent AD = 12", and the length of the secant
AC = 20". Calculate AB and BC.

308. THEOREM XVIII. — The radius of the circumscribed
circle of a triangle can be expressed in terms of the three sides
of the triangle.

State the hypothesis and conclusion.

Helps.— (1) Draw AE, OD, CD; (2) AABE ~ AACD;

(5) but pa = -

(6) .'. R =

-a)(s-b)(s -c);(255);
abc

— a)(s — b)(s— c)

208 PLANE GEOMETRY

EXERCISES

1. Calculate the radius of the circumscribed circle of a triangle of
which the sides are 10, 12 and 20.

2. Calculate the radius of the circumscribed circle of a triangle of
which the sides are 20, 20 and 10.

309. THEOREM XIX. — The square of the bisector of an
angle of a triangle equals the product of the sides including the
angle; less the product of the segments into which the third
side of the triangle is divided by the bisector.

State the hypothesis and conclusion.

Helps. — (1) Draw the circumscribed circle; produce
AD to E, dfaw CE; (2) AADB ~ AACE; (3)

(4) xy = bA'DE. In order to eliminate AE and DE; proceed
as follows; (5) AE = bA + DE; (6) Substitute from (5)

in (3); b + DE = y; (7) Solve for DE in (6) and sub-
stitute in (4); ^= b° ~ bl ; (8) .'. bl = be - xy.

OA OA

EXERCISES

1. The sides of a triangle are 10, 12 and 20. Calculate the length
of the bisector of the angle opposite the 12 side.

Help. — (1) Calculate the segments x and y by 246; (2) calculate
bA by the above formula.

2. Calculate by the above formula; the length of the bisector of the
vertex angle of an isosceles triangle of which the sides are 20,
20 and 10.

310. THEOREM XX. — The length of the bisector of an angle
of a triangle can be expressed in terms of the three sides of
the triangle.

CIRCLES 209

Helps.— (1) b\ = be - xy; (2) x = g^-j ,y = j-~>

(260, Theorem (26); (3) substitute from (2) in (1);

... . ,,
; (4) .Vti -

-c)2-a2-| r(fl + 6 + C)(-

(6+c)2 J= (6 + c)

EXERCISES

1. The sides of a triangle are 10, 12 and 20. Calculate the length
of the bisector of the angle opposite the 12 side, using the above
formula.

2. Calculate by this formula, the three angle bisectors of a triangle
of which the sides are 20, 20 and 10.

311. The Fourth Step in Classification.

1. Two chords which intersect within a circle (305).
Measure the chord segments to the nearest hundredth

of an inch. Test the proportion or the relation of products,
as discovered.

2. The principle of 305, Corollary 2.

3. Two secants which intersect outside a circle (306).
Measure the secants and their external segments. Test

the relation discovered.

4. A secant and a tangent which intersect outside a circle
(307).

5. The radius of the circumscribed circle of a triangle (308).
Draw any triangle and its circumscribed circle. Measure

the sides of the triangle and the radius of the circle. Test
the relation discovered.

6. Test the relation of 309 by measurement.

7. Test the relation of 310 by measurement.

312. The Position Assigned a Principle in Classification.
Notice in what widely separated positions apparently
closely related principles must be placed, if they are asso-
ciated with the group of principles by which they are

210 PLANE GEOMETRY

analyzed or explained. Thus the following special lines of
a triangle have been expressed in terms of the three sides :

1. Under similar triangles; an altitude (255) and a median
(256).

2. Under circles; an angle-bisector (309 and 310) and
the radius of the circumscribed circle (308).

3. Under areas; the radius of the inscribed circle, which
will be derived in Chapter XVII.

ADDITIONAL THEOREMS

313. (1) Analyze Theorem VIII (297) by drawing radii
perpendicular to the chord and to the tangent. The measure
of the required angle can be obtained from that of the cen-
tral angle thus formed.

(2) Analyze Theorem VIII (297) by drawing a chord
through the farther extremity of the given chord parallel
to the tangent. Apply 302, Theorem (2).

(3) Analyze Theorem IX (298) by drawing an auxiliary
chord through the extremity of one of the given chords
parallel to the other given chord. Apply 302, Theorem (2) .

(4) Analyze Theorem X (299) by the method of Theo-
rems (2) and (3).

(5) Analyze Theorem XI (300) by the method of the
preceding theorems.

(6) Analyze Theorem XII (301) by the method of the
preceding theorems.

(7) Analyze Theorem XI (300) by considering a tangent
as a special case of a secant whose points of intersection
coincide, thus deriving Theorem XI from Theorem X.

(8) Analyze Theorem II of Chapter V (119) by circum-
scribing a circle about the triangle.

(9) Show that the analysis of Theorem 8 is faulty logic
because the theorem itself enters into the analysis through
intermediate theorems.

(10) A chord subtending an arc of 120° bisects a radius
perpendicular to it.

(11) Tangents drawn from an exterior point a radius dis-
tant from the circumference, form an equilateral triangle with
their chord of contact.

CIRCLES

211

(12) Lines drawn from the vertices of a circumscribed
quadrilateral to the center of the circle, form central angles of
which each non-adjacent pair are supplementary.

(13) The angle between two tangents to a circle is double the
angle between the chord of contact and a radius drawn to a
point of tangency.

(14) A circumference whose diameter is the hypotenuse of
a right triangle, contains the vertex of the right angle.

(15) A circle drawn with one of the equal sides of an isos-
celes triangle as a diameter, bisects the base of the triangle.

Helps.— Draw AD; AD±BC, why? .*. AD bisects BC.

(16) A line is drawn through the point of contact of two
tangent circles (externally or internally tangent). Radii
drawn to the extremities of this line are parallel.

Helps. — Draw the line of centers. Show that alternate-
interior angles are equal.

(17) A line is drawn through the point of contact of two
tangent circles (two cases). Tangents drawn at the extremities
of this line are parallel.

Helps. — (a) Draw radii as in Theorem (16); or (6) Draw
a common tangent, and show that alternate-interior angles
are equal because their arcs have the same measure.

(18) The perpendicular from the center of a circle to aside

212

PLANE GEOMETKY

of an inscribed equilateral triangle equals one-half the radius
of the circle.

(19) The internal common tangent of two externally tangent
circles bisects the external common tangents.

(20) Tangents drawn to two tangent circles from a point on
their internal common tangent are equal (two cases).

(21) A circle is drawn with the radius of another circle as
its diameter. A chord of the larger circle drawn from the
point of tangency of the two circles is bisected by the circum-
ference of the smaller circle.

(22) All chords drawn from the point of tangency of two
circles (internally or externally tangent) are divided propor-
tionally by the circumference of the smaller circle.

Helps. — Draw a common diameter; AABF
AABD ~ AACE, etc.

AACGand

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213

(23) Two (unequal) circles intersect in points A and B;
a common secant is drawn through A intersecting the circles
in C and D; chords BC and BD are drawn. Then ^CBD
is the same for any position of the secant CD; or, /.CBD is
constant.

Helps. — Consider the measures of angles C and D and how
Z.CBD is related to these angles.

(24) The circles described on two sides of a triangle intersect
on the third side.

Helps. — Draw AD.LBC; show that each circle contains
point D.

(25) Of all triangles drawn with the same base and with
equal altitudes, the isosceles triangle has the largest vertex
angle.

Helps. — Circumscribe a circle about the isosceles triangle
ABC, what kind of line is AA'A"? .'. points A', A", etc.,
are outside the circle.

(26) A chord is a mean proportional between a diameter
drawn through one of its extremities and its projection on the
diameter.

(27) The sum of the three alternate angles of an inscribed
hexagon equals the sum of the other three alternate angles.

(28) Find in a way similar to that employed in the pre-
ceding theorem, that a relation exists between the sums of
the two sets of alternate angles of any inscribed polygon
of an even number of sides. State the theorem.

214

PLANE GEOMETRY

(29) The point of intersection of any pair of common
tangents of two circles divides the center sect into segments
proportional to the radii.

(30) Corollary of (29). The points of intersection of the
internal and external common tangents of two circles divide
the center sect harmonically.

(31) A pencil of chords, AB, AC, AD, etc., is drawn from
a point A in the circumference of a circle; a line parallel to
the tangent at A, intersects the chords in points B', C', Df,
etc. Then AB-AB' = AC-AC' = etc.

Helps. — Draw a diameter AX perpendicular to the tan-
gent at A; drsiwBX, CX, DX, etc.; &AB'X'~&ABX, etc.

(32) The common chord of two intersecting circles, if pro-
duced bisects their common tangents.

(33) If the altitudes of a triangle are produced to intersect
the circumscribed circle, the sects intercepted between the
orthocenter and the circumference are bisected by the sides of
the triangle.

Helps.— (1) x = y; (2) x = z; (3) :. y = z ; (4) hence
OG = GD.

(34) An internal common tangent of two equal circles,
which are externally out of contact, bisects the center sect.

(35) Secants drawn from the point of intersection of the
internal and external common tangents of two externally
tangent circles, to the centers of the circles, are perpendicular.

CIRCLES 215

(36) The perpendicular bisectors of the sides of an inscribed
polygon are concurrent at the center of the circle.

(37) The bisectors of the angles of a circumscribed polygon
are concurrent at the center of the circle.

(38) The perimeter of a circumscribed trapezoid equals
four times the sect which joins the middle points of the two
non-parallel sides.

(39) The sum of three alternate sides of a circumscribed
hexagon equals the sum of the other three alternate sides.

(40) Test the principles of Theorems (38) and (39) ex-
perimentally.

314. Locus CONSTRUCTIONS. — Construct the following
loci experimentally, by finding the positions of a sufficient
number of points to indicate the form of the complete locus.

1. The locus of the middle points of a system of parallel
chords of a circle.

2. The locus of the middle points of all equal chords of a
circle.

3. The locus of the middle points of all chords of a circle
which are concurrent in a point on the circumference.

4. The locus of middle points of all chords of a circle
which are concurrent in a point within the circle.

5. The locus of the middle points of all secants of a
circle which are concurrent in a point outside the circle.
Plot the bisection points both of the entire secants and of
their external segments.

6. The locus of the centers of all circles which are tangent
to a given straight line at a given point.

7. The locus of the centers of all circles tangent to a
given' circle at a given point. Draw both externally and
internally tangent circles.

8. From a given point on the given straight line of

216 PLANE GEOMETRY

Exercise 6, tangents are drawn to all the circles. Find
the locus of the points of tangency.

9. The locus of the vertices of all angles of a given size,
or value, whose sides pass through two given points. Mark
two points several inches apart, and use the 30° angle of a
drawing triangle for the given angle. In order to complete
the locus it is necessary to find positions of the angle where
one side produced through the vertex, contains one of the
given points.

315. REVIEW EXERCISES

1. A secant and a tangent intersect at an angle of 30°; the secant
intercepts an arc of 100° between its points of intersection with
the circle. Calculate the other two intercepted arcs.

2. Two angles of an inscribed triangle are 48° and 105°; calculate
the arcs subtended by the sides.

3. The vertex angle of an inscribed isosceles triangle is 50°; calcu-
late the arcs subtended by the sides.

4. Two secants intercept arcs of 150° and 95°; calculate the angle
between them.

5. Two secants form an angle of 30°, the nearer intercepted arc
being 50°; calculate the farther of the intercepted arcs.

6. The measure of the farther arc intercepted by two secants is
four times the measure of the angle formed by the secants; the
nearer arc is 30°; calculate the angle and remoter arc.

7. Two tangents form an angle of 25°; calculate the major and
minor intercepted arcs.

8. An inscribed angle subtends an arc of 75°; calculate its value.
What arc is subtended by an inscribed angle of 75°?

9. Two intersecting chords intercept, in order, the arcs 80°, 120°,
30°, 130°; calculate the angles formed by the chords.

10. One of the angles formed by two intersecting chords is 55°; an
arc opposite this angle is 78°, and an adjacent arc is_100°.
Calculate the other arcs.

11. A chord is drawn from a point of tangency making an angle of
42° with the tangent; calculate the values of the two subtended
arcs.

12. A chord subtends an arc of 75°; at what angle must a line inter-
secting it at an extremity be drawn in order to be a tangent to
the circle?

13. Three consecutive sides of an inscribed quadrilateral subtend

90°, 58°, calculate the angles of the quadrilateral,

CIRCLES 217

the angles between the diagonals, and the angles at which the
diagonals meet the sides.

14. Two tangents are drawn at the extremities of a chord which
subtends an arc of 155°; at what angle do the tangents intersect?

15. Two perpendicular chords intercept adjacent arcs of 40° and
70°; calculate the other intercepted arcs.

16. Two adjacent arcs AB and BC, measure 150° and 100°;
a tangent is drawn at B, and a secant through AC; calculate
the angle of intersection.

17. A chord 2" long is drawn perpendicular to a diameter of a circle
of 1.5" radius; at what point does the chord intersect the
diameter?

18. A chord of 1" is bisected by a chord on which a segment of
\" is cut off; calculate the radius of the circle.

19. From the end of a tangent sect of 2", a secant is drawn through
the center of a circle of 3" diameter; calculate the external seg-
ment of the secant.

20. The sides of an isosceles triangle are 4, 4, 3; calculate the three
angle-bisectors, and the radius of the circumscribed circle.

21. Calculate the radius of the circumscribed circle of a triangle
whose sides are 3, 4, 5.

22. The radius of a circle is 2"; calculate the chord subtending an
arc of 120°.

23. Two circles of 8" and 10" radii have a center sect of 20"; calcu-
late the tangent sects of internal and external common tangents.

24. In the formulas of 308 and 310, let a = b = c. Show that

R = ; 6 A = —~ — ; and that these values agree with values

3 ' 2

for the altitude and median of an equilateral triangle, derived
in a similar way from the formulas for pa and ma, 255 and 256
(see 260, Theorem (16); and with 254, Part (6).

25. Given an angle, and the length of a chord; construct a circle
such that the segment cut off by the chord contains inscribed
angles equal to the given angle.

26. Given a chord of a circle and the measure of its subtended arc;
construct the circle.

27. Divide a circle into two segments such that the angle inscribed
in one segment is double the angle inscribed in the other.

28. Construct a circle tangent to each of two parallel lines and to a
third line intersecting the parallels.

29. Construct a circle tangent to each of two concentric circles.
Write a formula for its radius.

30. Construct a circle tangent to each of two concentric circles
and also tangent to a line intersecting one or both of the given
circles.

218 PLANE GEOMETRY

31. Construct a circle of given radius, tangent to each of two given
circles which are externally out of contact; discuss all cases.

32. The sides of a triangle are 5", 6", 8". Calculate the segments
into which the sides are divided by the points of tangency of
the inscribed circle.

Helps. — AF = AE, etc. Form a system of equations, x -f y = 5,
etc.; solve for x, y, z.

33. Derive formulas for the segments into which the sides a, 6, c,
of a triangle are divided by the points of contact of the inscribed
circle.

34. Calculate the distance of the center of the inscribed circle of
the triangle of Exercise 32 from each vertex of the triangle.
Helps. — Calculate r; then d,A, etc.

35. Show that the general formula for d,A is,

= l

- a) + (s - b)(s -

36. Three circles are drawn externally tangent, each to the other
two. The center sects are 10, 12 and 15 inches. Find the
radii of the circles.

37. Two circles are externally tangent at A. One of the external
common tangents touches the circles at points B and C, respec-
tively. Show that a circle whose diameter is BC, contains point
A.

38. Show that two equal externally tangent circles intercept equal
chords on a common secant drawn through the point of tangency.

CIRCLES

219

39. The triangle ABC is formed by two fixed tangents AD and AE
and a movable tangent BC. Show that the perimeter of AABC
is constant, wherever tangent BC is drawn.

40. The altitude to the hypotenuse of a right triangle divides the
hypotenuse into segments of 10 and 40 inches respectively.
Calculate the altitude.

