• i
M
SOLID GEOMETRY
BY
FLETCHER DURELL, PH.D.
HEAD OF THE MATHEMATICAL DEPARTMENT, THE LAWRENCEVII LE SCHOOL
NEW YORK
CHARLES E. MERRILL CO.
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Copyright, 1904, by Charles K Merrill Co.
PREFACE
ONE of the main purposes in writing this book has
been to try to present the subject of Geometry so that the
pupil shall understand it not merely as a series of correct
deductions, but shall realize the value and meaning of its
principles as well. This aspect of the subject has been
directly presented in some places, and it is hoped that it per-
vades and shapes the presentation in all places.
Again, teachers of Geometry generally agree that the
most difficult part of their work lies in developing in
pupils the power to work original exercises. The second
main purpose of the book is to aid in the solution of this
difficulty by arranging original exercises in groups, each
of the earlier groups to be worked by a distinct method.
The pupil is to be kept working at each of these groups
till he masters the method involved in it. Later, groups
of mixed exercises to be worked by various methods are
given.
In the current exercises at the bottom of the page,
only such exercises are used as can readily be solved in
connection with the daily work. All difficult originals are
included in the groups of exercises as indicated above.
Similarly, in the writer's opinion, many of the nuineri-
(iii)
797957
iV PREFACE
cal applications of geometry call for special methods of
solution, and the thorough treatment of such exercises
should be taken up separately and systematically. [See
pp. 304-318, etc.] In the daily extempore work only such
numerical problems are included as are needed to make
clear and definite the meaning and value of the geometric
principles considered.
Every attempt has been made to create and cultivate
the heuristic attitude on the part of the pupil. This has
been done by the method of initiating the pupil into
original work described above, by queries in the course
of proofs, and also at the bottom of different pages, and
also by occasional queries in the course of the text where
definitions and discussions are presented. In the writer's
opinion, the time has not yet come for the purely heuristic
study of Geometry in most schools, but it is all-important to
use every means to arouse in the pupil the attitude and
energy of original investigation in the study of the subject.
In other respects, the aim has been to depart as little
as possible from the methods most generally used at
present in teaching geometry.
The Practical Applications (Groups 88-91) have been
drawn from many sources, but the author wishes to ex-
press his especial indebtedness to the Committee which has
collected the Real Applied Problems published from time
to time in School Science and Mathematics, and of which
Professor J. F. Millis of the Francis W. Parker School
of Chicago is the chairman. Page 360 is due almost en-
tirely to Professor William Betz of the East High School
of Rochester, K Y,
FLETCHER DURELL.
LAWRENCEVILLE, N. J., Sept. 1, 1904.
TO THE TEACHER
1. IN working original exercises, one of the chief dif-
ficulties o£ pupils lies in their inability to construct the
figure required and to make the particular enunciation,
from it. Many pupils, who are quite unable to do this
preliminary work, after it is done can readily discover a
proof or a solution. In many exercises in this book the
figure is drawn and the particular enunciation made. It
is left to the discretion of the teacher to determine for
what other exercises it is best to do this for pupils.
2. It is frequently important to give partial aid to tho
pupil by eliciting the outline of a proof by questions such
as the following: "On this figure (or, in these two tri-
angles) what angles are equal, and why?" "What lines
are equal, and why?" etc.
3. In many cases it is also helpful to mark in colored
crayon pairs of equal lines, or of equal angles. Thus, in
the figure on p. 37 lines AB and DE may be drawn with
red crayon, AC and DF with blue, and the angles A and D
marked by small arcs drawn with green crayon. If
colored crayons are not at hand, the homologous equal
parts may be denoted by like symbols placed on them,
thus : c F
or thus: . . „
P B D P E A B D
In solving theorems concerning proportional lines, it is
occasionally helpful to denote the lines in a proportion
vi TO THE TEACHER
(either given or to be proved) by fig-
ures denoting the order in which the
lines are to be taken. Thus, if OA:
OC=OD:OB, the relation may be
indicated as on the figure.
4. It is sometimes helpful to vary
the symbolism of the book. Thus, in dealing with in-
equalities a convenient symbol for "angle" is i£t
as 2f. A > 3! B.
5. Each pupil need be required to work only so many
originals in each group as will give him a mastery of the
particular method involved. A large number of exercises
is given in order that the teacher may have many to select
from and may vary the work with successive classes.
6. It is important to insist that the solutions of exer-
cises for the first few weeks be carefully written out; later,
for many pupils, oral demonstration will be sufficient and
ground can be covered more rapidly by its use.
7. In leading pupils to appreciate the meaning of
theorems, it is helpful at times to point out that not every
theorem has for its object the demonstration of a new and
unexpected truth (i. e., not all are "synthetic"), but that
some theorems are analytic, it being their purpose to re-
duce an obvious truth to the certain few principles with
which we start in Geometry. Their function is, therefore,
to simplify and clarify the subject rather than to extend
its content.
REFERENCES TO PLANE GEOMETRY.
DEFINITIONS AND FIRST PRINCIPLES.
41. Parallel lines are lines in the same plane which do
not meet, however far they be produced.
GENERAL AXIOMS.
1. Things which are equal to the same thing, or to equal
things, are equal to each other.
2. If equals be added to equals, the sums are equal.
3. // equals be subtracted from equals, the remainders are
equal.
4. Doubles of equals are equal; or, in general, if equals
be multiplied by equals the products are equal.
5. Halves of equals are equal; or, in general, if equals be
divided by equals the quotients are equal.
6. The whole is equal to the sum of its parts.
7. The whole is greater than any of its parts.
8. A quantity may be substituted for its equal in any
process.
9. // equals be added to, or subtracted from, unequals, the
results are unequal in the same order; if unequals be added
to unequals in the same order, the results are unequal in that
order.
10. Doubles, or halves, of unequals are unequal in the
same order.
11. // unequals be subtracted from equals, the remainders
are unequal in the reverse order,
vii
Vlll KEFERENCES TO PLANE GEOMETRY.
12. //, of three quantities, the first is greater than the. second,
and the second is greater than the third, then the first is greater
than the third.
GEOMETRIC AXIOMS.
1. Through two given points only one straight line can be
passed.
2. A geometric figure may be freely moved in space with-
out any change in form or size.
3. Through a given point one straight line and only one
can be drawn parallel to another given straight line.
Geometric figures which coincide are equal.
71. At a given point in a straight line but one perpen-
dicular can be erected to the line.
75. The complements of two equal angles are equal; tlu
supplements of two equal angles are equal.
76. The sum of all the angles about a point equals four
right angles.
BOOK I.
78. // one straight line intersects another straight line,
the opposite or vertical angles are equal.
79. //, from a point in a perpendicular to a given line,
two oblique lines be drawn cutting off on the given line equal
segments from the foot of the perpendicular, the oblique lines
are equal and make equal angles with the perpendicular.
81. A triangle is a portion of a plane bounded by three
straight lines.
92. The sum of any two sides of a triangle is greater than
the third side.
94. The perpendicular is the shortest line that can be
drawn from a given point to a given line.
95. If, from a point within a triangle, two lines be drawn
to the extremities of one side of the triangle, the sum of the
REFERENCES TO BOOK I. IX
other two sides of the triangle is greater than the sum of the
two lines so drawn.
96. Two triangles are equal if two sides and the included
angle of one are equal, respectively, to two sides and the in-
cluded angle of the other.
98. Two right triangles are equal if the hypotenuse and an
acute angle of one are equal to the hypotenuse and. an acute
angle of the other.
99. In an isosceles triangle the angles opposite the equal
sides are equal.
100. // two angles of a triangle are equal, the sides oppo-
site are equal, and the triangle is isosceles.
101. Two triangles are equal if three sides of one are equal
to three sides of the other, respectively.
102. Two right triangles are equal if the hypotenuse and
a leg of one are equal to the hypotenuse and a leg of the other.
108. // two sides of a triangle are equal, respectively, to
two sides of another triangle, but the third side of the first is
greater than the third side of the second, then the angle opposite
the third side of the first triangle is greater than the angle oppo-
site the third side of the second.
109. Of lines drawn from the same point in a perpen-
dicular, and cutting off unequal segments from the foot of the
perpendicular, the more remote is the greater.
112. I. Every point in the perpendicular bisector of a line
is equally distant from the extremities of the line; and
II. Every point not in the perpendicular bisector is un-
equally distant from the extremities of the line.
113. Two points each equidistant from the extremities of
a line determine the perpendicular bisector of the line.
115. T1i-e perpendicular bisector of a line v'.s tJtc locus of
all points equidistant from the extremities of the line.
120. Parallel lines are lines in the same plane which
do not meet, however far they be produced.
X REFERENCES TO PLANE GEOMETRY.
121. Two straight lines in the same plane, perpendicular
to the same straight line, arc parallel.
122. Two straight lines parallel to a third straight line
are parallel to each other;
Lines parallel to parallel lines are parallel;
Lines perpendicular to parallel lines are parallel;
Lines perpendicular to non-parallel lines are not parallel.
123. If a straight line is perpendicular to one of two given
parallel lines it is perpendicular to the other also.
125. // two straight lines are cut by a transversal, mak-
ing the alternate interior angles equal, the two straight lines
are parallel.
127. // two straight lines are cut by a transversal, making
the exterior interior angles equal,, the two straight lines are
parallel.
130. Two angles whose sides are parallel, each to each, are
either equal or supplementary.
134. The sum of the angles of a triangle is equal to two
right angles.
142. Two right triangles are equal if a leg and an acute
angle of one are equal to a leg and the homologous acute angle
of the other.
147. A parallelogram is a quadrilateral whose opposite
sides are parallel.
149. A rhombus is a rhomboid whose sides are equal.
150. A rectangle is a parallelogram whose angles are
right angles.
151. A square is a rectangle whose sides are equal.
155. The opposite sides of a parallelogram are equal, and
its opposite angles are also equal.
156. A diagonal divides a parallelogram into two equal
triangles.
157. Parallel lines comprehended between parallel lines are
equal,
REFERENCES TO BOOK II. XI
180. // two sides of a quadrilateral are equal and parallel,
the other two sides are equal and parallel and the figure is a
parallelogram.
1 62. Two parallelograms are equal if two adjacent sides and
the included angle of one are equal, respectively, to two adjacent
sides and the included angle of the other.
163. Two rectangles which have equal bases and equal
altitudes are equal.
174. In an equiangular polygon of n sides each angle
, (n-2)2rf.Z8 2n-4
equals , « or rt. Z s.
n n
175. The sum of the' exterior angles of a polygon formed
by producing its sides in succession equals four right angles.
176. // three or more parallels intercept equal parts on one
transversal, they intercept equal parts on every transversal.
177. The line which joins the midpoints of two sides of a
triangle is parallel to the third side, and is equal to one-half
the third side.
178. The line which bisects one side of a triangle and is
parallel to another side bisects the third side.
179. The line which joins the midpoints of the legs of a
trapezoid is parallel to the bases and equal to one-half their
sum.
BOOK II.
197. A circle is a portion of a plane bounded by a curved
line, all points of which are equally distant from a point
within called the center.