41. Show that, if the perpendicular sides of a right triangle are in
the ratio of 1 to 2, the altitude to the hypotenuse divides it in
the ratio of 1 to 4.

42. A surveyor measured a chord AB of a circular railroad curve,
equal to 240 feet, and the perpendicular distance CD at the
middle of chord AB, equal 40 feet. Show how to calculate the
radius of the curve.

43. The altitude to the hypotenuse of a right triangle of which
one acute angle is 30°, divides the hypotenuse in the ratio
of 1 to 3.

44. Two chords of 6 and 10 inches respectively, intersect so as to
divide the shorter chord into segments of 2 and 4 inches. Find
the segments of the longer chord.

45. A secant and a tangent are drawn from the same external point.
The chord intercepted on the secant is 8 inches and the external
segment of the secant is 14 inches. Find the length of the
tangent.

46. The radius of a circle is 21 inches. Find the length of a tangent
drawn from a point 35 inches from the center of the circle.
Find also the length of a secant drawn from the same point if
the length of its external segment is 20 inches.

47. Tangents drawn to two intersecting circles from any point on
their common chord produced are equal.

220 PLANE GEOMETRY

48. The common chords of each pair of three intersecting circles
are concurrent.

Helps. — (1) AB and CD intersect in 0; draw chord EO which
intersects circle's x and z in F and G, respectively; (2) A O'OB
= EO-OF, etc.; (3) .*. OF = OG, and therefore points G and F
coincide.

49. The center line of two circles, which are externally out of con-
tact, intersects the circles in the successive points A, B, C, D,
and intersects their common external tangent in point E.
Show that EA'ED = EB'EC.

Helps. — (1) Mark the points of tangency, T and T'; draw chords

jf> A FC1
AT, BT, CT', DT'; (2) J^ = |^,, etc.

50. In Exercise 49, draw a secant from point E intersecting the
circles in points F, G, H, J. Show that EF-EJ = EG- EH.

51. The radii of two eircles are 10 and 4 inches, respectively, and
their center sect is 20 inches. Find the points of intersection
of the internal and external common tangents with the center
sect.

4 x ,4 y

To = 20=~x and 10 - 20+7

52. The radii of two concentric circles are 10 and 24 inches respec-
tively. Find the length of a chord of the larger circle that is
tangent to the smaller circle.

53. The radius of a circle is 24 inches. Find the length of the
shortest chord that can be drawn through a point 20 inches
from the center of the circle.

APPLICATIONS

316. FAMOUS PROBLEMS. — The Trisection of an Angle —
This is one of the famous problems of antiquity. There is
no known solution by use of rule and compass alone. The
following method requires the use of a scale, and would not
be considered a solution by those old Greek geometers who

CIRCLES

221

regarded the science chiefly as a splendid intellectual
recreation. C

/.ABC is the given angle. Draw any circle with center
B; produce AB; find by trial, using a scale, a sect ED such
that EF = BF. Then x = %/.ABC, and can be laid off
at ABG, etc.

EXERCISES

1. Prove that /x = I/ABC.

2. A method of trisecting an angle, which has been frequently
suggested by students is as follows : Draw BD

= BE; draw DE; trisect DE by some purely
"geometrical" method, as in 261, Exercise 31;
draw BF and BG. Prove, that /x does not
equal /y by showing that the bisector of /.DBG
does not bisect DG, 245. D A

3. Can the error in the construction of Exercise 2 be detected by
measurement?

The Division of a Sect in Mean and Extreme Ratio. — This
is such a division of the sect that one segment is a mean
proportional between the given sect and the other segment.
There are two points of division, one internal and the other
external.

C E B

— x. >|«-,y-x — W

— 5 H

AB is the given sect. Draw circle D, tangent to AB at
, with radius = \ AB; draw AD. Then E is the required

222

PLANE GEOMETRY

internal point of division, if AE = AF; and H is the required
external point if AH = AG.

EXERCISES

4. Prove

s — x

Helps. — (1) -£g = -£Q; (2) change by the Law of Division (240).

5. For the external point of division H, the segments of sect AB
sue HA = s + x and HB = 2s + x. Prove 2s + x = lit?

S + X S

6. Another construction for the division of a sect into mean and

s — x x

extreme ratio is shown. Prove

7. Calculate the values of the segments of a sect divided in mean
and extreme ratio. Solve the equation of Exercises 4 and 6 by

-i ±V8

algebra. Show that x

8. In Fig. 1, let HA = xir- then

•s = +.618s or -1.618s.
= — • Solve for x\;

' s. Show that x\ = s + x or — x, and there-

fore gives the points of division H and E, which were found in
Exercise 7.

9. Measure the segments of the sects AB of the preceding figures.
Do the measurements agree with the calculated values?

Artistic Value of Mean and Extreme Ratio. — This division
of a sect possesses a remarkable artistic and psychological

CIRCLES

223

interest. It has been called the " divine ratio" or the
" golden section."

EXERCISES

10. Draw several equal vertical lines, side by side, and divide them
at different points. It will be found that many persons possess a
marked preference for the "divine ratio," in which the point
of division is above the center; that is, x = .62s.

11. Measure the most pleasing proportioned book covers, platters,
"oval" (elliptical) picture frames, etc. It will be found that
the width approximates .62 of the length.

A Mean Proportional.

EXERCISES

12. Explain the construction of the figure and prove that x is the
required sect.

13. Calculate the value of x and check the geometrical solution by
measurement.

The Geometrical Solution of a Quadratic Equation.—
C

Let the given equation be x2 — 5x + 4 =0. Then a; (5 —

x 2

x) = 4; .'. ~ = ^ — - • Draw a circle with AB = 5 units as
z o — x

diameter; draw AC JLA5, equal 2; draw CD\\AB, draw
DE±AB. Then A# and EJ5 are the roots of the equation.

EXERCISES

14. Solve the equation x2 — 9x + 14 =0, geometrically.

15. Solve the equation xz — Wx + 25 = 0, geometrically.

16. Solve the equation x2 — 6x = 0, geometrically.

17. Check the geometrical solutions by algebraic solutions of the
preceding equations.

224

PLANE GEOMETRY

317. Terrestrial Measurements. — The curvature of a
sphere in a distance d, is expressed in terms of the intercept
c, on a radius, included between the arc and the tangent.

EXERCISES

1. In the right triangle BAG, OB2 = GA2_ __ A d B

+ ZB2. Substitute the lengths r, c^
and d; omit the term c2 which will be
very small in comparison with the other

terms; solve for c. Show that c = »-•

Solve also for r and d. In these form-
ulas, d' may be substituted for d, when
d is small in comparison with r.

2. Let r = 100 inches and d = 20 inches.
Find c.

3. Let d = 20 inches and c = 1 inch. Find r.

4. Let r = 120 inches and c = 4 inches. Find d.

The Size of the Earth. — Careful measurements prove that
the curvature of the earth is approximately 8 inches in 1
mile.

EXERCISE

5. Find the radius of the earth from the preceding data, using the
formula for r.

The Distance of the Horizon at Sea. — This depends upon
the height of the observer above the mean surface of the
water.

EXERCISES

6. Find the distance of the horizon to an observer who stands upon
a boardwalk so that his eye is 30 feet above the water level of
the ocean.

30
Helps. — c = £^7; miles; let r = 3960 miles, d = \/2cr miles.

7. Find the distance at which the topmast
of a vessel is visible to the observer of
Exercise 6, if the topmast is 80 feet
above the water line.

80
Helps. — For the vessel, c' =^280' The

distance required = d of Exercise 6 +
d' of Exercise 7.

CIRCLES

225

8. How high must a war balloon ascend at the sea coast in order to
sight an enemy fleet 100 miles distant?

Help. — A more exact result will be obtained in this problem by
solving the equation (c + 3960) 2 - 39602 = 1002.

9. How far out at sea may a lighthouse be visible if the light is
120 feet above water level, and the watch on the vessel is 60 feet
above the water level? The effective distance will be consider-
ably less than that calculated except under most favorable
weather conditions.

318. Navigation. The "Danger Angle." — The reefs or
bars C, D, E, etc., are shown on a sailing chart, A and B

' 1

\ / \ - \ / '

\ ,\' ^} // J>

r-..*£K \ C&! °'

being the positions of two prominent objects also marked
on the chart. The navigator draws circles passing through
A and B (or they are already shown on the chart), in such
a way as to include or exclude the danger region. He
measures any inscribed angles x and y. If he then lays his
course on the line FG so that the sextant reading of the angle
ASB is less than x and greater than y, the course is safe.
Angles x and y are the " danger angles."

EXERCISE

1. Explain why the method employed ensures a safe course.

The Three Point Problem. — This name is given to a
method used for accurately locating a ship which is near
enough to the shore to distinguish three prominent objects.

226

PLANE GEOMETRY

The angles x and y are measured with a sextant on the
ship. The circles 0 and 0' can then be constructed, their

intersection on the chart determining the position of the
ship at S.

EXERCISES

2. What relation exists between angles x and R; T and yf

3. What relation exists between angles z and R; w and Tl

4. Prove therefore that z = 90° — x, and w = 90° — y.

5. Plot three points A, B, C; AB = 20,000 ft., BC = 15,000 ft.,
Z.ABC = 160°. Find the position of the ship S if the angle x
is measured 40°, and angle y is measured 35°.

319. Surveying. Location of Soundings. — The " three
point problem'7 is generally used to locate the boat which
makes the soundings along a shore line, or within a lake or

bay. The series of points A, B, C, D, etc., are located and
plotted on the shore. At each observation point, 1, 2, 3,
etc., angles x, y, x', y'; etc., are measured with a sextant,
and simultaneously a sounding is taken. A record is kept,
from which the soundings can be plotted on the map or
chart.

CIRCLES

227

Point

1st angle

2nd angle

Sounding, feet

AB 75° 10'

BC 42° 15'

12.5

AB 50° 5'

5C61° 3'

24.3

BC 31° 52'

CD 35° 37'

27.8

Etc.

EXERCISE

1. Plot the points 1, 2 and 5, from the record.

To Lay Out a Semicircular Curve by Ordinates.'

QDEFG

The diameter AB is divided into any number of equal parts,
AD, DE, EF, etc., with a smaller part AC = KB at the
ends. The ordinates CC', DD', etc., are calculated by 30 6
Cor. 2.

EXERCISES

2. If AB = 400 feet, AC = CD = 20 feet, DE = EF = FG = etc.
= 40 feet, calculate CC', DD', etc., to OO'.

3. Plot the diameter and points of division, draw the ordinates,
measure off the lengths of the ordinates. Test the accuracy of
the calculated lengths by drawing a circle with center at O.

320. Astronomy. The Height of the
Moon's Mountains. — Galileo, about
the year 1600, measured the heights
of some of the mountains on the
moon by the following method. He
measured the distance AM at the
instant that the sun's ray SA, first
iluminated the top of the mountain M .
the moon must also be known.

The diameter of

228

PLANE GEOMETRY

EXERCISE

1. If AB = 2160 miles, and AM is measured 120 miles, calculate
the height of the mountain M .

Help. — (a) Use the formula of 317, Exercise 1; and (6) use
the method of 317, Exercise 8.

Eclipses of the Sun and Moon. — The eclipse of the sun is
caused by the converging cone of total shadow of the moon
falling upon the earth's surface at P (Fig. 1). The eclipse
of the moon is caused by the entrance of the moon into the
cone of total shadow of the earth (Fig. 2).

FIG. 2.

EXERCISES

2. Take the distance from center to center of the sun and earth as
93,000,000 miles; the distance from the earth to the moon as
241,500 miles; the radius of the sun 432,000 miles; the radius of
the earth 3960 miles; the radius of the moon as 1080 miles. (The
distances from the earth to the sun and to the moon are variable.)
Calculate the length MF of the moon's shadow, Fig. 1. Does
the moon's shadow reach the surface of the earth? Calculate
the radius of the shadow at P.

SE - EB

3. In Fig. 2, show that EG = ^ _ jg *

4. Calculate the length of the earth's total shadow, using the values
given in Exercise 2 and the formula of Exercise 3.

CIRCLES

229

5. Calculate the diameter of the earth's shadow, Fig. 2, at the dis-
tance at which the moon enters it. If the moon enters the
shadow centrally, is the shadow sufficiently broad to entirely
obscure the moon?

321. THE CIRCLE IN DESIGN. — Ornamental designs in-

FIG. 10.

volving circles and arcs of circles are found in railings,
brackets, cut glass, window glass, rugs and linoleums,

230

PLANE GEOMETRY

and in innumerable other familiar objects. A few sugges-
tive illustrations are given.

Figure 1 is called the curved equilateral triangle. Figures
5 and 6 are art glass designs; Fig. 8 is a column base; Fig.
10 shows various forms of mouldings and cornices.

EXERCISES

1. Make other designs for some definite application, as a piece of
scroll work to fill in the top of an arched doorway; a piece of
embroidery or a sofa cushion; an arm of a built-in bench for a
porch or hall; etc.

2. Name some familiar objects which are always or frequently
circular in form.

Architectural Designs Containing Tangent Circles. — These
forms occur in endless variety. Constructions may be
made by trial, but interesting applications of algebraic
calculation may be made, and should be made in laying
out exact work. Construct the figures shown, from the
calculated values.

EXERCISES

3. Express the radius r, of one of the circles in terms of the dis-
tance, d.

Helps.— CD = r\/3; d = CD + 2r = r(2 + \/3);

= d(2 - V3) = -27d.
4. Express r in terms of d.

2+ x/3

CIRCLES

231

Helps.— CD = rV3; d = r(I + \/3);
5. Express r in terms of d.

Help.— r =

d( V5 - 1)

6. Express r and AB in terms of d.

Helps.— In &ABD,AB2 = (d - r)2 -
In AABC, AB2 = (r + ^) * - gV =

= -r - 2rfr + r2.

and A£ = .49^.
7. Express r and AB in terms of d.

Helps.— r = -A

232 PLANE GEOMETRY

8. Express r and A B in terms of d.

Helps.— r = ~ d and AB = d.
9. Express r and BC in terms of d.

Helps.— In ACDE, DE =

+ DE = (d+V/3);^C = d _ r. In

, (d - r)'
2

4±3\/2

= (-4 ± 3V2)

VAC2- CD2 =
.'. BC = .30d.

BC =

- .24

= AD - DG;

; ZX? = r\/3

NOTE. — The preceding problems are found in "A Source
Book of Problems for Geometry," by Mable Sykes; Allyn
and Bacon.

322. Various Useful Applications. To Draw an Arc
Through Three Given Points When the Radius Is Large. — A

B
A

pencil point at B will describe the required arc if the arms

CIRCLES

233

BA and BC slide along pins inserted at points A and C.
Why?

To Find the Diameter of a Column. — A carpenter may be

required to cut a circular column along a diametral plane.
Explain how the diameter is determined by laying the
square on the column as shown.

To Make a Core Box Truly Cylindrical.—
Explain how a pattern maker uses his

square, and why the vertex of the square ~-

must scrape the hollow as it is slid around.

To Find the Radius of an Arc. — Find the center by the

method of constructing a circle which will contain three
given points.

The Length of a Chord. — Show how r is ex-
pressed in terms of the chord c, and its dis-
tance d, from the center. When d = $r, show
that c — r\/3' Solve this last formula also
for r in terms of c.

The Rise of an Arc. — The rise of the arc AB from the
chord c, of the arc is a. Show that
2r±

c)(2r-c)

_ d ± V(d + c)(d - c)

Solve also for r or d in terms of c and a. r =

4a2

234

PLANE GEOMETRY

EXERCISES

1. Find the length of a chord 6 inches from the center of a circle of
radius 12 inches.

2. Find the diameter of a circle if a chord drawn 10 inches from
the center is 20 inches long.

3. A circular arch bridge is to span a stream 52 feet wide. If the
rise of the crown of the arch is 13 feet, what is the radius of the
arch?

4. A doorway is 6 feet wide. The rise of
the circular top is 1 foot above the
straight sides. Calculate the radius of
the arc.