198. A radius of a circle is a straight line drawn from
the center to any point on the circumference.
208. Radii of the same circle, or of equal circles, are
equal.
216. In the same circle, or in equal circles, equal arcs
subtend equal angles at the center.
Xli REFERENCES TO PLANE GEOMETRY.
217. In the same circle, or in equal circles, of two unequal
central angles the greater angle intercepts the greater arc, and,
conversely, of two unequal arcs, the greater arc subtends the
greater angle at the center.
218. In the same circle, or in equal circles, equal chords \
subtend equal arcs.
219. In the same circle, or in equal circles, equal arcs are
subtended by equal chords.
220. In the same circle, or in equal circles, the greater
of two (minor) arcs is subtended by the greater chord; and,
CONVERSELY, the greater of two chords subtends the greater
(minor) arc.
226. In the same circle, or in equal circles, equal chords
are equidistant from the center; and, CONVERSELY, chords
which are equidistant from the center are equal.
229. A straight line perpendicular to a radius at its ex-
tremity is tangent to the circle.
230. The radius drawn to the point of contact is perpen-
dicular to a tangent to a circle.
237. The two tangents drawn to a circle from a point out-
side the circle are equal, and make equal angles with a line
drawn from the point to the center.
241. // two circles intersect, their line of centers is per-
pendicular to their common chord at its middle point.
253. Properties of variables and limits.
1. The limit of the sum of a number of variables equals
the sum of the limits of these variables.
2. The limit of a times a variable equals a times the limit
of the variable, a being a constant.
3. The limit of —th part of variable is —th part of the limit
1 a a
of the variable, a being a constant.
254. // two variables are always equal, and each ap-
proaches a limit, their limits are enual.
REFERENCES TO BOOK III. Xlll
257. The number of degrees in a central angle equals the
number of degrees in the intercepted arc; that is, a central
angle is measured by its intercepted arc.
273. From a given point without a given line to draw a
perpendicular to the line.
274. At a given point in a given line erect a perpen-
dicular to that line.
279. Through a given point without a given straight line
to draw a line parallel to a given line.
286. To circumscribe a circle about a given triangle.
BOOK III.
303. The mean proportional between two quantities is
equal to the square root of their product.
305. // the antecedents of a proportion are equal, the con-
sequents are equal.
307. // Jour quantities are in proportion, they are in pro-
portion by alternation ; that is, the first term is to the third as
the second is to the fourth.
310. // four quantities are in proportion, they arc in pro-
portion by division ; that is, the difference of the first two is to
the second as the difference of the last two is to the last.
312. In a series of equal ratios, the sum of all the ante-
cedents is to the sum of all the consequents as any one ante-
cedent is to its consequent.
314. Like powers, or like roots, of the terms of a propor-
tion are in proportion.
317. A line parallel to one side of a triangle and meeting
the other two sides, divides these sides proportionally.
318. If a line parallel to the base cut the sides of a triangk,
a side is to a segment of that side as the other side is to the cor-
responding segment of the second side.
321. Similar polygons are polygons having their ho-
XIV REFERENCES TO PLANE GEOMETRY.
mologous angles equal and their homologous sides propor-
tional.
323. // two triangles are mutually equiangular, they are
similar.
326. // two triangles have their homologous sides propor-
tional they are similar.
327. // two triangles have an angle of one equal to an angle
of the other, and the including sides proportional, the triangles
are similar.
328. // two triangles have their sides parallel, or perpen-
dicular, each to each, the triangles are similar.
329. // two polygons are similar, they may be separated
into the same number of triangles, similar, each to each, and
similarly placed.
342. In a right triangle,
I. The altitude to fhe hypotenuse is a mean proportional
between the segments of the hypotenuse;
II. Each leg is a mean proportional between the hypote-
nuse and the segment of the hypotenuse adjacent to the given
leg.
343. The perpendicular to the diameter from any point
in the circumference of a circle is a mean proportional between
the segments of the diameter; and the chord joining the point
to an extremity of the diameter is a mean proportional be-
tween the diameter and the segment of the diameter adjacent to
the chord.
346. In a right triangle, the square of the hypotenuse is
equal to the sum of the squares of the legs.
347. In a right triangle, the square of either leg is equal
to the square of the hypotenuse minus the square of the other
leg.
351. // the square on the side of a triangle equals the sum
of the squares on the other two sides, the angle, opposite the first
side is a right angle.
REFERENCES TO BOOK IV. XV
352. //, in any triangle, a median be drawn to one side,
I. The sum of the squares of the other two sides is equal
to twice the square of half the given side, increased by twice
the square of the median upon that side; and
II. The difference of the squares of the other two sides is
equal to twice the product of the given side by the projection
of the median upon that side.
BOOK IV.
383. The area of a rectangle is equal to the product of its
base by its altitude.
385. The area of a parallelogram is equal to the product
of its base by its altitude.
389. The area of a triangle is equal to one-half the prod-
uct of its base by its altitude.
390. Triangles which have equal bases and equal alti-
tudes (or which have equal bases and their vertices in a line
parallel to the base) are equivalent.
391. Triangles which have equal bases are to each other as
their altitudes;
Triangles which have equal altitudes are to each other as
their bases.
392. Any two triangles are to each other as the products
of their bases and altitudes.
397. // two triangles have an angle of one equal to an
angle of the other, their areas are to each other as the products
of the sides including the equal angles.
398. The areas of any two similar triangles are to each
other as the squares of any two homologous sides.
399. The areas of two similar polygons are to each other
as the squares of any two homologous sides.
XVI REFERENCES TO PLANE GEOMETRY.
BOOK V.