5. A surveyor wishes to find the
radius of a railroad curve. He
measures a chord of the curve =
100 feet, and the offset at the
center of the chord = 2 feet. What is
the radius of the curve?

6. A chord measured on a piece of a broken
wheel is 20 inches and the rise of the arc
is 5 inches. What is the radius of the
wheel?

To Lay Out a Circular Curve by Offsets from the Tangent

at the Beginning Point of the Curve. — Show that the value of
the perpendicular offset p, at a distance d, from the point
of tangency T, is p = r — \/r2 — d2.

CIRCLES 235

EXERCISE

7. Calculate the offsets from the tangent of a curve of 1000 feet
radius, at distances from the point of beginning of the curve of
100, 200 and 300 feet.

The Strongest Rectangular Beam That

Can Be Cut from a Circular Log. — The
beam is laid out on the end of the log by
drawing a diameter AB, and trisecting at
points C and D, and erecting perpendicu-
lars CE and DF, etc.

EXERCISES

8. Calculate the sides AE and BE of the beam in terms of the
diameter d. Show that ^^ = ^ approximately.

9. What are the dimensions of the strongest beam that can be cut
from a log 2 feet in diameter?

323. PROBLEMS FOR FIELD WORK. — (1) To Lay Out a
Circular Curve by Ordinates. — Use the data of 319, Exercise
2, or calculate the ordinates for a curve of any other radius.
Stake out points A, C, D, - —B; erect perpendiculars
by the method of 237; measure on each perpendicular the
length of the calculated ordinate; drive stakes at points
C", £>', E', etc.

2. To Lay Out a Circular Curve by Offsets from the Tan-
gent.— Take the radius of the curve = 200 feet. Select a
point A on a straight line where the curve is to begin.
Calculate the offsets for values of d = 40, 80, 120 etc! feet.
See 322, Exercise 7. Stake out these distances along the
tangent from point A, and erect perpendiculars to the tan-
gent; measure on each perpendicular the length of the
calculated offset until the curve has turned a quarter circle;
extend the last offset which is the tangent at the other
extremity of the curve.

CHAPTER XIV

REGULAR POLYGONS. CIRCLES ASSOCIATED
WITH POLYGONS

PRINCIPLES DETERMINED EXPERIMENTALLY

324. Previous Mention. — Regular polygons have been
defined and briefly considered in Chapter VI; inscribed and
circumscribed polygons in Chapter XII.

325. Experiment I. Equilateral Inscribed Polygons.—
Draw several circles; draw equilateral inscribed polygons

of 4, 5, 6, 7 sides by dividing the circumferences by trial
into equal arcs.

Observe that the polygons are equiangular.

EXERCISES

1. How can the division points of the circumference be found with-
out trial in the case of the equilateral quadrilateral?

2. How can the division points be found without trial in the case
of the hexagon?

326. Experiment II. Equiangular Circumscribed Poly-
gons.— Draw several circles; divide the circumferences
into 4, 5, 6, 7 equal arcs; draw polygons whose sides are
tangent at the division points. Such polygons will be
equiangular by 301.

Observe that the polygons are also equilateral.

327. Experiment III. — Properties of Regular Polygons.—
Calculate each interior angle of regular polygons of 4, 5, 6,

7 sides, by 191, Theorem (5). Draw the polygons by draw-
ing alternately a side and an angle of each polygon.

(a) Bisect all the interior angles of the polygons. Ob-
serve that the bisectors are concurrent.

236

REGULAR POLYGONS 237

(b) Draw the perpendicular-bisectors of all the sides of
the polygons.

Observe that they are concurrent.

Observe (1) that the circles of each polygon are concen-
tric; (2) that the inscribed circle is tangent to each side at
its middle point; (3) that the triangle formed by a half-side
of a polygon, a radius of the circumscribed circle and a
radius of the inscribed circle is a right triangle.

328. Definitions. — The common center of the inscribed
and circumscribed circles of a regular polygon is the center
of the polygon.

The radius of the circumscribed circle of a regular polygon
is the radius of the polygon.

The radius of the inscribed circle of a regular polygon is
the apothem of the polygon.

The right triangle formed by a half-side, radius and apo-
them is an element of the polygon.

EXERCISES

1. Calculate the central angle of an element of a regular polygon of
4, 5, 6, 7, sides.

2. Write a general formula for the central angle of an element of a
regular polygon of n sides.

3. Write a formula for a half -side s, expressed in terms of the per-
imeter p.

4. Express a general relation between a half -side s, a radius r, and
an apothem a.

5. Write a general formula for one of the equal angles of a triangle
formed by two adjacent sides of the polygon and a diagonal join-
ing the extremities of the two sides.

Help. — (1) Write the value of one interior angle of the polygon;

180°
(2) subtract from 180°; (3) divide by 2. Value is •

329. PROBLEMS RELATING TO REGULAR POLYGONS. — In

these problems only an element of the polygon is required,
except in those which involve diagonals.

238 PLANE GEOMETRY

1. Find by measurement the radius and apothem of a
regular pentagon of which the side is 1.4 inches.

2. Find by measurement the side and radius of a regular
hexagon of which the apothem is 1.2 inches.

3. Derive a formula for the value of the side and
radius of an equilateral hexagon, expressed in terms of
the apothem.

Help.— Refer to 260, Theorem (18).

4. Check the measured values of Exercise 2 by calcu-
lating the same values by the formula of Exercise 3.

5. Find by measurement the apothem and radius of a
regular pentagon of which the perimeter is 30 feet.

6. Find by measurement the apothem and side of a
regular decagon of which the radius is 2 inches.

7. Find the number of sides of a regular polygon of which
the side is 2 inches and radius 2.61 inches.

Helps. — Construct the element; measure the central
angle; divide it into 360°.

8. Find the number of sides of a regular polygon of
which the side is 20 feet and the apothem 37.32 feet.

9. The least diagonal of a regular decagon is 2 inches.
Find by measurement the side, apothem and radius.

Helps. — (1) Calculate the angles of the triangle formed
by diagonal and two adjacent sides, by 328, Exercise 5; (2)
Construct this triangle. (3) Construct the element of the
polygon on the side as thus determined.

10. The least diagonal of a regular polygon is 2.83 inches
and a side is 2 inches. Find the number of sides of the
polygon.

11. Calculate the radius, apothem and perimeter of a
square of which the side is 20 inches.

12. Calculate the radius, apothem and perimeter of an
equilateral triangle of which a side is 20 inches.

13. Calculate the apothem and radius of a regular octagon
of which the side is 6 inches.

Helps.— (1) 62 = r2 + r2 - 2r(.707r) (252 and 254); (2)
.586r2 = 36; r = \/61.43 = 7.84; (3) a2 = (r + s} (r - s).

REGULAR POLYGONS

239

330. EXPERIMENT IV. — THE RATIO OF THE CIRCUMFER-
ENCE TO THE DIAMETER OF A CIRCLE.

(a) Draw a number of circles of different, size; say 1.0",
1.5", 2.0", 2.5", 3.0" in diameter. Measure the circumfer-
ences as carefully as possible with a flexible scale. A narrow
uniform strip of paper placed on edge may be used. Enter
the values in column c' of the accompanying table.

c'"

Average
value
of c

It"

Average
value
of*

1.0

1.5

2.0

2.5

3.0

(6) Rankine's Approximation for Finding the Length of a
Circular Arc. — AB is the given arc with center C. Draw

A F GO

radius CA and tangent AD. Find AE = J arc AB, and
draw arc EF with center A. Draw arc BG with center F.
Then sect AG = arc AB} approximately. The error in this
method almost vanishes for small arcs and reaches ¥^ of
the length of the arc for an arc of 60°. It should not be used
for arcs exceeding 90°.

Use the circles of part (a) and find the lengths of arcs of
60° for each circle. Multiply by 6 and enter the values
in column c" of the table.

240

PLANE GEOMETRY

BF = arc AB = J circumference. Find the values of the
circumferences of the circles used in Parts (a) and (6) by this
method. Enter in column c"r of the table.
B

E A
Complete the table as indicated.

EXERCISES

1. Calculate the circumference of a circle of which the diameter is
20 feet.

2. Calculate the diameter of a circle of which the circumference is
50 feet.

3. What is the diameter of a cask if the circumference measures
10 feet? Will the cask pass through a door 3 feet wide?

4. WTrite a formula for the length of an arc of F degrees, in a circle
whose diameter is d.

331. THE LENGTH OF AN ARC DETERMINED FROM ITS
CHORD. — A formula that gives the length of an arc not ex-
ceeding 60°, with close approximation is I = — « — / where

C is the chord of the arc and c is the chord of J the arc.

To use the formula, construct the arc to scale and measure
the chords.

EXERCISES

1. Find the length of an arc of 45° in a circle whose radius is 20
inches.

2. Calculate the length of the arc of Exercise 1 by using the formula
of 330, Exercise 4.

3. The chords of Exercise 1 may be calculated as follows: (1) C* =

(C\ 2
2 ) > where a

REGULAR POLYGONS 241

is the apothem of a regular octagon whose side is C; (3)

40 c

— = 2Q _ — Substitute these values in the formula I =

8c-C
— n — » and solve for I.

332. Exact Constructions of Regular Inscribed Polygons.
The problem of finding rule and compass constructions
for regular inscribed polygons, has been a favorite one
with mathematicians. Methods have been given in
preceding chapters for the rule and compass construction
of inscribed equilateral triangles, squares and regular
hexagons. By successively bisecting the arcs subtended
by the sides of these polygons, regular inscribed polygons
of 8, 12, 16, 24, 32, etc., sides may be constructed.

A Rule and Compass Construction of a Regular Inscribed
Decagon. — Draw a radius OA; divide
it into mean and extreme ratio at
point B (316); OB is the required side
of the decagon. A regular inscribed
pentagon, and regular polygons of
20, 40, etc., sides can be obtained
from the decagon.

For all practical purposes, the
trial and error method used in 325
is the best, except for polygons of 3, 6, 12, etc., and 4, 8,
16, etc., sides.

In 1796, Gauss, the greatest mathematician of the
Eighteenth century, then nineteen years of age, discovered
a method of constructing all regular inscribed polygons
the number of whose sides is given by the formula

n = 2*(22" + 1)
where p and q are integers.

EXERCISE

Find the value of n in Gauss' formula for q = 0, p = 0; q = 0,
p = 1; q = 1, j) = 0: a = 1, p = 1; etc., to q = 3, p = 3.
16

242 PLANE GEOMETRY

APPLICATIONS

333. (1) An hexagonal nut, one inch on each side, is to be
cut from a round bar of iron. What diameter of stock must
be taken?

2. A square nut is to be made for a bolt 1J inches in
diameter. The least thickness of metal around the hole
equals the radius of the bolt. What diameter of round
stock must be taken?

3. A gardener plans a summer-house to be laid out as a
regular octagon. The radius is 10 feet. /,

He drives a stake C, at the center and a £>
second stake A at the center of the arc
subtended by the side parallel to the path.
Measure from a drawing of an element
made to scale, the chords AB and (side)
BD. The stakes B, D, E, etc., can then
be placed.

4. In the description of Solomon's Temple (1 Kings,
VII, 23) the dimensions of the " Molten sea," which
appears to have been a great circular brazen vessel, are
given as "ten cubits from the one brim to the other"
(diameter), and "a line of thirty cubits did compass it
round about" (circumference). What was the writer's
estimate of the value of TT?

CHAPTER XV
REGULAR POLYGONS AND CIRCLES

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

334. Constants, Variables and Limits. — A constant is a
quantity which possesses a fixed value.

A variable is a quantity whose value changes continually.

The limit of a variable is the constant which the variable
continually approaches, but may never quite equal.

EXERCISES

1. One half of a pile of sand is removed in a day, one half of the
remainder is removed the second day, and so on. The whole
pile of sand is a constant value; the total amount removed up
to any given tune is a variable. What is the limit of the variable?

2. The sum of the infinite geometrical series, $ + i + i +, etc.,
for n terms, is a variable. What is the limit of this variable as
n is indefinitely increased?

3. What is the limit of the repeating decimal, .333 . . . . ?

Analysis by the Principles of Limits. — Some analyses
involving metrical relationships (that is, where numerical
values, or measures, are involved) are made by obtaining a
relation under certain given conditions, and then considering
the figure to undergo successive modifications until it
merges into another figure. If the property which was
shown to exist in the primary figure holds true during the
successive modifications of that figure, it holds true in the
final figure. This final figure is the limit of the changing
or variable figure; the values of sects, angles, arcs, areas,
ratios, in the final figure are the limits of the variable sects,
angles, etc., of the variable figure.

243

244 PLANE GEOMETRY

EXERCISE

4. If a regular polygon possesses some property which remains un-
changed in principle as the number of sides of the polygon is
indefinitely increased, what limiting figure must also possess
this same property?

Abbreviation. — Approaches as a limit, ==.
THE FIRST STEP IN CLASSIFICATION

335. Postulates of Regular Polygons and Circles.—

(1) An equilateral inscribed polygon is regular.

2. A circumscribed polygon whose points of tangency
divide the circumference of the circle into equal arcs, is
regular.

3. The perimeter of a regular inscribed polygon (a) is
less than the circumference of the circle; (6) increases as the
number of sides is indefinitely increased; (c) approaches the
circumference of the circle as a limit.

4. The perimeter of a regular circumscribed polygon
(a) is greater than the circumference of the circle; (6) de-
creases as the number of sides is indefinitely increased; (c)
approaches the circumference of the circle as a limit.

5. Two regular polygons of the same number of sides are
similar (see 234).

6. Corresponding sects of two regular polygons of the
same number of sides are proportional.

Limit Axioms. — (1) If two variables are proportional to
two constants, the limits of the variables are proportional
to the constants.

2. If two variables approaching limits are always equal,
their limits are equal.

3. If a variable is approaching a constant as a limit, the
variable multiplied by or divided by any number approaches
the constant multiplied by or divided by the same number,
as a limit.

EXERCISES

1. How much less is the perimeter of a regular hexagon inscribed
in a circle whose radius is 10 inches, than the circumference of
the circle? Use the value of TT = 3|.

REGULAR POLYGONS AND CIRCLES

245

2. How much greater is the perimeter of a regular quadrilateral
circumscribed about a circle whose radius is 10 inches, than the
circumference of the circle?

3. If the diagonal of a square whose side is 10, is 14.142, what is
the diagonal of a square whose side is 20?

4. The apothem of a regular hexagon whose side is 10, is 8.66.
What is the apothem of a regular hexagon whose side is 30?
What is the side of a regular hexagon whose apothem is 34.64?

THE SECOND STEP IN CLASSIFICATION

336. Theorem I. — -The perimeters of two regular polygons
of the same number of sides are proportional to the radii of the
polygons.

Hypothesis. — Regular polygons 0 and 0', of the same
number of sides.

Conclusion. — —. = ->•
p' r'

Analysis.

STATEMENT
AB BC CD
A'B'~ B'C' ~ C'D'

REASON
335; Postulate 5

AB + BC + CD +etc.

A'B' + B'C' + C'D' + etc.
3- Or?/ = -i7D>

AB
A'B1

* . P _r
5* " p'-r'

64; Axioms 11
and 13

335; Postulate 6
64; Axiom 1

246

PLANE GEOMETRY

337. Theorem II. — The ratio of the circumference to the
radius is constant for all circles. Or: Circumferences of
circles are proportional to their radii.

Hypothesis. — 0 and 0' are circles with circumferences
c and c', and radii r and rr.

c c'

Conclusion. = — ,•

Analysis.

STATEMENT

1. Inscribe regular polygons of the
same number of sides in the two
circles, the perimeters being p
and p1 '.

2. The polygons are similar.

3. Let the number of sides of the
polygons be indefinitely in-
creased by doubling the num-
ber of sides; then the successive
polygons are regular, and simi-
lar at each step.

p r

4- p> = ?•

5. But p = c, and p' = c'.

REASON

— . = -. or - =
c' r' r

335; Postulates 1
and 5

336; Theorem I
335 ; Postulate 3
335 ; Limit Axiom 1

KEGULAR POLYGONS AND CIRCLES

247

Corollaries. — (1) The ratio of the circumference to the

c cr

diameter is constant for all circles. That is; -3 = -r, = etc. = TT.

2. In any circle; (a) c = nd; (b) c = 2irr.

3. The length of an arc of n degrees of a circle whose radius

is r, is a = - 2irr.

THE THIRD STEP IN CLASSIFICATION

338. Theorem III.— The side of a
regular inscribed polygon can be ex-
pressed in terms of the side and radius
of the regular inscribed polygon of half
the number of sides.

Hypothesis. — AC is the side x, of a
regular inscribed polygon of n sides;
A B is the side s of a regular inscribed

polygon of sides; R is the radius.

Conclusion. — x may be expressed in terms of s and R.
Analysis.

STATEMENT

1. Draw CD, OA, AD

2. CD J_ AB

3. AE = |-

4. Z.CAD is a right Z.

5. x2 = CD-CE

6. x2 = 2R(R - OE)

R -

7. x2 = R(2R -

8. x =VR(2R -
Corollary.— If R = 1, x

- s2).

248

PLANE GEOMETRY

EXERCISE

The side of a regular hexagon inscribed in a circle of unit radius is 1.
Calculate the side of a regular dodecagon (12 sides) inscribed in the
same circle.

339. Theorem IV.— The side of a
regular circumscribed polygon can be
expressed in terms of the side and
apothem of the regular circumscribed
polygon of half the number of sides.

Hypothesis— FC, CD, are the sides
x, of a regular circumscribed polygon
of n sides; AB is the side s of a

regular circumscribed polygon of ~
sides; r is the apothem.

Conclusion. — x may be expressed in terms of s and r.

Analysis.

STATEMENT
1. Draw OE, OC, OB

2. OC bisects
EC = OE
CB OB
EC

Z.EOE.

L' EB - EC

5 * -

\/OE2 +
r

EB*

2 ~ 2 V
6. * -

"+©'

2rs

2r + V 4r2 + s2
Corollary . — If r = 1, x =

2 + -v/4 + s2
EXERCISE

The side of a regular quadrilateral (square) circumscribed about a
circle of unit radius is two. Calculate the side of a regular octagon
circumscribed about the same circle.

REGULAR POLYGONS AND CIRCLES

249

340. Theorem V. — The value of IT can be calculated with
as close an approximation as desired.

Hypothesis. — Circle 0, with radius r = 1; circumference
c; diameter d = 2r = 2.

C C

Conclusion. — The value of r = -3 or ~~ can be calculated,

etc.
A nalysis.

STATEMENT REASON

1 . Draw a regular inscribed hexagon
with side AB = s, and a circum-
scribed square with side CD — s'.

2. Side AB = s = r = 1; side CD =
s' = 2r = 2

3. The perimeter of the hexagon =
6; the perimeter of the square = 8

4. .*. c >6 and <8; .'. TT>! or 3 335; Postulates 3 and 4
and < f or 4

5. The perimeter of a regular in-
sribed polygon of 12 sides =

12[\/2^ A/4^T| =6.211657 338; Theorem III
The perimeter of a regular cir-

250

PLANE GEOMETRY

cumscribed polygon of 8 sides

2 2

627418 339; Theorem IV

1
'] = 6-'

f A/4 + 22
6. .'. c > 6.211657 and < 6.627418

6<2101657or 3. 105829 and <

or 3.313709

6.627418

335; Postulates 3 and 4

7. If the doubling of the sides of
inscribed and circumscribed poly-
gons is continued indefinitely, the
following tables are obtained:

REGULAR INSCRIBED POLYGONS

Ratio of

No. of
sides

Perimeter

perimeter
to diameter

No. of
sides

Perimeter

perimeter
to diameter

of circle

6.211657

3.105829

6.627418

3.313709

6.265257

3.132629

6.365196

3.182598

6.278700

3.139350

6.303450

3.151725

6.282064

3.141032

6.288226

3.144118

192

6.282905

3.141453

128

6.284448

3.142224

384

6.283115

3.141558

256

6.283500

3.141750

512

6.283264

3.141632

REGULAR CIRCUMSCRIBED
POLYGONS

8. The perimeters of both inscribed
and circumscribed regular poly-
gons approach the circumference
c as a limit:

and *L

-j or
a

' d" 4 d

> 3.141558 and < 3.141632
TT = 3.1416, approximately.

335; Limit Axiom 3

REGULAR POLYGONS AND CIRCLES* 251

341. — The History of TT. — The determination of TT is one
of the famous geometric problems. It is attempted in the
MS. of Ahmes, now preserved in the British Museum.
Other later attempts took the form of a determination of
the area of a circle, or of the side of a square of equal area.
Such a value involves that of TT, and thus the problem has
been known as "squaring the circle."

Archimedes (250 B.C.), and Ptolemy (150 A. D.), obtained
values of TT almost as accurate as that calculated in 340.
Metrus and Romanus, of Holland, both belonging to the
16th century, calculated the value of TT by the method of
340, the latter using polygons up to 1,073,741,324 sides,
and obtaining a value correct to sixteen decimals. Lam-
bert (1750) proved TT incommensurable, that is, having no
exact value; and Lindemann (1882) proved it transcendental,
that is, not expressible as a radical nor as a root of an alge-
braic equation with integral coefficients. There are other
methods of calculating TT, which belong to the realm of
higher mathematics. Mr. Shanks, in 1873, carried the
calculation to 707 decimals. Ten decimals are necessary
for the most exact astronomical calculations, and four to
seven decimals are sufficient for those of mechanics and
engineering.

The establishment of the theorem of 340 may justly be
regarded as one of the greatest triumphs of the logical
system.

The number r enters into a great many important mathe-
matical expressions, or formulas, which have nothing what-
ever to do with a circle. The fact that TT equals the ratio

^ is really not its definition, but should be regarded merely

as an incidental — and the most elementary — property
(Ball's " Mathematical Recreations and Problems").

ADDITIONAL THEOREMS

342. (1) The apothem and radius of a regular triangle
can be expressed in terms of the side.

252 PLANE GEOMETRY

Helps.— (a) r = = .2887s;

(6) R = = .5774s.

(2) The side of a regular triangle can be expressed in terms
of (a) the apothem, and (b) the radius.

Helps.— (a) s = 2r\/3 = 3.4642r:
(6) s = #V3 = 1.7321#.

(3) The central angle subtended by a side of a regular
polygon is double the angle between a side and a least diagonal.

(4) The perimeter of the circumscribed equilateral triangle
of a circle has a fixed ratio to the perimeter of the inscribed
equilateral triangle.

(5) Find the ratio of the sides and perimeters of the inscribed
and circumscribed squares of a circle.

(6) Find the ratio of the sides and perimeters of the in-
scribed and circumscribed regular hexagons of a circle.

(7) The side of a regular decagon whose radius is r, is,

Helps— By 332, — — = -• See also 316.
s r

343. THE FOURTH STEP IN CLASSIFICATION.

EXERCISES

1. Draw a regular inscribed polygon of 5 sides, in a circle whose
radius is about 3 inches. Bisect one of the arcs subtended by
a side of the polygon and join the bisection point with an end
of the nearest side,thus obtaining a side of the regular inscribed
polygon of 10 sides. Measure a side of each polygon. Calcu-
late the side of the decagon from the side of the pentagon by
using the formula of 338. Does the calculated result agree
with the measured value?

2. Use the figure of Exercise 1 and draw a regular circumscribed
pentagon and one side of the regular circumscribed decagon.
Measure a side of each polygon. Calculate the side of the
decagon from that of the pentagon by using the formula of 339.
Does the calculated result agree with the measured value?

REGULAR POLYGONS AND CIRCLES 253

3. Construct accurate figures for all the Theorems of 342 and test
the correctness of the formulas and other principles of the
section by careful measurements of the figures. The figure of
Exercise 1 may be used for the test of Theorems (3) and (7).

344. REVIEW EXERCISES

.1. If a string 6 feet long is laid so as to form an equilateral triangle,
what are the radii of the inscribed and circumscribed circles?

2. Can a cask which measures 9 feet in circumference be taken
through a 3-foot door?

3. Calculate the length of an arc of 75° in a circle whose radius is
1200 feet.

Helps. — (a) Use the formula of 337, Corollary 3; (b) use the for-
mula of 331, the chords C and c being measured from a figure
drawn to scale.

4. The length of an arc of 60° is 12.35 inches. Calculate the diame-
ter of the circle.

5. The circumference of a circle is 10 feet. What is the circum-
ference of a circle of double the radius?

6. The chord C of an arc of 60° in a circle whose radius is 10 feet,
is 10. Calculate the length of an arc of 60° by the formula of
337, Corollary 3. Also (a) calculate the chord c of one-half the
arc by the formula of 331 ; and (6) calculate this chord by the
formula of 338 ; and (c) find some other method of calculating
this chord.

7. Find the diameter of a baseball or tennis ball by measuring the
circumference.

8. Calculate the side of an equilateral triangle of which the
apothem is 30.

9. Find the length of an arc of 60° in a circle whose radius is 30
inches (337, Corollary 3).

10. Find the length of an arc of 45° in a circle whose radius is 20
inches.

11. Find the length of an arc of 115° 35' in a circle whose diameter
is 50 inches.

lt,35

n 60 _ 6,935

(eLp' 360 ~ 360 21,600

12. The chord C of an arc of 60° in a circle whose radius is 30 inches
is 30 inches. Calculate the chord of \ the arc from this value
and the result obtained for the length of the arc in Exercise 9;
(a) using the formula of 331 ; (b) using the formula of 338.

13. If the diagonal of a square whose side is 5, is 7.07, what is the
diagonal of a square whose side is 8? What is the side of a
square whose diagonal is 8 (335, Postulate 5) ?

254

PLANE GEOMETRY

APPLICATIONS

345. The Design of a Running Track. — The best and
usual shape of a running track is that of a rectangle with
semi-circular ends (Fig. 1). The form of Fig. 2 may be
used when the available ground space does not permit the
use of the better form. The standard line on which the
length of the track is calculated is indicated by the dotted
line, and is 1| feet from the inner edge of the track.

FIG. 1.

FIG. 2.

EXERCISES

1. Calculate the total length of a track if AB = 200 feet and BC =
200 feet (Fig. 1).

Help. — The two ends form a complete circle 203 feet in diameter.

2. Calculate the total length of a track of the form of Fig. 2, if
AB = 200 feet, BC = 200 feet, and the radius of the corner
turns, measured to the standard line, = 50 feet.

Helps.— EF = 203 - 2 X 50 = 103 feet.

3. What must be the length of AB for a quarter-mile track if GK
(Fig. 1) = 200 feet?

4. What must be the value of BC for an eight-laps-to-the-mile track,
in which AB = 100 feet (Fig. 1)? Find the over-all length,
/«/, to the standard line.

5. The over-all length, /«/, of a quarter-mile running track is 400

feet. Find AB and GK (Fig. 1).
Helps.— Let x = GK; then AB = 400 - x; .'. length of track is

2(400 - a) + 3.1416* = ^~-

6. How many feet more will a runner travel in one lap of the quar-
ter-mile track of Exercise 3, if he runs 3 feet from the inner edge
of the track, than he would travel on the standard line?

346. POWER TRANSMISSION AND SHOP PRACTICE. — The
simplest forms of power transmission is by gear wheels and
belts. The velocity ratio of the driving shaft A, and the

KEGULAR POLYGONS AND CIRCLES

255

driven shaft B, is the ratio of the cir-
cumferences of the pitch circles of the
two gear wheels; since the point of
contact of the teeth of the gear wheels
is at the point of tangency of these
circles; Fig. 1. The velocity ratio of
the two shafts C and D is the ratio of
the circumference of the driving pulley
C and^the driven pulley D, around
which the belt passes; Figs. 2 and 3.

FIG. 1.

EXERCISES

1. The diameter of gear wheel A (Fig. 1) is 12 inches, and that
of gear wheel B is 8 inches. The number of revolutions per
minute of shaft A is 60. How many revolutions per minute
does shaft B make?

2. What is the velocity ratio of the first and third shafts of a train
of three gears of 10, 12 and 8 inches diameter, respectively?

FIG. 2.

FIG. 3.

3. The driving gear of a bicycle, carried on the pedal shaft, is
8 inches diameter; the rear gear on the wheel axle is 3 inches
diameter; the tire of the rear wheel is 28 inches diameter.
How many revolutions per minute must the pedals make in
order to drive the bicycle at the rate of 15 miles per hour?
Help. — Number of revolutions per minute of pedal shaft =

5280 X 15 X 12 1 3

60 X 28 X 3.1416 8*

4. If the diameters of the pulleys on both shafts in Fig. 2 are 24
inches, and the shafts are 10 feet apart, center to center, calcu-
late the length of belt required.

5. In Fig. 2, the diameters of the pulleys are 36 inches and 18 inches,
and the distance of the shafts is 8 feet, center to center. Draw
the pulleys and the distance between centers to scale, measure
the lengths SM = PQ, and the angles MDP and QCS, and

256

PLANE GEOMETRY

calculate the arcs MNP and QRS by the method of 337.
Find the total length of the belt.

6. Use the dimensions of Exercise 5 to apply to the crossed belt
of Fig. 3. Calculate the length of the belt.

The cutting speed of a tool or the velocity of some moving
part of a machine must be known by those who design
machinery.

EXERCISES

7. If the allowable cutting velocity of a tool when a piece of iron is
turned in a lathe, is 40 feet per minute at the surface of the
revolving piece, at how many revolutions per minute should a
lathe be driven in order to produce this surface velocity for a
piece of iron which is 5 inches in diameter? d inches in
diameter?

•8. At what velocity does the rim of a 10-foot diameter fly-wheel
travel, if it makes 80 revolutions per minute?

9. If pulley C in Figs. 2 and 3 is 24 inches diameter and makes
50 revolutions per minute, what is the linear velocity of the belt
in feet per minute?

10. If a belt is moving at a linear velocity of 500 feet per minute,
and is driving a shaft by a 3-foot pulley, how many revolutions
per minute is the driven shaft making?

Draughting and other applications.

EXERCISES

11. Show that in drawing a side view of a hexagonal
nut, that AB = CD = Id, and BC = \d. In
practice, d is made 2di.

12. Draw a bolt whose diameter di is 1£ inches, and
a hexagonal nut.

13. ABCDis a square. With centers at A, B, C, D,
arcs are drawn with a radius = AO3 intersect-
ing the sides in points E, F, G, etc.
Show that the figure EFGHIJKL is a
regular octagon. This method is very
useful in practice as a quick and accur-
ate construction.

14. Draw a square side of which is 3
inches. Construct a regular octagon
by the method of Exercise 13. Calcu-
late the side of the octagon.

REGULAR POLYGONS AND CIRCLES 257

15. Show that a side of a square from which a regular octagon of
side s, may be cut by the method of Exercise 13, is given by
the formula, s' = s(\/2 + 1). Also s = s'(\/2 — 1).

16. Find the side of a square from which a regular octagon whose
side is 2 inches may be cut.

347. Astronomical Problems. — All sections of the earth
by a plane may be considered (except in the most accurate
work) as circles.

EXERCISES

1. Taking the diameter of the earth as 41,780,000 feet, calculate
the length in feet of one minute of arc on the equator or on a
meridian, or on any "great circle."

2. The earth rotates on its axis in 24 hours. Calculate the velocity
of a point on the equator in miles per hour.

3. The earth moves about the sun in an orbit that is approximately
circular with a mean radius of 90,000,000 miles, in about 365
days. Calculate its velocity in the orbit in miles per hour.