444. Formula for the circumference in terms of the
radius.
449. The area of a circle is equal to one-half the product
of its circumference by its radius.
486. Two points symmetrical with respect to a line or
axis are points such that the straight line joining them is
bisected by the given line at right angles.
487. A figure symmetrical with respect to an axis is a
figure such that each point in the one part of the figure has
a point in the other part symmetrical to the given point,
with respect to an axis.
489. Two points symmetrical with respect to a point or
center are points such that the straight line joining them is
bisected by the point or center.
490. A figure symmetrical with respect to a point or
center is a figure such that each point in the figure has an-
other point in the figure symmetrical to the given point with
respect to the center.
TABLE OF CONTENTS
PAGE
BOOK VI. LINES, PLANES AND ANGLES IN SPACE 319
BOOK VII. POLYHEDRONS 360
BOOK VIII. CYLINDERS AND CONES . . . 403
BOOK IX. THE SPHERE 425
NUMERICAL EXERCISES IN SOLID GEOMETRY 4(59
APPENDIX I. MODERN GEOMETRIC CONCEPTS 485
II. HISTORY OF GEOMETRY 490
III. REVIEW EXERCISES. PLANE GEOMETRY .... 491
SOLID GEOMETRY 506
PRACTICAL APPLICATIONS OF SOLID GEOMETRY 510
FORMULAS . 516
(rvii)
SYMBOLS AND ABBREVIATIONS
H- plus, or increased by. Adj. . . adjacent.
— minus, or diminished by. Alt. . . alternate.
X multiplied by. Art. . . article.
-s- divided by. Ax. ... axiom.
= equals; is (or are] equal to. Constr. . construction0
— approaches (as a limit). Cor. . . corollary.
=c= is (or are) equivalent to. Def. . . definition.
> is (or are) greater than. Ex. . . . exercise.
< is (or arc) Zess Ma?i. Ext. . . exterior.
.'. therefore. Fig. . . figure.
J_ perpendicular, perpendicular to, Hyp. . . hypothesis.
or, is perpendicular to. Ident. . identity.
.k perpendiculars. Int. . . interior.
|| parallel, or, is parallel to. Post. . . postulate.
\\sparallels. Prop. . .proposition.
/. , A angle, angles Rt. . . . right.
A, A triangle, triangles. Sug. . . suggestion.
/ 7 , £Z7 parallelogram, parallelograms. Sup. . . supplementary.
G, © cirete, circles. St. ... straight.
Q. E. D. quod erat demonstrandum; that is, which was to be
proved.
Q. E. F. quod erat faciendum; that is, which was to be made.
A few other abbreviations and symbols will be introduced and
their meaning indicated later on.
SOLID GEOMETRY
BOOK VI
LINES, PLANES AND ANGLES IN SPACE
DEFINITIONS AND FIRST PRINCIPLES
497. Solid Geometry treats of the properties of space of
three dimensions.
Many of the properties of space of three dimensions are determined
by use of the plane and of the properties of plane figures already
obtained in Plane Geometry.
498. A plane is a surface such that, if any two points
in it be joined by a straight line, the line lies wholly in the
surface.
499. A plane is determined by given points or lines, if
no other plane can pass through the given points or lines
without coinciding with the given plane.
500. Fundamental property of a plane in space. A
plane is determined by any three points not in a straight line.
For, if through a line con- .c
necting two given points, A
and B, a plane be passed, the
plane, if rotated, can pass
through a third given point,
(7, in but one position.
The importance of the above principle is seen from the fact that it
reduces an unlimited surface to three points, thus making a vast
economy to the attention. It also enables us to connect different
planes, and treat of their properties systematically.
(319)
320 BOOK VI. SOLID GEOMETRY
501. Other modes of determining a plane. A plane may
also be determined by any equivalent of three points not in
a straight line, as by
a straight line and a point outside the line; or by
two intersecting straight lines; or by
two parallel straight lines.
It is often more convenient to use one of these latter methods of
determining a plane than to reduce the data to three points and use
Art. 500.
\
502. Representation of a plane in geometric figures. la
masoning concerning the plane, it is often an advantage
to have the plane represented in all directions. Hence, in
drawing a geometric figure, a plane is usually represented
to the eye by a small parallelogram.
This is virtually a double use of two intersecting lines, or of two
parallel lines, to determine a plane (Art. 501).
503. Postulate of Solid Geometry. The principle of
Art. 499 may also be stated as a postulate, .thus:
Throtigh any three points not in a straight line (or their
equivalent) a plane may be passed.
504. The foot of a line is the point in which the line
intersects a given plane.
505. A straight line perpendicular to a plane is a line
perpendicular to every line in the plane drawn through its
foot.
A straight line perpendicular to a plane is sometimes called a normal
to the plane.
LINES AND PLANES 321
506. A parallel straight line and plane are aline and
plane which cannot meet, however far they be produced.
507. Rarallel planes are planes which cannot meet,
however far they be produced.
508. Properties of planes inferred immediately.
1. A straight line, not in a given plane, can intersect the
given plane in but one point.
For, if the line intersect the given plane in two or more
points, by definition of a plane, the line must lie in the
plane. Art. 498.
2. The intersection of two planes is a straight line.
For, if two points common to the two planes be joined
by a straight line, this line lies in each plane (Art. 498) ;
and no other point can be common to the two planes, for,
through a straight line and a point outside of 4t only one
plane can be passed. Art. 501.
Ex. 1. Give an example of a plane surface; of a curved surface;
of a surface, part plane and part curved; of a surface composed of
different plane surfaces.
Ex. 2. Four points, not all in the same plane determine how
many different planes ? how many different straight lines ?