The Size of the Earth. — The most accurate method of
determining the size of the earth is by the measurement of
an arc of a " great circle" on the earth's surface. A and
B are two places on the same meridian; AC and BD are
the horizontal planes at these points. Observations are
made on a star S in the meridian plane of arc AB, the angles
of elevation, CAS and DBS, being measured simultaneously.
Eratosthenes made a fairly accurate calculation of the
diameter of the earth by this method (using the sun as the
object of observation) near Alexandria, Egypt, about 250
B.C. S

258 PLANE GEOMETRY

EXERCISES

1. Show that Z.AOB = /.DBS ~ /.CAS, BS and AS being
parallel.

2. If /.DBS is measured 67° 31', and /.CAS is measured 65° 43'
find /.AOB.

3. If arc AB is measured 124.5 miles, calculate the circumference
and diameter of the earth in miles.

348. THE RADIAN. — This is a new unit of angle and arc.
It is the central angle of a circle (or its arc) whose arc is the
length of the radius. Do not confuse this with the central
angle subtended by a chord equal to the radius. The
Radian is not used in elementary geometry, but is an im-
portant unit in higher mathematics.

EXERCISES

1. Calculate the number of degrees in a radian.

Q£JA°

Help. — Radian = -7, —

2. What central angle is subtended by a chord equal to the radius?

3. How many radians are contained in an arc of 137°?

349. The Circle and Regular Polygon in Design. — Many
ornamental designs for rugs, carpets, wall paper, china and
glass-ware decorations, lace and embroidery, tiling, stone-
tracery, etc., are based upon the regular polygon and its
inscribed and circumscribed circles. Refer to a ''Source
Book of Problems for Geometry," by Sykes; Allyn and
Bacon; for an analysis of numerous designs of this kind.
See also 195 and 321.

EXERCISES

1. Lay out some of these designs on squared paper, on a larger scale.

2. Make some original designs based upon regular polygons and
circles.

3. Sketch some designs which you see in use. Draw the principle
lines of the designs and omit details and scrolls.

REGULAR POLYGONS AND CIRCLES 259

FIELD WORK

350. Design and lay out a running track which is 4, 6 or 8
laps to the mile. Refer to 345, Fig. 1. The available
rectangular plat should first be measured approximately
and laid out on paper to scale. The shape and dimensions
of the standard line is then determined and laid out on the
plat. The point A is then located on the ground from a
measurement scaled from the map; AG = 1J feet. Lay
out A B parallel to the side of the field. Erect a perpendi-
cular to AB at point A, and set stakes at points 0 and 0\.
In the same way set stakes at points D and C. The per-
pendiculars are most accurately laid out with a transit, but
a tape may be used. Measure DC as a check. The half
circles AMD and BNC are laid out by holding one end of the
tape on stakes 0 and 0\ and carrying the end around, say,
4 paces at a time and driving stakes at each point on the
curves' thus found. Set stakes on lines AB and DC about
10 feet apart. The outside line of the track is laid out, say,
10 feet from the inside line. Points opposite A, B, C, D are
first located. The arcs may be laid out from centers 0 and
Oij or if the radius is too great for the length of tape, stakes
may be set by sighting the stakes which are already placed
on the inner arcs, in line with the center hub stakes 0 and
0,.

CHAPTER XVI
AREAS

MEASUREMENT OF AREAS. PRINCIPLES DETERMINED
EXPERIMENTALLY

351. An area is a definite part of a surface. A new unit
is required for the measurement of an area.
The unit of area is the area, or surface, or extent,
of a square, each side of which is a unit sect.
Thus unit A is a square centimeter, and unit B is
a square inch.

Equivalent areas contain the same area
measure, but may be different in shape.

A segment and a sector of a circle are
certain parts of the area of a circle.
Refer to 267.

An annulus is the area included be-
tween two concentric circles.

Only plane areas are considered in this text.

Abbreviation. — Equivalent; is equivalent to; =c=.

EXERCISES

1. What geometric units have been previously considered, for
measuring: (a) sects; (6) angles; (c) arcs? Recall their defini-
tions.

2. What unit, or units, are used to measure areas the size of this
page? the area of a room? of a farm? of a country?

3. Write a table of equivalents between different area units.

4. Are equivalent areas necessarily congruent?

5. Draw a square, a circle, an oblong and an equilateral triangle
which are approximately equivalent.

6. Add geometrically, a square 2 inches on a side and an oblong
2 inches by 3 inches.

260

AREAS

261

7. Subtract geometrically, the areas of Exercise 6.

8. If two equal rectangles are added to or subtracted from two equal
triangles, are the results equivalent? Are they necessarily
congruent?

9. Draw an annulus of which the radii of the bounding circles are
1.2 and 1.7 inches, respectively.

352. Experiment I. — The area of a rectangle.

Draw rectangles; (a) 2 inches
by 3 inches, full size; (6) 2|
inches by 3 inches; (c) 2J inches
by 3J inches. Divide into square
inches and fractions thereof;
count the area measures in
units and fractions. Observe

the relation between the area measure of each rectangle
and the sect measures of length and width. Squared
paper may be used in this and other experiments in areas.
State result.

EXERCISES

1. State a formula for the area R of a rectangle, in terms of the
width w, and the length I.

2. Solve the formula of Exercise 1 for w and for I.

3. Draw a square foot to one-half linear scale, and divide it into
square inches to one-half linear scale. How many square inches
are contained in the square foot?

4. Calculate the number of square inches (that is, the area) in a
rectangle which is 15 inches by 8 inches.

5. How many square feet are in a rectangle 18 inches by 24 inches?

6. What is the length of a rectangle whose area is 750 square inches
and width 15 inches?

7. Draw a square inch A BCD; cut it A £ B

along a line DE; place the portion

DBA, cut off, so that AD coincides
with BC. Is the resulting figure
equivalent to the original square
inch? May this new figure also be
called a square inch, since it contains
an area equal to that of the given
square inch?

262

PLANE GEOMETRY

353. Experiment II. — The area of an oblique parallelo-
gram.

Draw rhomboids: (a) with base 3 inches and altitude 2
inches (full size); (6) with base 3J inches and altidude 2J

inches. Make the angle x equal the angle x of 352, Exercise
7. Divide the rhomboids into units and fractions thereof,
of the shape of the unit referred to in 352, Exercise 7.
Observe the relation between the area measures and the
dimensions of the figures. State result.

EXERCISES

1. State a formula for the area P of a rhomboid in terms of the base
6, and the altitude a.

2. Solve the formula of Exercise 1 for a and 6.

3. Calculate the area of an oblique parallelogram of which the
base is 25 feet and altitude is 12 feet.

4. Perform Experiment II by cutting the rhomboid on some
altitude, and placing the portion cut off in such a position
as to form a rectangle. Obtain the area measure of this rec-
tangle, which must equal that of the given parallelogram.

5. If a rhomboid is drawn with the sides including any other angle
than the angle x selected for the figures of Experiment II, would
the result be the same?

354. Experiment III. — The area of a triangle.
Draw triangles of different shapes. Trace on thin paper;
revolve the traced figures on the points A,B,C, respectively,

AREAS

263

which are the middle points of one side of the triangles, and
draw the symmetrical impressions of the triangles as indi-
cated in the second figure. What figure is formed in each

case by two adjoining congruent triangles? What is the
area of the complete figure? What is the area of the single
given triangle? State result.

EXERCISES

1. Write a formula for the area T, of a triangle. Solve for each
letter.

2. Calculate the area of a triangle of which a = 8", b = 10".

3. Calculate the base and altitude of a triangle whose area = 40
square inches, and of which the base exceeds the altitude by
2 inches.

4. Two sides of a triangle are 20 and 36, and the included angle is
45°. Calculate the area. Help. — First find the altitude.

5. Perform Experiment III by dividing the given triangles into
square units and fractions thereof. Use squared paper. This
method employs no analysis and is purely experimental.

355. Experiment IV. — The area of a trapezoid.
Draw trapezoids; trace; revolve on the middle point of
one of the non-parallel sides. What figure is formed by two

adjoining trapezoids? What is the area of this figure?
What is therefore the area of the trapezoid?

264

PLANE GEOMETRY

EXERCISES

1. State the result of Experiment IV as a formula. Solve for each
letter.

2. The dimensions of a trapezoid are 61 = 10, bz = 18, a = 8.
Calculate the area, S.

3. The bases of a trapezoid are 15 and 25; one of the non-parallel
sides is 10; and its projection on a base is 6. Calculate the area.

4. In a trapezoid, 61 = 120, 62 = 190, a side = 100, the angle in-
cluded between the 100 side and the 190 base = 60°. Calcu-
late the area.

5. The area of trapezoid is 400 square units, the altitude is 30 units,
one base is 10 units. Find the other base.

6. Perform Experiment IV by dividing the given trapezoids into
square units and fractions thereof.

356. Experiment V.— The area of a circle.

Draw a circle about 4 inches in diameter and cut it out.
Cut it on a diameter AB. Cut each semicircle into the
same number of equal sectors; say into 8 sectors. Place
the sectors as shown in Fig. 2. How may the resulting
figure be made to approximate more and more closely to a
rectangle as in Fig. 3? What are the base and altitude of
this rectangle? What is therefore the area of a circle?

246

FIG. 2.

FIG. 1.

FIG. 3.

EXERCISES

1. State the area C as a formula in terms of r (or d) and c.

2. State the value of C in terms of r alone (except that IT will
enter); in terms of d; in terms of c. Refer to 337 Corollary 2.

3. Solve the formulas of Exercise 2 for r, d and c, respectively.

4. Calculate the area of a circle whose radius is 10".

5. Calculate the area of a circle whose circumference is 10".

AREAS 265

G. Calculate the radius of a circle whose area is 7854 square inches.

7. Perform Experiment V by drawing a circle on squared paper
and counting the area measure.

8. Show by a method similar to that employed in Experiment V,
that the area of a sector equals one-half the product of the arc
and the radius. Express this as a formula.

9. Express the area of a sector in terms of its
central angle, n degrees, and its radius r.

Help. — Arc a = =•=?. 2irr.

OOU

10. Calculate the area of a sector whose radius is 10"
and of which the length of arc is 12".

11. Calculate the area of a sector whose radius is 20" and central
angle 70°.

357. EXPERIMENT VI. — THE RATIO OF Two AREAS.
Draw two rectangles having different bases and altitudes.

Divide them into square inches and fractions
thereof. Find the area measure of each rec-
tangle and express as a ratio.

Divide the two rectangles into any other
arbitrary square unit s, and again find their area measure
in terms of this unit, and express as a ratio. Observe
how the ratios compare. State result.

EXERCISES

1. A's farm contains twice as many square feet as B's farm con-
tains. Does A's farm also contain twice as many square kilo-
meters as B's farm 1

2. If two figures are drawn to some part-size scale, will the ratio of
the areas of the figures be the same as in the full-size figures?

358. EXPERIMENT VII. — THE RATIO OF THE AREAS OF
SIMILAR FIGURES.

(a) Draw a rectangle. Draw other similar rectangles
with ratios of similitude to the first rectangle of 2 : 1 ; 3 : 1 ;
J£ : 1; respectively: that is, the length and width of the
second rectangle are each double the corresponding dimen-
sions of the first rectangle; etc. Divide these rectangles
into smaller rectangles of the size and shape of the first
rectangle, and fractions thereof. Observe how the ratios

266 PLANE GEOMETRY

of the areas compare with the ratios of similitude of the
figures.

(6) Draw similar triangles. Divide the larger triangles
into smaller triangles of the size and shape of the smallest
triangle of the series. With this triangle as a unit of
measure, observe how the ratios of the areas of the triangles
compare with their ratios of similitude. State the general
result.

EXERCISES

1. State the result as a formula in terms of the areas S and S', and
a ratio of similitude — or ^7*

2. The area of a rectangle is 5 square inches. What is the area of
a similar rectangle whose ratio of similitude to the first is 10 : 1?

3. The area of one rectangle is 36 times the area of a similar rec-
tangle. The base of the larger rectangle is 12 feet. What is
the base of the smaller rectangle?

4. The area of an equilateral triangle is 10 square inches. What
is the side of an equilateral triangle whose area is 40 square inches,
expressed in terms of the side s, of the first triangle?

5. Show by drawings that the law of the ratio of the areas of two
similar figures holds for, (a) parallelograms, (6) trapezoids.
Help. — In the figures of trapezoids, a cutting and re-uniting of
the figure is necessary.

369. REVIEW EXERCISES

1. Is it correct to say that a number of inches multiplied by a
number of inches results in a number of square inches? Can
two denominate numbers be multiplied? If a denominate
number is multiplied by an abstract number, what kind of
number results?

2. Show that the curved (convex) surface of a circular cylinder,
bounded by parallel planes perpendicular to the axis, is a rec-
tangle. State a formula for the area of the convex surface.
For the total area.

3. Show that the convex surface of a right circular cone is a sector.
What is its arc? Its radius? State a formula for the area of
the convex surface. For the total area.

4. The circumference of a tree is 80 inches. Find the area of
cross-section.

5. A horse is tied to a stake by a rope 50 feet long. Over what
area can he graze?

AREAS

267

6. A horse is tied to a stake in the corner of a square court by a
rope 50 feet long. Over what area can he graze?

7. A horse is tied to a stake at the corner of a square building.
The rope is 100 feet long, and the building is 50 feet square.
Over what area can he graze?

8. How many square feet will be enclosed by a rope 120 feet long,
if it is laid down on the ground (a) in the form of a square; (6)
in the form of an equilateral triangle; (c) in the form of a circle?

9. What may we conclude is the greatest (or maximum) area for
a given perimeter, based upon the results of Exercise 8?

10. What part of a circle is the area of a sector of 110°? Of 86° 32'?

APPLICATIONS

360. Surveying. — Land is generally valued and sold in
terms of its area measure. City lots are sometimes rated
in terms of the "front foot."

If it is sold for $1200,

EXERCISES

1. A town building lot is 120 feet by 80 feet,
what price is obtained per square foot?

2. A farm is bounded by four
straight lines. A diagonal AC
is 560.75 feet; the distances BE
and DF, perpendicular to AC, are
respectively 93.44 feet and
612.20 feet. Find the area of
the farm in square feet. (The
figure is not drawn to scale.)

361. Estimating. — Estimates for painting, roofing, pav-
ing, flooring, wall paper, carpets, etc., are made in terms of
the areas involved.

EXERCISES

1. Find the cost of laying a cement pavement
on the court of which the plan is given, at
$2.50 a square yard. The dimensions are
in feet.

2. A room is 20 feet long, 12 feet wide, and
10 feet high. Find the cost of plastering
the walls and ceiling of the room at 20
cents a square foot, deducting 25 percent

of the area for doors, windows and woodwork.

268 PLANE GEOMETRY

3. What is the cost of laying an asphalt pavement per mile of road,
14 feet wide, at $4.50 per square yard?

4. A barn roof is 32 feet wide at the eaves, 12 feet
high to the ridge, and 40 feet long. Allow two
feet for overhang at the eaves. Find the cost
of roofing with tar paper at 5 cents a square |
foot, which allows for overlapping and waste.

5. How many sheets of paper 8 by 11 inches, can be cut from 100
sheets 24 by 44 inches?

362. PROBLEMS FOR FIELD WORK. — (1) Find the area
covered by a building all of the angles of which are right
angles, by measuring all horizontal dimensions. Draw a
plan and divide the area into rectangles.

2. Find the area covered by any existing or proposed
paths, roads, pavements, etc. The cost may be estimated
by procuring a trade price of the required pavement per
square foot or square yard.

CHAPTER XVII
AREAS

CLASSIFICATION AND EXPLANATION OF PRINCIPLES

363. The First Step in Classification. Area Postulates.—
(1) If an area is divided and the portions are placed in any
other relative position without overlapping, the resulting
area is equivalent to (that is, equal in measure to) the
original area.