Ex. 3. Three parallel straight lines, not in the same plane, deter-
mine how many different planes ?
Ex. 4. Four parallel straight lines can determine how many differ-
ent planes ?
Ex. 5. Two intersecting straight lines and a point, not in their
plane, determine how many different planes f
322 BOOK VI. SOLID GEOMETRY
•
PROPOSITION I. THEOREM
509. If a straight line is perpendicular to each of two
other straight lines at their point of intersection, it is per-
pendicular to the plane of those lines.
Given AB _L lines EC and ED, and the plane MN pass-
ing through EG and ED.
To prove AB _L plane MN.
Proof. Through E draw EG, any other line in the plane
MN.
Draw any convenient line CD intersecting EC, EG and
ED in the points C, G and D, respectively.
Produce the line AB to F, making BF=AB.
Connect the points C, G, D with A, and also with F.
Then, in the A ACD and FCD, CD=CD. Ident.
A C= CF, and AD=DF. Art. 112.
/. &ACD=AFCD. (Why?)
/. Z ACD=Z. FCD. (Why?)
Then, in the A ACG and FCG, CG=CG, (Why?)
AC= CF, and Z ACG = Z JW#. (Why ?)
.'. A J.C#=A ^(7O. (Why?)
LINES AND PLANES 323
/. AG=GF. (Why?)
.*. B and O are each equidistant from the points A and F.
:. BO is J_ AF; that is, AB J_ BO. Art. 113.
/. AB _L plane MN, Art. 505.
(for it is J_ any line, EG, in the plane MN, through its foot}.
Q. E. D.
PROPOSITION II. THEOREM
510. All the perpendiculars that can be drawn to a
given line at a given point in the line lie in a plane perpen-
dicular to the line at the given point.
Given the plane MN and the line #<7both J_ line AB at
the point B.
To prove that BC lies in the plane MN.
Proof. Pass a plane AF through the intersecting lines
AB and BC. Art. 503.
This plane will intersect the plane MN in a straight
line BF. Art. 508, 2.
But AB J_ plane MN (Hyp.) .-. AB J_ BF. Art. 505.
Also AB _L BC. Hyp.
.*. in the plane AF, BC and BF _L AB at B.
:. BC and BF coincide. Art. 71.
But BF is in the plane MN.
:. BC must be in the plane MN,
(for BC coincides with BF) which lies in the plane MN}.
Q. £• D*
324 BOOK VI. SOLID GEOMETKY
511. COR. 1. At a given point B in the straight line AB,
to construct a plane perpendicular to the line AB. Pass a
plane AF through AB in any convenient direction, and in
the plane AF at the point B construct BF J_ AB (Art.
274). Pass another plane through AB, and in it construct
BP ± AB. Through the lines
BF and BP pass the plane
MN (Art. 503). MN is the
required plane (Art. 509).
512. COR. 2. Through a
given external point, P, to pass
a plane perpendicular to a
given line, AB. Pass a plane through AB and P (Art.
503), and in this plane draw PB _L AB (Art. 273). Pass
another plane through AB, as AF, and in AF draw BF JL
AB at B (Art. 274). Pass a plane through BP and BF
(Art. 503). This will be the plane required (Art. 509).
513. COR. 3. Through a given point but one plane can
be passed perpendicular to a given line.
Ex. 1. Five points, no four of which are in the same plane, deter-
mine how many different planes ? how many different straight lines ?
Ex. 2. A straight line and two points, not all of which are in the
same plane, determine how many different planes ?
Ex. 3. In the figure, p. 322, prove that the triangles GAD and
GDF are equal.
Ex. 4. In the same figure, if AB=S and BC=Q, find FC.
LINES AND PLANES 325
PROPOSITION III. PROBLEM •
514. At a given point in a plane, to erect a perpendicu-
lar to the plane.
p —
Given the point A in plane MN.
To construct a line perpendicular to MN at the point A.
Construction. Through the point A draw any line CD in
the plane MN.
Also through the point A pass the plane PQ JL CD
(Art. 511), intersecting the plane MN in the line RS. Art. 508, 2.
In the plane PQ draw AK J_ line RS at A. Art. 274.
Then AK is the J_ required.
Proof. CD _L plane PQ. Constr.
/. CD J_ AK. Art. 505.
Hence AK J_ CD.
But AK _L #£. Constr.
/. AK JL plane MF. Art. 509.
Q. E. F.
515. COR. At a given point in a plane but one perpen-
dicular to the plane can be drawn. For, if two Ja. could be
drawn at the given point, a plane could be passed through
them intersecting the given plane. Then the two Ja would
be in the new plane and J_ to the same line (the line of
intersection of the two planes, Art, 505), which is im-
possible (Art, 71),
326 BOOK VI. SOLID GEOMETBY
PROPOSITION IV. PROBLEM
516. From a given point without a plane, to draw a line
perpendicular to the plane.
A
^
,B
Given the plane MN and the point A external to it.
To construct from A a line JL plane MN.
Construction. In the plane MN draw any convenient
line EC. Pass a plane through EG and A (Art. 503), and
in this plane draw AD _L EC. Art. 273.
In the plane MN draw LD J_ EC. Art. 274.
Pass a plane through AD and LD (Art. 503), and in
that plane draw AL J_ LD. Art. 273.
Then AL is the _L required.
Proof. Take any point C in EC except D, and draw LC
and AC.
Then A J.DC, AD£ and LDC are right A. Constr.
. Art. 400.
+ ~D&. Art. 400, Ax. 8.
.'. AC2 = AL2 + LC2. Art. 400, Ax. 8.
/. /.ALC is a right Z . Art. 351.
But AL JL LD. Constr.
/. AL X JOT. Art. 509.
Q. E. F.
517r COR. But one perpendicular can be drawn from a
given external point to a given plane.
LINES AND PLANES
327
PROPOSITION V. THEOREM „
518. I. Oblique lines drawn from a point to a plane,
meeting the plane at equal distances from the foot of the
perpendicular, are equal;
II. Of two oblique lines drawn from a point to a plane,
but meeting the plane at unequal distances from the foot of
the perpendicular, the more remote is the greater.
Given AB i. plane MN, BD = BC, and BH > EG.
To prove AD = AC, and AH > AC.
Proof. I. In the right A ABD and ABC,
AB = AB, and BD=BC.
/. AABD=AABC.
(Why?)
(Why?)
(Why?)
II. On .RET take BF=BG, and draw AF.
Then A F= AC (by part of theorem just proved)
But AH > AF.
:. AH > AC.
(Why ?)
Ax. 8.
Q. E. D.
519. COR. 1. CONVERSELY: Equal oblique lines drawn
from a point to a plane meet the plane at equal distances
from the foot of the perpendicular drawn from the same point
to the plane; and, of two unequal lines so drawn, the greater
line meets the plane at the greater distance from the foot of
the perpendicular.
328 BOOK VI. SOLID GEOMETRY
520. COR. 2. The locus of a point in space equidistant
from all the points in the circumference of a circle is a
straight line passing through the center of the circle and
perpendicular to its plane.
521. COB. 3. The perpendicular is the shortest line that
can l)e drawn from a given point to a given plane.
522. DEF. The distance from a point to a plane is the
perpendicular drawn from the point to the plane.
PROPOSITION VI. THEOREM
523. If from the foot of a perpendicular to a plane a
line l)e drawn at right angles to any line in the plane, the line
drawn from the point of intersection so formed to any point in
the perpendicular, is perpendicular to the line of the plane.
\A-7P
Given AB J_ plane MN, and EF _L CD, any line in MN.
To prove AF _L CD.
Proof. On CD take FP and FQ equal segments.
Draw AP, BP, AQ, BQ.
Then BP=BQ. Art. 112.
Hence AP=AQ. Art. 518.
/. in the line AF, the point A is equidistant from P
and Q, and F is equidistant from P and Q.
:. AF i. CD. (Why?)
Q. E. D.
Ex. In the above figure, if AS=Q, AF=8, and ^£=10, find QF,
BF and £.
LINES AND PLANES 329
PROPOSITION VII. THEOREM
524. Two straight lines perpendicular to the same plane
are parallel.
A C
\
•F
D
Given the lines AB and CD J. plane MN.
To prove AB \\ CD.
Proof. Draw BD, and through D, in the plane
draw FH J_ BD.
Draw AD.
Then BD _L FIT. Constr.
AD J_ FH. Art. 523.
CD J_ J^ff. Art. 505.
/. BD, AD and CD are all J. FH at the point D.
.'. BD, AD and CD all lie in the same plane. Art. 510.
.'. AB and CD are in the same plan^. (Why?)
But AB and CD are J_ #Z). Art. 505.
/. AB and CD are ||. Art. 121.
Q. £. D.
525. COR. 1. If one of two parallel lines is perpendicu-
lar to a plane, the other is perpendicular to the plane also.
For, if AB and CD be ||, and AB _L
plane PQ, a line drawn from C J_ PQ
must be || AB. Art. 524.
But CD must coincide with the line
so drawn (Art. 47, 3); .-. CD X PQ.
330 BOOK VI. SOLID GEOMETKY
526. COE. 2. If two straight lines are each parallel to
a third straight line, they are parallel to each other. For,
if a plane be drawn _L to the third line, each of the two
other lines must be _L to it (Art. 525), and therefore be ||
to each other (Art. 524).
PROPOSITION VIII. THEOREM
527. If a straight line external to a given plane is paral-
lel to a line in the plane, then the first line is parallel to the
given plane.
Given the straight line AB \\ line CD in the plane MN.
To prove AB \\ plane MN.
Proof. Pass a plane through the || lines AB and CD.
If AB meets MN it must meet it in the line CD.
But AB and CD cannot meet, for they are ||. Art. 120.
.'. AB and MN cannot meet and are parallel. Art. 506.
Q. E. D.
528. COR. 1. If a straight line is parallel to a plane ,
the intersection of the plane with any plane passing through
the given line is parallel to the given line.
LINES AND PLANES
331
529. COR. 2. Through a given line
(CD) to pass a plane parallel to another
given line (AB) .
Through P, any point in CD, draw
QR || AB (Art. 279) . Through CD and
QR pass a plane (Art. 503). This will be the plane
required (Art. 527).
If AB and CD are not parallel, but one plane can be
drawn through CD\\AB.
PROPOSITION IX. THEOREM
530. Two planes perpendicular to the same straight line
are parallel.
M
Given the planes MN and PQ _L line AB.
To prove MN\\ PQ.
Proof. If MN and PQ are not parallel, on being pro-
duced they will meet.
We shall then have two planes drawn from a point per-
pendicular to a given line, which is impossible. Art. 513.
.*. MN and PQ are parallel.
Art. 507.
Q. £. D.
332 BOOK VI. SOLID GEOMETRY
• PROPOSITION X. THEOREM
531. // two parallel planes are cut by a third plane, the
intersections are parallel lines.
Given MN and PQ two 1 1 planes intersected by the plan*
RS in the lines AB and CD.
To prove AB \\ CD.
Proof. AB and CD lie in the same plane R8.