2. The unit of area is a unit of the same kind, and is a
square (or its equivalent) each side of which is a unit sect.

3. The area of a rectangle is the unit of measure multi-
plied by the product of the length and width of the rec-
tangle; or, it equals the product of the length and width
(or base and altitude) .

4. The area of a circle is the limit which the areas of
inscribed and circumscribed regular polygons approach, as
the number of sides is indefinitely increased.

5. The areas of two similar rectangles are proportional
to the squares of two homologous sides.

THE SECOND STEP IN CLASSIFICATION

364. Theorem I. — The area of an oblique parallelogram
equals the product of the base and altitude.

D E C

Hypothesis. — An oblique parallelogram A BCD, with
base DC = b, and altitude AE = a.
Conclusion. — Area A BCD = ab.

269

270 PLANE GEOMETRY

Analysis.

STATEMENT REASON

1. Extend DC and draw Postulate
BF±DC

2. &ADE ^ ABCF Postulate: 2 sides and in-

clude angle =, etc.

3. DC = EF = b Axiom.

4. Area ABFE =c= Area 363; Postulate 1.
ABCD

5. Area ABFE = ab 363; Postulate 3.

6. .'. Area ABCD = ab Axiom.

Corollaries. — (1) The areas of two parallelograms which
have the same base and different altitudes are proportional
to the altitudes.

2. State a similar corollary for parallelograms which
have different bases and the same altitude.

EXERCISES

1. Are all rectangles which have their bases and altitudes equal,
each to each, congruent?

2. Are all oblique parallelograms having their bases and altitudes
equal, each to each, congruent? Are they equivalent?

3. Write a formula expressing the ratio of the areas, P and P', of
two rectangles or two oblique parallelograms, whose bases and
altitudes are respectively b and a, and 6' and a'.

4. Write a formula expressing the ratio of the areas of any two
parallelograms whose bases are each equal to 6, and whose
altitudes are respectively a and a'.

5. Write a formula similar to that of Exercise 4 if the parallelo-
grams have equal altitudes a, and bases b and b' respectively.

6. Write a formula for the area S, of a square whose side is s.

7. Write a formula for the side s of a square whose area is S.

8. The base and altitude of two parallelograms are respectively
12" and 36", and 12" and 18". What is the ratio of their areas ?

9. Find the area of a rectangle of which the base is 10 inches and
the diagonal is 12 inches.

10. The sides of an oblique parallelogram are 10" and 15"; the
projection of the shorter side on the longer side is 6". Calcu-
late the area. Construct the parallelogram.

AREAS 271

11. Show that the area of the oblique parallelogram of Theorem I
equals the side AD multiplied by the altitude drawn to that
side.

12. Prove by similar triangles that the product ab = the product
of AD by the altitude drawn to AD.

365. Theorem II. — The area of a triangle equals one-half
the product of the base and altitude.

Helps.— (1) Draw BE\\AC and CE\\AB; (2) area EEC A
= ab; (3) /. area A ABC = J ab.

Corollaries. — (1) The areas of two triangles which have
the same base and different altitudes are proportional to the
altitudes.

2. State the Corollary which corresponds to 364, Corol-
lary 2.

EXERCISES

1. Show that the area of the triangle ABC may be expressed as one-
half the product of any side by the altitude drawn to that side.

2. Show that the area of a right triangle equals one-half the product
of the perpendicular sides.

3. Write formulas expressing the ratios between the areas of two
triangles T and T', the relations between the dimensions of
which correspond to those stated for parallelograms in 364,
Exercises 3, 4 and 5.

4. Find the area of a triangle of which two sides are 12" and 20",
and the projection of the shorter side on the longer is 8".

5. The mathematical treatise of Ahmes gives the area of an isosceles
triangle whose sides are 10, 10, 4, as 20. What formula did
Ahmes employ? What is the correct area? When is Ahmes'
formula approximately correct?

6. Find the area of a triangle of which two sides are 20 and 35 and
the included angle is 90°.

272 PLANE GEOMETRY

7. Find the area of a triangle of which two sides are 20 and 35 and
the included angle is 45°.

8. Find the area of a triangle of which two sides are 20 and 35 and
the included angle is 135°.

366. Theorem III. — The area of a trapezoid equals one-
half the product of the sum of the bases by the altitude.

h, 6z

i
i

Write out the analysis.

EXERCISES

1. Find the area of a trapezoid whose bases are 20" and 35" and
whose altitude is 15".

2. Find the other base of a trapezoid of which the area is 750 square
feet, the known base 30 feet, and altitude 40 feet.

3. Find the altitude of a trapezoid of which the area is 1200 square
feet, and the bases are 30 and 20 feet, respectively.

4. The method used by the ancient Egyptians for finding the area
of a piece of land in the form of a trapezoid was to multiply
half the sum of the parallel sides by one of the other sides.
Is this method ever correct? When is it approximately correct?

THE THIRD STEP IN CLASSIFICATION

367. Theorem IV. — The area of a triangle can be expressed
in terms of the three sides.

Helps. — Combine the result of Theorem II, 365, and the
value of an altitude, 265.
T =\A(s - a)(s - b)(s^- c).

EXERCISES

1. Find the area of a triangle whose sides are 10, 12 and 14.

2. Find the area of the isosceles triangle of 365, Exercise 5, by the
formula of this section.

3. Find the area of a right triangle whose sides are 30, 40 and 50
by the formula of this section. Check by the shorter method
of 366.

AREAS

273

368. Theorem V. — The area of a regular polygon equals
one half the product of the perimeter by the apothem.

Write out the analysis.

Corollary. — // s = a side, n = the number of sides, a

the apothem, then P = —=-•

EXERCISES

1. Find the area of a regular octagon of which a side is 24.85 inches
and the apothem 30 inches.

2. Find the area of a regular hexagon whose side is 40.

3. Use the formula of Theorem V to find the area of a square whose
side is 40. Check by the formula S = ab (363, Postulate 3).

369. Theorem VI. — The area of any circumscribed polygon
equals one-half the product of the perimeter by the radius of the
inscribed circle.

Write out the analysis.

274

PLANE GEOMETRY

370. Theorem VII. — The area of a circle equals one-half
the product of the circumference and radius.

Hypothesis. — Circle 0, with radius r, and circumference c.

Conclusion. — Area of circle, S = -*•

Analysis.

STATEMENT

1. Circumscribe any regular poly-
gon A BCD . . . , about circle
0, with an area P, and a peri-
meter p.

REASON

2.P-f

3. Indefinitely increase the num-
ber of sides of polygon A BCD .'.

4. Area P = Area S '

5. And p = c

. rp re
' 2 2

7. But P always = ~, and

• s -rc

Corollaries.

1. S = irr2 or 3.1416r2.

2. S = ~ or .7854d2.

369; Theorem VI.

363; Postulate 4
335; Postulate 4 (c)

335 ; Limit Axiom 3
335; Limit Axiom 2

3. S = -A- or .0796c2.
4?r

AREAS 275

4. The areas of two circles are proportional to the squares
of their radii, diameters or circumferences.

EXERCISES

1. The formula given by Ahmes for the area of a circle is S = ^.dz.

Show that this is equivalent to a value of TT = 3.16.

2. Is it probable from the evidence of the formula itself that Ahmes
arrived at this measure by comparing areas experimentally, or
by deductive reasoning?

3. Calculate the radius, diameter, and circumference of a circle
whose area is 78.54 square inches; of a circle whose area is 1200
square inches.

4. Find the area of a sector of 10° in a circle of radius 30.

5. Find the area of a sector whose arc equals twice the radius, in a
circle of 12" radius.

371. Theorem VIII. — The radius of the inscribed circle
of a triangle may be expressed in terms of the three sides.

Hypothesis. — AABC; the inscribed circle with center 0
and radius r.

Conclusion. — r may be expressed in terms of the sides a,
b, c, of the triangle. •

Analysis.— (1) Draw OA, OB, OC; (2) Area ABOC =

^, etc.; (3) Area A ABC = q + k + c. r = s - r; (4) But

area A ABC = \/s(s - a) (s - b) (s - c); (5) .'. r =
(s — a) (s — b) (s — c)
s

EXERCISES

1. Calculate the radius of the inscribed circle of a triangle whose
sides are 3, 4 and 5.

2. Calculate the area of the triangle of Exercise 1, using the formula
S = sr; and the value of r found in Exercise 1.

ADDITIONAL THEOREMS

372. (1) Two triangles of which two sides are equal, each
to each, and the included angles supplementary, are equivalent;
that is, equal in area.

276 PLANE GEOMETRY

(2) A median divides a triangle into two equivalent
triangles.

(3) The area of a trapezoid equals the product of the
altitude by the median.

(4) The area of any quadrilateral equals one-half the
product of a diagonal by the sum of the perpendiculars drawn
to the diagonal from the two opposite vertices. Write the
formula.

(5) // the diagonals of a quadrilateral are perpendicular,

the area is S = — „-, where di and dz are the diagonals.

(6) The area of a rhombus equals one-half the product of
the diagonals.

(7) The area of a square is S = « / where d is the diagonal.

(8) The diagonal of a square may be expressed in terms of
the area.

(9) The area of a circumscribed square is double that of the
inscribed square of the same circle.

(10) The area of an equilateral triangle is T = — j— /

where s = a side of the triangle.

Helps. — (a) Derive the formula from the area formula,

ab
T =-%, (365) ; (6) Derive it from T = \/s(s-a)(s- 6)(s-c),

(367), where s = one-half the sum of the sides of the triangle.

(11) The side of an equilateral triangle may be expressed
in terms of the area.

Help.-s =?VfV3
o

(12) The area of a regular hexagon is H = — v~*i where s

is a side of the hexagon.

(13) The side of a regular hexagon may be expressed in
terms of the area.

(14) The area of an annulus may be expressed in terms of
the radii, or of the diameters, of the bounding circles.

AREAS 277

Helps.— (1) S = Si - S2; (2) Si = irrS and S2 = 7rr22;
(3) Subtract and factor; S = 7r(ri + r2)(ri — r2).

(15) The area of a triangle may be expressed in terms of the
three sides and the radius of the circumscribed circle.

Helps.— Refer to 308 and 367.
_obc
" 4R

(16) The radius of the circumscribed circle of a triangle
may be expressed in terms of the area and sides of the triangle.

(17) The area of a minor segment of a circle is S = ™-

irr2 — ~; where x is the central angle of the segment, r the

radius of the circle, c the length of the chord of the segment, and
p the perpendicular upon the chord from the center of the
circle.

Help. — Area of segment = area of sector — area of
triangle.

(18) State the formula for the area of a major segment of a
circle.

(19) The difference of the areas of a circle and the circum-
scribed square may be expressed in terms of the radius or
diameter of the circle.

(20) The different of the areas of a circle and the inscribed
square may be expressed in terms of the radius or diameter of
the circle.

373. EXERCISES

1. If the area of a triangle of which two sides and the included
angle are respectively 37", 51" and 65°, is approximately 216
square inches. What is the area of a triangle of which two
sides and the included angle are respectively 37", 51", and 115°?

2. Check the value given in Exercise 1 by constructing one of the
triangles and measuring an altitude.

3. A diagonal of a quadrilateral is 27" and the perpendiculars
drawn to this diagonal from the other two vertices are respec-
tively 10" and 45". Calculate the area of the quadrilateral.

4. The diagonals of a rhombus are 12" and 20". Calculate the
area.

278 PLANE GEOMETRY

5. Find the diagonal of a square whose area is 3200 square inches.

6. The area of a square circumscribed about a circle is 240 square
inches. Find the area of a square inscribed in the same circle.

7. Find the area of an equilateral triangle whose side is 20 inches,
using the formula of 372, Theorem (10.)

8. The area of an equilateral triangle is 100 square inches. Find
the side of the triangle.

9. Find the area of a regular hexagon inscribed in a circle whose
radius is 12 inches.

10. Find the area of an annulus whose radii are 10" and 20"
respectively.

11. The radius of one of the bounding circles of an annulus is 24
inches and the area of the annulus is 1200 square inches. Find
the radius of the other bounding circle (a) if the given circle is
the smaller of the two circles; and (b) if the given circle is the
larger of the two circles.

12. Find the area of the triangle of 308, Exercise 1, using the for-
mula of 372, Theorem (15.)

13. Find the area of a segment of 120° in a circle whose radius is
12 inches.

14. Find the difference in the areas of a circle whose radius is 20
inches, and its inscribed square, using the formula of 372.
Theorem (20.)

THE FOURTH STEP IN CLASSIFICATION

374. EXERCISES

1. Construct the triangle of 367, Exercise 1, to a smaller scale.
Measure an altitude. Calculate the area by the formula
S = %ab (365). Compare the result with the value obtained in
the Exercise referred to.

Exercises 2 and 3, 367, also are checks on the formula of 367.

2. Construct the triangle of 371, Exercise 1, and the inscribed circle.
Measure the radius. Compare with the value calculated in
the Exercise referred to.

3. Construct the inscribed and circumscribed squares of the same
circle. Measure a side of each, and calculate the areas. Does
the result agree with the principle of 372, Theorem (9) ?

4. Construct an equilateral triangle whose side is 20. Draw and
measure an altitude and calculate the area by the formula of
365. Does the result agree with the value obtained in 373,
Exercise 7?

5. Construct a triangle whose sides are 10, 12 and 14 (Exercise 1).
Construct the circumscribed circle. Measure the radius. Cal-

AREAS

279

culate the area by the formula of 372, Theorem (15). Does the
result agree with the values obtained in 367, Exercise 1, and in
Exercise 1 of this section?

RELATIONS BETWEEN AREAS
THE SECOND STEP IN CLASSIFICATION

375. Theorem IX. — The areas of two similar triangles are
proportional to the squares of any two homologous sides; or,
to the square of their ratio of similitude.

Hypothesis. — Similar triangles ABC and A'B'C'.
Area AABC a2

Conclusion. —

Area AA'B'C' a'2

Analysis.

STATEMENT

1. Area AABC = ~*

2. Area AA'B'C' = *

Area AABC

ha-a

Area AA'B'C' ha'-a'
ha _ a
17 ~ a'

Area AABC _ a2
Area AA'B'C' ~ a/2

REASON
Theorem

Theorem
Axiom
Theorem of
Axiom

Corollaries. — (l)The areas of two similar triangles are
proportional to the squares of homologous altitudes, medians,
or of any two homologous lines.

2. Any homologous lines (sides, altitudes, etc.) of two
similar triangles are proportional to the square roots of the
areas of the triangles.

280

PLANE GEOMETRY

EXERCISES

1. The sides of two triangles are respectively 7, 10 and 14, and 10 5,
15 and 21. What is the ratio of the areas of the triangles?

2. The area of a triangle is four times the area of a similar triangle.
What is the ratio of two homologous sides of the triangles?

THE THIRD STEP IN CLASSIFICATION
376. Theorem X. — The areas of two similar polygons are
proportional to the squares of two homologous sides.

Hypothesis. — Similar polygons X and Y.
!r = T\ = etc.

Conclusion. — -

Analysis.

STATEMENT

REASON

1. Divide -X" and Y into triangles
similarly placed, A\, B\ . . .

5. ±1 = *i =

tri-

Theorem of
angles.

375; Theorem IX.

7 ^ — a*2 _ ^l2

F := a? = S2 =
State corollaries similar to those of 375.

AREAS

281

EXERCISES

1. The area of a pentagon is 50 square inches; the ratio of similitude
of a second pentagon similar to the first pentagon, is 9:1. Find
the area of the second pentagon.

2. The areas of two similar quadrilaterals are 50 and 100. A side
of the first quadrilateral is 10; what is the homologous side of the
second quadrilateral?

377. THEOREM XI. — The areas of two triangles which have
an angle of one triangle equal to an angle of the other, are
proportional to the products of the sides including the equal
angles.

Hypothesis.— A ABC and DEF with

Area AABC AB-AC
C°ncluSWn'- Are* ADEF -=D'
Analysis.

STATEMENT

1. Place A ABC in the position
B'DC' and draw EC'.
Area ADB'C' DB' AB
= DE = DE
DF DF

REASON

0
2'

Area

Area ADEF

Area

Area A ABC

AB - AC

OCK ^ ..
365; C°r°llary
oc_ ^ „
365; C°r°llary

Axl°m

Area A DE- DF

Corollary. — The areas of two triangles which have an
angle of one supplementary to an angle of the other, are
proportional, etc.

EXERCISES

1. The area of a triangle, two of whose sides are 10 and 12 and of
which the included angle is 60°, is 103.92. What is the area of
a triangle, two of whose sides are 5 and 30 and of which the in-
cluded angle is 60°?

282

PLANE GEOMETRY

2. What is the area of a triangle, two of whose sides are 15 and 25
and of which the included angle is 120°?

378. Theorem XII. — The area of the square drawn on the
hypotenuse of a right triangle equals the sum of the areas of
the squares drawn on the two perpendicular sides.

K H

State the hypothesis and the conclusion.

Helps.— (I) Rectangle CDEF =c= 2 times &BCD; (2)
Square Z ~ 2 times AACH; (3) ABCD ± AACH; (4)
.'. Rectangle CDEF =c= square Z; etc.

This theorem has been analyzed before under similar tri-
angles, but there is an essential difference in the method
employed here and that of the former analysis. The analy-
sis here given was employed by Euclid.

More than three hundred different analyses of the theo-
rem have been discovered.

THE FOURTH STEP IN CLASSIFICATION

379. EXERCISES

1. Draw an irregular hexagon; draw a second hexagon similar to
the first, with the ratio of similitude 1 : 2. Divide the hexagons
into triangles in any way, and measure a side and an altitude to
that side in each triangle. Calculate the areas of each hexagon
by adding the areas of the triangles into which it is divided.
How does the ratio of the areas of the hexagons compare with
their ratio of similitude? Does this agree with the Theorem
of 376?

AREAS 283

2. Draw any oblique scalene triangle; draw a second triangle with
an angle equal to an angle of the first triangle. Measure a side
and an altitude to that side in each triangle and find the areas
by the formula of 365. Find also for each triangle the products
of the sides including the equal angles. Compare the ratio of
the areas with the ratio of these products, by reducing each
ratio to a decimal. Does the result agree with the principle of
377?

ADDITIONAL THEOREMS

380. — (1) The square on the base of a right isosceles tri-
angle is equivalent to four times the triangle.

(2) The four triangles into which the diagonals divide a
parallelogram are equivalent.

(3) Straight lines drawn from the middle point of a diag-
onal of a quadrilateral to opposite vertices divide it into two
equivalent parts.

(4) The parallelogram formed by joining the middle points
of each two adjacent sides of a quadrilateral is equivalent to
one-half the quadrilateral.

(5) A line drawn through the middle point of the median
of a trapezoid, cutting the bases, divides the trapezoid into
two equivalent parts.

(6) The triangle formed by joining the middle point of one
of the non-parallel sides of a trapezoid to the extremities of
the other non-parallel side is equivalent to one-half the trape-
zoid.

(7) The sects joining the middle points of the three sides
of a triangle divide it into four equivalent (congruent) triangles.

(8) The areas of two parallelograms having an angle of one
equal to an angle of the other are proportional to the products
of the sides including the equal angles.

(9) The sum of the areas of the squares on the sects drawn
from any point within a rectangle to two opposite vertices
equals the sum of the areas of the squares on the sects drawn
from the point to the other two opposite vertices.

Helps. — Draw lines through the point parallel to the sides
of the rectangle, forming right triangles.

284

PLANE GEOMETRY

(10) The area of the equilateral triangle drawn on the
hypotenuse of a right triangle equals the
sum of the areas of the equilateral triangles
drawn on the other two sides.

Helps. -(1) = ;

b*'

Y+Z a2 + c2

= 1; (3) :.X = Y +Z.

X b

(11) State and analyze a general
theorem for similar polygons, sug-
gested by Theorem (10).

(12) The area of a triangle of which one angle is 30° equals
one quarter the product of the including sides. Express as a
formula.

(13) The area of a triangle of which one angle is 45° equals
one quarter the product of the including sides times the square
root of two. Express as a formula.

(14) Find the formula for the area of a triangle of which
one angle is 60° and the including sides are b and c.

(15) Write the formulas for the areas of triangles of
which two sides are b and c and the included angle is 150°,
135° or 120°.

(16) // semicircles are drawn on the three sides of a right
triangle, as in the figure, the sum of the

crescents X and Y equals the area T of
the given triangle.

Helps. — (1) Write the formulas for
the semicircles in terms of their diame-
ters: (2) write the sum of the areas X
and Y as an algebraic sum of T and the semicircles.

(17) // lines are drawn from any point within a parallelo-
gram to the four vertices, the sum of the

areas of two opposite triangles thus formed
is equal to the sum of the areas of the other
two opposite triangles.

(18) A parallelogram is divided into two
equivalent areas by a line drawn through
the point of intersection of the diagonals.

AREAS 285

(19) A trapezoid is divided into two equivalent areas by a
line which joins the middle points of the bases.

(20) The area of a trapezoid equals the product of one of the
non-parallel sides by the perpendicular drawn to that side
from the middle point of the other of the non-parallel sides.

(21) Lines which join the middle points of the sides of a
parallelogram, in order, form four equivalent triangles, and
the sum of the areas of these triangles equals the area of the
parallelogram formed by the four lines.

(22) Lines which are drawn through any point on a diagonal
of a parallelogram, parallel to the sides of the parallelogram,
form with the sides of the given parallelogram four smaller
parallelograms, two of which are equivalent.

381. REVIEW EXERCISES

1. Find the non-parallel sides and the diagonals of an isosceles
trapezoid of which the area is 144 square units and the altitude
is 8 units and one base is 12 units.

2. Two sides of a triangle are 12" and 15" and the altitude to the
longer of the two sides is 10". Find the altitude to the other
of the two sides.

3. The perpendicular sides of a right triangle are 4 and 6. Find
the area of the triangle.

4. The hypotenuse of a right triangle is 20 and its base is 12.
Find the area and the altitude to the hypotenuse.

5. Find the side of a rectangle of which the area is 120 square
inches and the other side is 6 inches.

6. Find the diagonal of a rectangle of which the area is 48 square
inches and base 8 inches.

7. The base and altitude of a rhomboid are 12" and 10" respec-
tively. Find the area.

8. Two sides of a rhomboid are 8" and 14" and the altitude to the
longer side is 7". Find the altitude to the shorter side.

9. Two sides of a parallelogram are 50" and 30" and the included
angle is 3010. Find the area.

10. Calculate the area of a quadrilateral of which one diagonal is
120, and the perpendiculars upon this diagonal from two oppo-
site vertices are 50 and 75.

11. Two perpendicular diagonals of a trapezium are 81 and 44.
Calculate the area.

286 PLANE GEOMETRY

12. The side of a rhombus is 10 and a diagonal is 16. Calculate
the other diagonal and the area.

13. Calculate the area of a square whose diagonal is 24. Check
by finding the area from a side.

14. Calculate the areas of the circumscribed and inscribed squares
of a circle whose diameter is 8.

15. The area of a square is 72; calculate the diagonal without
finding the side. Check by finding the side and then
the diagonal.

16. The side of an equilateral triangle is 12. Calculate the area.
Check by first finding the altitude.

17. Calculate the radius of the circle in which a regular inscribed
hexagon has an area 180.

18. The difference of the areas of a circle and the circumscribed
square is 85.84 square inches. Calculate the radius of the
circle. Check by finding each area from the radius.

19. The sum of the four segments included between a circle and the
inscribed square is 456.64 square inches. Calculate the radius
of the circle.

20. The difference between the areas of a circle and the regular
inscribed hexagon is 1000 square inches. Calculate the radius
of the circle and its area.

21. The area of a circumscribed square is 25 square inches. What
is the area of the square inscribed in the same circle?

22. Find the area of the bases of a cylinder which is 6 inches long
and 2 inches in diameter.

23. How many square inches are in the convex surface of a cylinder
12 inches long and 10 inches diameter?

24. A cylindrical boiler is 10 feet long and 4 feet diameter. How
many square feet of sheet metal are required to construct its
convex surface?

25. Find the area of the base of a cone which is 6 inches in altitude
and 2 inches in diameter at the base.

26. Calculate the total area of the surface of a cone 16 inches alti-
tude and 24 inches diameter of base.

Help. — Calculate the slant height; convex surface; area
of base.

27. The sides of a rhombus and the shorter diagonal are each 10".
Calculate the longer diagonal and the area.

28. Calculate and arrange in order of length, the circumfer-
ence of a circle, the perimeters of a square, an equilateral
triangle, a regular hexagon; all of the same area. Let the
area = 1. In which figure is the ratio of perimeter to area
a minimum?

AREAS

287

29. Show by a figure that (a + 6)2 = a2 + 2ab + 62.
Help. — Show the equivalence of the areas.

30. Show by a figure that (a - 6)2 = a2 -
2ab + 62.

31. Show by a figure that (a + 6) (a — 6) =
a2 - 62.

32. Show by a figure that (a + 6) (a + c) =
a2 + a& -f ac + be.

33. Show by a figure that (a — 6) (a — c) = a2 — db — ac + be.

34. Show by a figure that (a + b + c)2 = a2 + b2 + c2 + 2a6 +
2ac + 26c.

Euclid gave the geometric construction for Exercises 29 to 31 in
the "Elements." Algebra was almost unknown at that time,
but many familiar algebraic relationships were discovered in
geometric form.

35. If two polygons are similar, and one is four times the other
in area, what is their ratio of similitude? What if one is double
the other?

36. Divide the altitude of a triangle into
four equal parts by lines parallel to
the base. What parts of the entire
triangle are the areas of the strips?
Help. — Calculate the areas of &ADE,
AFG, etc., in terms of A ABC.

37. The homologous sides of two similar polygons are 5 and 10.
Find the side of a similar polygon equivalent to their sum; to
their difference.

38. The area of an equilateral triangle is 1200 square inches. Calcu-
late the side.

39. The cross-section of a cut through a hill is a trapezoid, 20 feet

wide at the bottom, 10 feet deep, the angles of slope x, 45°.
Calculate the area of cross-section.

40. Write a formula for the cross-section of Exercise 39, when the
cut is w feet wide at bottom, c feet deep, slope 45°.

41. The scale of a map is 100 miles to the inch. What area is in-
closed by a square inch? What area is inclosed by a rectangle
\ inch by 1£ inches?

42. If a rectangular area 2 by 3 inches is known to represent on a
certain map an area of 15,000 square miles, what is the linear
scale of the map?

288

PLANE GEOMETRY

8 C

43. The following is found in many popular lists of "trick" prob-
lems. Draw a square 8 inches on a side; cut it as shown;
re-assemble the parts to form a

rectangle 5 by 13 inches. How
many square inches are ap-
parently gained by the opera-
tion? If this were true what
axiom of areas would be con-
tradicted? Show the fallacy of the apparent result by
calculating the length of DE if AC and BC are straight lines.
If DE = 3 inches, can both these lines be straight?

44. The base, 6, of a triangle is increased by an amount b' ; how
much must the altitude a be decreased in order that the area
shall remain unchanged?

45. The altitude and base of a triangle are a and b; the altitude and
base of a similar triangle are a' and b'. Calculate the altitude
and base of a< similar triangle equal to their sum.

Helps.

+ a'6')

+ r2; b" = (ab + a'6')

= b\/l + r2; where a" and 6" are the required base and alti-
tude, and r is the ratio of similitude of the given triangles.

46. The altitude and base of a triangle are 10 and 13; the altitude
and base of a similar triangle are 20 and 26. Calculate the
altitude and base of a similar triangle equal to their sum.

47. Calculate the area of an isosceles triangle whose equal sides are
24 and vertex angle is 30°; 60°; 135°; 90°.

48. Two similar fields contain together 579 square feet and their
ratio of similitude is rV Calculate the area of each field.

Si = T> m Si +£2 = 193
02 \.2t 01 4*7

49. Calculate the area of a parallelogram whose diagonals are 75
and 200, and one of their included angles 30°.

(a + 6) Vab

50. Show that T

where T is the area of a right

triangle, and a and b are the segments of the hypotenuse formed
by the altitude. Calculate the area of the triangle if these
segments are 4 and 6.

51. When Dido and her followers landed on the coast of Africa,
"they asked of the natives only so much land as they could
enclose with a bull's hide" (Bullfinch, "Age of Fable"). It is
told that the hide was cut into strips. If this hide contained
50 square feet and the strips were -fa inch wide, what area circle
was obtained? What area square could be enclosed?

AREAS

289

52. A square is 2" on a side. Find the side of a second square
whose area is 1^ times the area of the first square.

53. Find the side of a square equivalent to the sum of two squares
whose sides are 3" and 5" respectively.

54. Calculate the altitude of a triangle whose base is 20", and
which is equivalent to a triangle whose altitude is 12" and
base 16".

55. Calculate the altitude of a triangle whose base is 160"
and which is equivalent to the sum of two triangles which
measure respectively, altitude 50", base 120", and altitude 75",
base 60".

56. Find the area of the segment ACB cut off by a chord which
subtends an arc of 120°, in a circle whose

radius is 20 inches.

Helps. — Area of segment ACB = area of
sector OACB - area of AOAB. Alti-
tude OD of AOAB = i radius; why?
Find the base of AOAB; etc.

57. A stake is driven on the farther edge of
a stream 50 feet wide, and a horse on the
nearer side of the stream is tied to the

stake by a rope 100 feet long. Over what area can the horse
graze without crossing the stream ?

58. Find the area of the minor segment cut off by a chord which
subtends an arc of 90° in a circle of 10 inches radius.

59. What fractional parts of the area of the circle
are included between each pair of equidistant
parallel chords in this figure?

60. What area is included between a circle of 10
inches radius and the sides of an inscribed
square?

61. Find the area of an annulus whose radii are 12 and 20 inches
respectively?

62. Find the area of a triangle whose sides are 102, 104, 106.
Help.— Result is S = 23'32-13'5 = 4680.

63. Find the side of a square which is equivalent to the sum of two
triangles whose bases and altitudes are respectively 10 and 20
inches, and 8 and 20 inches.

64. Write a formula expressing the area of a parallelogram of which
two sides are a and b and the included angle is 60°.

65. Write a formula expressing the area of a parallelogram of which
two sides are a and 6 and the included angle is 45°.

66. The perimeter of a rectangle is 480 feet and the area is 8000
square feet. Find the dimensions of the rectangle.

290

PLANE GEOMETRY

67. Two homologous sides of two similar pentagons are 6 and 10
inches respectively. The area of the smaller pentagon is 720
square inches. What is the area of the larger pentagon?

68. If the areas of two similar quadrilaterals are 1620 and 288
square inches, respectively, what is the ratio of any two homolo-
gous sides of the quadrilaterals?

69. A line is drawn parallel to a side of a triangle and divides the
other two sides into segments whose ratio is 2 to 3. What is
the ratio of the areas into which the triangle is divided ? There
are two positions of the line.

70. Where must a line be drawn, parallel to one side of a triangle,
in order to bisect the area of the triangle?

APPLICATIONS

382. Construction Problems in Areas. — These construc-
tions are made by the rule and compass method, and are
chiefly interesting because of their association with the
early development of geometry. Almost all problems in
combining and dividing areas, equivalence of areas, etc.,
which occur in practice are better performed by geometrical
or trigonometrical calculation. Such constructions as are
here given may sometimes be valuable as checks on
calculation.

1. To construct a square equivalent to the sum of two given
squares.

A and B are the given squares. Prove that C =c= A + B.

2. To construct a square equivalent to the difference of two
given squares.

Show how this construction can be made.

3. To construct a square equivalent to the sum of any
number of given squares.

Show how the required square is obtained.

AREAS

291

4. To construct a triangle equivalent to a given triangle and
with a given base.

Explain the construction.