Also AB and CD cannot meet; for if they did meet
the planes MN and PQ would meet, which is impossible.
Art. 507.
.*. AB and CD are parallel. Art. 41.
Q. E. D.
532. COR. 1. Parallel lines included between parallel
planes are equal. For, if AC and BD are two parallel lines,
a plane may be passed through them (Art. 503), intersect-
ing MN and PQ in the || lines AB and CD. Art. 531.
/. ABDC is a parallelogram. Art. 147.
.*. AC=BD. Art. 155.
LINES AND PLANES
333
533. COR. 2. Two parallel planes are everywhere equi-
distant.
For lines _L to one of them are || (Art. 524). Hence
the segments of these lines included between the || planes
are equal (Art. 532).
PROPOSITION XI. * THEOREM
534. If two intersecting lines are each parallel to a given
plane, the plane of these lines is parallel to the given plane.
D
K
Given the lines AB and CD, each || plane PQ, and inter-
secting in the point F\ and MN a plane through AB and
CD.
To prove MN\\PQ.
Proof. From the point F draw FH _L PQ.
Pass a plane through FC and FH, intersecting PQ in
HK; also pass a plane through FB and Fit, intersecting
PQ in HL.
Then HK \\ FC, and HL \\ FB. Art. 528.
But FH _L HK and HL. Art. 505.
/. FH J_ FC and FB. Art. 123.
/. FH _L M^. Art. 509.
.-. MN II P#. Art. 530.
Q. B. ».
334
»
BOOK VI. SOLID GEOMETRY
PROPOSITION XII. THEOREM
535. A straight line perpendicular to one of two parallel
planes is perpendicular to the other also.
\ T"
Given the plane MN \\ plane PQ, and AB JL PQ.
To prove AB J_ MN.
Proof. Through AB pass a plane intersecting PQ and
MN in the lines BO and AF, respectively; also through
AB pass another plane intersecting PQ and MN in BJ) and
AH, respectively.
Then BO 1 1 AF, and J5D || AJ3". Art. 531.
But AB _L BG and J9D. Art. 505.
.*. AB J_ AJP and AH. Art. 123.
.*. A-B _L plane M2V. Art. 509.
Q. £. D.
536. COR. 1. Through a given point to pass a plane
parallel to a given plane.
Let the pupil supply the construction.
537. COR. 2. Through a given point but one plane can
be passed parallel to a given plane,
LINES AND PLANES 335
PROPOSITION XIII. THEOREM
538. If two angles not in the same plane have their
corresponding sides parallel and extending in the same direc-
tion, the angles are equal and their planes are parallel.
\
\
Given the /.BAG in the plane MN, and the £B'A'C>
in the plane PQ; AB and A'B' \\ and extending in the
same direction; and AC and AC' \\ and extending in the
same direction.
To prove £BAC = /.B'A'C', and plane MN \\ plane PQ.
Proof. Take AB = ABr, and AC=ACf.
Draw AA, BB't CC', BC, B'C'.
Then ABB' A is a ZZ7 , Art. 160.
(for AB and AfB' are=and \\ ).
/. BB' and A A' are = and IK Art. 155.
In like manner CC' and AA' are = and ||.
/. BB' and CC' are = and ||. (Why?)
.'. BCC'B' is a C3 , and BC=B'Cf. (Why ?)
/. A ABC= A A'B'C'. (Why ?)
/. £A = /.A. (Why?)
Also AB || A'B', :. AB \\ plane PQ. Art. 527.
Similarly AC \\ plane P^.
:. plane JOT || plane PQ. Art. 534.
o. E. B.
336 BOOK VI. SOLID GEOMETRY
A PROPOSITION XIV. THEOREM
539. // two straight lines are intersected by three paral-
lel planes^ the corresponding segments of these lines are
proportional.
Mr
wrp.
Given the straight lines AB and CD intersected by the
|| planes MN, P^/and ES in the points A, F, B, and C
H, D, respectively.
AF GH
To prove
Proof. Draw the line AD intersecting the plane PQ in G.
Draw FG, BD, GH, AC.
Then FG \\ BD, and GH \\ AC. Art. 531.
Art 317
And
AF CH
(Why?)
" FB HD
Q. E. D.
Ex. 1. In above figure, if AF=2, FB=5, and CH=3, find CD.
Ex 2. If CH=3, HD=±, and AB=IQ, find AF and BF.
DIHEDKAL ANGLES 337
DIHEDRAL ANGLES
540. A dihedral angle is the opening between two in-
tersecting planes.
From certain points of view, a dihedral angle may be regarded as
a wedge or slice of space cut out by the planes forming the dihedral
angle.
541. The faces of a dihedral angle are the planes form-
ing the dihedral angle.
The edge of a dihedral angle is the straight line in which
the faces intersect.
542. Naming dihedral an-
gles. A dihedral angle may be
named, or denoted, by naming
its edge, as the dihedral angle
AB; or by naming four points,
two on the edge and one on
each face, those on the edge
coming between the points on
the faces, as P-AB-Q. The
latter method is necessary in naming two or more dihedral
angles which have a common edge.
543. Equal dihedral angles are dihedral angles which
can be made to coincide.
544. Adjacent dihedral angles are dihedral angles hav-
ing the same edge and a face between them in common.
545. Vertical dihedral angles are two dihedral angles
having the same edge, and the faces of one the prolonga-
tions of the faces in the other.
546. A right dihedral angle is one of two equal adja-
cent dihedral angles formed by two planes.
338
BOOK VI. SOLID GEOMETRY
547. A plane per-
pendicular to a given
plane is a plane form-
ing a right dihedral
angle with the given
plane.