5. To construct a triangle equivalent to a given rectangle
and with a given base.

Show how this construction can be made.

6. To construct a square equivalent to a given rectangle.
Help. — Use the principle of 316, Exercise 12, to obtain

the side of the required square.

7. To construct a triangle equiv-
alent to a given polygon.

Show that AFCG =0= polygon
ABCDE.

8. To construct a square equiv-
alent to a given triangle.

Helps. — Let a and b be the altitude and base of the given
triangle, and s be the side of the required square.

F A

Then

ab , a s

~2 = 8 ' and ~8 = T"

Use the method of Problem 6 to complete the construction.

EXERCISES

1. Construct a square equivalent to the sum of two given triangles.

2. Construct a circle equivalent to the sum of two given circles.

3. Construct an equilateral triangle equivalent to a given triangle.

4. Bisect a triangle by a line parallel to a side.

5. Bisect a quadrilateral by a line drawn from a given vertex.

292

PLANE GEOMETRY

6. Construct an isosceles triangle of which the vertex angle equals
an angle of a given triangle, and which is equivalent to the given
triangle.

Helps.— (1) c2 = ab; (2) to find c, refer to 316, Exercise 12, or
Exercise 13.

383. Surveying. — When the shape of a piece of land is
rectangular, triangular, circular, etc., its area is found by
methods already given in this chapter. The shape of a
parcel of land is generally irregular, and special methods
have been devised for computing the areas of such plots
from linear and angular measurements.

1. Areas Bounded by Straight Lines, (a) Area by
Coordinates. — Take vertex A, as the origin of coordinates.
By measuring the sides and
interior angles of the field
ABCDE, it is possible to calcu-
late by trigonometry the coor-
dinates of all the vertices. The
lines B'B, C'C, D'D, E'E, and
AB', B'C', C'D', D'E', E'A, can
then be found. These lines are
the dimension lines of the part
areas, ABB', B'BCC', C'CDD',
D'DEE', E'EA, and from them the values of these areas
can be found. The polygon ABCDE is an algebriac sum
of the part areas.

EXERCISES

1. The coordinates (abscissa and ordinate) of the vertices of a
field ABCDE, are A, (0, 0); B, ( + 100, +80); C, ( + 150, +90);
D, ( + 175, +20); #,( + 120, -50). Plot the vertices on squared
paper and draw the field.

2. Find the values of the lines B'B, C'C, D'D and E'E.

3. Find the values of the lines AB', B'C', C'D', D'E' and E'A.

4. Find the areas ABB', B'BCC', C'CDD', D'DEE' and E'EA.

AREAS

293

5. Find the area of the field ABCDE from the algebraic sum of the
part areas.

Help.— Polygon A BCDE = - AABB' - trapezoid B'BCC' +
trapezoid C'CDD', etc.

(6) Area by Triangles. — By this method the field is
divided into triangles, (a) by diagonals, (6) by lines drawn
to every vertex from some interior point, (c) in any
convenient way.

EXERCISES

6. The measurements
ZBOC = 60°

of a field ABCDE
= 90°, /.DOE =

are Z.AOB = 60

60°, /.EOA = 90°; OA = 120 feet, OB
= 120 feet, OC = 80 feet, OD = 150
feet, OE = 1 00 feet. Find the area by
finding the sum of the areas of the tri-
angles. Trigonometry is required for
this calculation when the central angles
are other than 30°, 45°, 60°, 90°, and their supplements.
7. Find the area of a four-sided field A BCD, of which the measure-
ments are AB = 400 feet, BC = 1500 feet, CD = 1200 feet,
DA = 800 feet, and diagonal BD = 1000 feet. Find the areas
of triangles ABD and BCD by the formula of 367.

2. Areas Bounded by Irregu-
lar Curved Lines. — The methods
used give the area approxi-
mately, but with sufficient exact-
ness. They apply to plots of
either of these forms.

(a) The Ordinate Method —
Measurements are taken in the
field, AD, DF, FH, etc., and per-
pendiculars AC, DE, etc. The
field is then plotted accurately to
scale, and the line AB is divided
into equal parts and the ordinates
Oi, 02, etc., are drawn at the mid-
dle points of the equal segments.
The area equals the average ordi-
nate multiplied by the uniform segment x.

FIG. 1.

FIG. 2.

294 PLANE GEOMETRY

EXERCISES

8. The measurements of a field (Fig. 1) are AD = DF = FH =
etc. = 40 feet; ordinates ylt y2, etc., are 80, 90, 120, 140, 140,
110, 75, 70, 30 feet. Plot the field on squared paper.

9. Find the area of the field of Exercise 8 by the ordinate method.

(6) The Trapezoid Method. — Each part area, AC ED,
DEGF, etc., is approximately a trapezoid.

EXERCISES

10. Show that the area of a field of either form, when AD, DF, etc.
are all equal to x, is given by the formula,

S = \ h/l+2/n +2(7/2 +2/3+ .... •</«-!)]•

11. Find the area of the field of Exercise 8 by the trapezoid method.
(c) Simpson's Rule. — This is considered the most

accurate method. The formula for the area is,

S = [y\ + yn + 4(i/2 + 2/4 + etc.) + 2(y3 + yb + etc.)]

= [A + 4B + 2C]-

EXERCISES

12. For what do the letters A, B, C, stand in the final formula?

13. Calculate the area of the field of Exercise 8 by Simpson's Rule.

3. 'Special Methods of Finding Areas, (a) Area by
Squared Paper. — The field is plotted and the area measure
is counted.

EXERCISE

14. Find the area of the field of Exercise 8 by this method.

(b) Area by Planimeter. — An area may be measured
automatically by this instrument, by tracing the bounding

line with the point A while the point B is fixed. The re-
volving counter C records the area. Explanations of the

AREAS 295

planimeter may be found in texts on surveying and
engineering.

(c) Area by Approximate Rectangle, Triangle, Circle, etc. —
The area is plotted accurately, and straight lines are drawn
which approximate as closely
as possible to the bounding
curve, while the form of the
rectilinear figure is made a
rectangle, triangle. or trape-
zoid; or in some cases an approximate circle or semi-circle
may be used. The dimensions of this approximate figure
are scaled from the drawing.

EXERCISE

15. Find the area of the field of Exercise 8 by drawing an approxi-
mate rectangle.

384. Dividing Land. — Problems of dividing land occur
constantly in surveying. They are solved by methods of cal-
culation, and may be checked by geometrical constructions.

EXERCISES

1. Divide a rectangular field 120 feet by 240 feet into three equiva-
lent parts by lines parallel to the shorter side.

2. Divide a rectangular field 120 feet by 240
feet, into three equivalent parts by lines
drawn from a vertex of the field. Find
BF and DE. ^

3. The sides of a triangular field are AB =

150 feet, BC = 200 feet, CA = 210 feet. Divide it into two
equivalent parts by a line drawn from the vertex A.

4. Divide the field of Exercise 3 into two equivalent parts by a
line DE parallel to BC. Calculate AD and AE.

5. Divide the field of Exercises 3 and 4 into two equivalent parts
by a line drawn from a point F, which is 20 feet from B on the
side AB.

385. Estimating and Engineering.

EXERCISES

1. A room is 30 feet long, 20 feet wide, 10 feet high. Calculate
the areas of the four walls and ceiling. There are 4 windows

296

PLANE GEOMETRY

3| feet by 6 feet, and 2 doors 4 feet by 1\ feet. What area is
covered by plaster? Estimate its cost at 95 cents per square
yard.

2. A barn is 40 feet wide, 70 feet long, and 24 feet high to the
eaves; the rise of the roof is 12 feet.

Estimate the cost of painting the
barn two coats if 1 gallon of paint
will cover 400 square feet one coat,
and paint costs $2.75 per gallon.

3. Shingles average 6 by 24 inches. When laid on a roof, \ of the
shingle is exposed. Estimate the cost of shingling the roof of
the barn of Exercise 2 at $20.00 per 1000 shingles.

4. A rectangular piece of sheet metal 4 inches by 5 inches, weighs
\ a pound. What is the weight of a rectangular piece of the
same sheet metal 10 inches by 12 inches?

5. Find the number of square yards covered by a rectangular
building 120 feet by 55 feet. Each yard in depth of cellar ex-
cavation forms a cubic yard with a square yard of surface.
Estimate the number of cubic yards excavated in digging the
cellar 6 feet deep; and the cost at 75 cents per cubic yard.

6. Make a similar cost estimate for the excavation of a cellar 6 feet

2~<v

<\4

deep for the building of which the floor plan is here given.

7. The number of gallons of water flowing past a given line across
a river is found by measuring its "cross-section." Find the
cross-section of a river, the soundings being taken from a
bridge at intervals of 50 feet from shore to shore. Depths in
feet: 10, 20, 25, 30, 20, 18, 15, 10, 5. Total width 500 feet.
If the velocity of current is 200 feet per- minute, how many
cubic feet of water flow past the bridge per minute?

8. If 10 per cent of the water of Exercise 7 can be pumped from
the river to supply a city, is there sufficient water for a city of
1,000,000 inhabitants, allowing 200 gallons per person per day
of 24 hours?

AREAS

297

9. A playground is trapezoidal in shape, the parallel sides being
300 feet and 240 feet, and a third side perpendicular to them
250 feet. How many children can be accommodated if 200
square feet is the estimated allotment per child?

S',5

o, ',

0 ;

120 \

v/;;////////;//;///v////;;/#7.

10. A cement walk 4 ft. wide is laid along two sides of a corner lot.
Calculate the cost at 25 cents per square foot.

11. A cement walk 5 feet wide runs straight for 80 ft., then follows
a circular curve of an inner radius of 50 feet for a quarter circle,
then runs straight for 100 feet. Estimate the cost at 20 cents
per square foot.

12. Parquet floors of white oak are to be laid in rooms as follows:
two rooms 30 by 12 feet; three rooms 22 by 18 feet; one room
35 by 15 feet; one hall 6 by 24 feet. Estimate cost if flooring
2 inches wide costs $25 per 1000 linear feet, and an additional
charge is made of 10 cents per square foot for laying and
finishing.

13. A circular rod in a roof truss must sustain a safe tension of
20,000 pounds per square inch of cross-section. What must
be its diameter if the total tension in the rod is 60,000 pounds?
What if the tension is 12,000 pounds?

14. How many boiler tubes 3 inches in diameter are required to give
a total cross-section equal to that of a 3-foot diameter chimney?

15. Three pipes, 10 inches, 12 inches, 8 inches, diameter, respec-
tively, empty into a basin. What diameter must be given to
a single discharging pipe whose cross-sectional area equals the
combined areas of the three pipes?

16. What total pressure is exerted by the steam in the cylinder of a
steam engine against a piston 12 inches in diameter if the pres-
sure is 120 pounds per square inch? The horsepower of the

in which p is the

average steam pressure, I the length of stroke, a the area of the
piston, n the number of strokes per minute. Find the horse-
power of the engine considered if the length of stroke is 20
inches and number of strokes is 160 (number of revolutions 80)
per minute.

17. A transit telescope magnifies 7 diameters. How many times
is the area magnified?

engine is given by the formula h.p. = oo QQQ

298 PLANE GEOMETRY

18. Show that the formula for the area of an annulus may be ex-
pressed, A = TT-— —'t; where di and dz are the outer and inner

diameters and t is the thickness (width of ring). This is often
used as a shop formula.

19. The cross-section area of an I-beam is required in order to find

<-<* ^

the weight per yard of length. Show that the area of cross-
section is given by the formula A = de + 2c(a + 6).
20. Find the cross-section area of an I-beam if a = I", b = 1|",
c = li", d = *", e = 10".

PROBLEMS FOR FIELD WORK

386. (1) Find the area of a plot of ground bounded by
straight lines, by measuring all of the sides of the plot and
as many diagonals as are necessary to divide it into tri-
angles. The diagonals should be selected so as to form as
nearly as possible equilateral triangles. Use the formula
of 367.

2. Find the area of a piece of land entirely bounded by
a curved line, or by straight and curved lines. Use the
method of 383, Part 2(6).

INDEX OF PRINCIPAL FORMULAS

Arc

length of, 330, 331

radius of, 315, Exercise 42, 322

rise of, 322
Areas

of an annulus, 372, Theorem (14)

of a circle, 370

of a circumscribed polygon, 369

of an equilateral triangle, 372, Theorem (10)

of a figure bounded by straight and curved lines, 383

of a parallelogram, 364

of a polygon, 383

of a rectangle, 352, 363

of a regular hexagon, 372, Theorem (12)

of a regular polygon, 368

of a segment of a circle, 372, Theorem (17)

of a square in terms of the diagonal, 372, Theorem (7)

of a trapezoid, 366

of a triangle, 365, 367

of a triangle in terms of the sides and the radius of the circum-
scribed circle, 372, Theorem (15)

Chord, length of, 322
Circle

angle formed by a tangent and a chord, 297

angle formed by two chords, 298

angle formed by a secant and a tangent, 300

angle formed by two secants, 299, 302, Theorem (8)

angle formed by two tangents, 301

area of, 370

inscribed angle, 296, 302, Theorem (7)

length of an arc, 330, 331

length of a chord, 322

relation between a half chord and the segments of a diameter
perpendicular to the chord, 260, Theorem (20), 305

relation between the segments of a secant and an intersecting
tangent, 307

relation between the segments of two intersecting chords, 305

relation between the segments of two intersecting secants, 306

299

300 INDEX OF PRINCIPAL FORMULAS

Circle

relations between radius, diameter and circumference, 337

value of TT, 340
Curvature of a sphere, 317
Curves

compound curves, 282, Exercise 9

laying out a circular arc, 282

laying out a circular arc by using two transits, 282

laying out a circular arc by ordinates, 319

laying out a circular arc by offsets from the tangent, 322

7-curves, 282, Exercise 12

Horizon at sea, distance of, 317

Polygon

area of, 383

sum of the interior angles of, 187

sum of the exterior angles of, 188
Proportion, laws of, 240
Pythagorean

numbers, 261, Exercise 14

theorem, 250

Sect

division into mean and extreme ratio, 316
harmonic division, 260, Theorem (33)
projection of, 254, 265, Exercises 22, 23, 24

Square

diagonal in terms of a side, 260, Theorem (14), 265, Exercise 28

side in terms of the diagonal, 260, Theorem (17)

side of an octagon obtained from, 261, Exercises 61, 62

Trapezoid

area of, 366

relation between lengths of median and bases, 184
Triangle

altitude, length of, 255

altitude of an equilateral triangle, 260, Theorem (16), 265,
Exercise 31

area of, 365, 367

bisector of an interior angle, length of, 309, 310

circumscribed circle, radius of, 372, Theorem (16)

circumscribed circle of an equilateral triangle, radius of, 315,
Exercise 24

inscribed circle, radius of, 371

INDEX OF PRINCIPAL FORMULAS 301

Triangle

median, length of, 256

medians, point of concurrence of, 185

segments into which a bisector of an interior angle divides the
opposite side, 245, 260, Theorem (26)

segments into which a bisector of an exterior angle divides the
opposite side, 246

side of an equilateral triangle expressed in terms of the altitude,
260, Theorem (18)

side opposite an acute angle, length of, 252, 265, Exercise 25

side opposite an obtuse angle, length of, 253

sides of a right triangle, relation between, 250

solution of, by trigonometry, 265
Trigonometry

distance across a pond, etc., 265, Exercise 17

distance across a river, etc., 265, Exercise 4

distance to an island, etc., 265, Exercise 18

height of an aeroplane, etc., 265, Exercise 7

height of a pole, etc., 265, Exercise 5

height of a stack, etc., 265, Exercise 20

rise of a hill, 265, Exercise 6

solution of an oblique triangle when two sides and the included

angle are given, 265, Exercises 25, 27
Trisection of an angle, 316

1C 49582

541281

UNIVERSITY OF CALIFORNIA LIBRARY

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