Many of the properties of dihedral angles are obtained
most conveniently by using a plane angle to represent the
dihedral angle.
548. The plane angle of a dihedral an-
gle is the angle formed by two lines drawn
one in each face, perpendicular to the edge
at the same point.
Thus, in the dihedral angle C-AB-F,
if PQ is a line in the face AD perpendicu-
lar to the edge AB at P, and PR is a line
in face AF perpendicular to the edge AB
at P, the angle QPR is the plane angle of the dihedral
angle C-AB-F.
549. Property of plane angles of a dihedral angle.
The magnitude of the plane angle of a dihedral angle is
the same at every point of the edge. For let EAC be the
plane Z of the dihedral Z E-AB-D at the point A.
Then PR \\ AE, and PQ \\ AC (Art. 121.)
= /.EAC (Art. 538).
D
550. The projection of a point upon a plane is the foot of
a perpendicular drawn from the point to the plane.
ritt
551. The projection of a line
upon a plane is the locus of the pro-
jections of all the points of the line
on the plane. Thus A'B' is the
projection of AB on the plane MN.
DIHEDKAL ANGLES
PROPOSITION XV. THEOREM
339
552. Two dihedral angles are equal if their plane angle*
are equal.
Given /.DBF the plane Z of the dihedral £ C-AB-F,
£D'B'F' the plane Z of the dihedral /.C'-A'B'-F', and
/.DBF = £DfBfFf.
To prove Z C-AB-F = Z Cf-A'B'-F'.
Proof. Apply the dihedral Z Cf-A'B'-F' to Z C-AB-F
so that /.D'B'F' coincides with its equal, /.DBF.
Greom. Ax. 2.
Then line A'Bf must coincide with AB, Art. 515.
(/or A'B' and AB are loth ± plane DBF at the point B).
Hence the plane A'B'Df will coincide with plane ABD,
Art. 501.
(through two intersecting lines only one plane can be passed).
Also the plane AfB'Ff will coincide with the plane ABF,
(same reason).
.'. Z G'-A'B'-F' coincides with Z C-AB-F and is equal
to it. Art. 47.
Q. E. D.
553. COR. The vertical dihedral angles formed by two
intersecting planes are equal.
In like manner, many other properties of plane angles
are true of dihedral angles.
340 ' BOOK VI. SOLID GEOMETRY
V PROPOSITION XVI. THEOREM
554. Two dihedral angles have the same ratio as their
plane angles.
Fig.l
Fig,2
Jig. 3
Given the dihedral A C-AB-D and C'-A'B'-D' having
the plane A CAD and C'A'D', respectively.
To prove Z C'-A'B'-D'
/.CAD.
Z C-AB-D = Z C'A'D'
CASE I. When the plane A C'A'D' and CAD (Figs. 2
and 1), are commensurable.
Proof. Find a common measure of the A C'A'D' and
CAD, as Z CAR, and let it be contained in Z C'A'D' n
times, and in Z CAD m times.
Then Z C'A'D' : Z CAD = n : m.
Through A'B' and the lines of division of Z C'A'D' pass
planes, and through AB and the lines of division of
Z CAD pass planes. These planes will divide the dihedral
Z C'-A'B'-D' into n, and /.C-AB-D into m parts, all
equal. Art. 552.
/. /.C'-A'B'-D' : /.C-AB-D = n : m.
Hence Z C'-A'B'-D' : Z (J-AB-D = Z C'A'D' : Z CAD.
(Why?)
DIHEDKAL ANGLES 341
CASE II. When the plane angles C'A'D' and CAD (Figs.
3 and 1) are incommensurable.
Proof. Divide the Z CAD into any number of equal
parts, and apply one of these parts to the Z C'A'D1. It
will be contained a certain number of times with a remain-
der, as £LA'D', less than the unit of measure.
Hence the A G'A'L and CAD are commensurable.
/. Z C'-A'B'-L\tC-AB-D=. Z C'A'L:Z.CAD. Case I-
If now we let the unit of measure be indefinitely dimin-
ished, the /.LA'D', which is less than the unit of measure,
will be indefinitely diminished.
.-. Z C'A'L = Z C'A'D' as a limit, and
Z G'—A'B'—L = Z C'—A'B'—D' as a limit. Art. 251.
Z C' — A'Bf — L
Hence — — — — — — becomes a variable, with
Z. O~-A-D ~ JJ
Z C'-A'B'-D' .. .. .,
— . as its limit; Art. 253, 3.
Z. 0 ^4,-D JL/
, Z
Also . ~ . j becomes a variable with as its
limit. Art. 253, 3.
tO-A'B'-L Z C'A'L
But the variable . . = the variable
Z \jA.jJ
always. Case I.
, Z C'-A'B'-D' ., . Z
•"• the hmit =the hmlt
Q. E. D.
Ex. 1. How many straight lines are necessary to indicate a dihe-
dral angle (as /.E-AB-D, p. 338)? How many straight lines are
necessary to indicate the plane angle of a dihedral angle ? Hence,
what is the advantage of using a plane angle of a dihedral angle
instead of the dihedral angle itself ?
Ex. 2. Give three additional properties of dihedral angles anal-
ogous to properties of plane angles given in Book I,
342
X
BOOK VI. SOLID GEOMETKY
PROPOSITION XVII. THEOREM
555. If a straight line is perpendicular to a plane, every
plane draivn through that line is perpendicular to the plane.
Given the line AB _L plane MN, and the plane PQ
passing through AB and intersecting MN in RQ.
To prove PQ J_ MN.
Proof. In the plane MN draw BG JL RQ at B.
But AB _L RQ. Art. 505.
.*. Z ABC is the plane Z of the dihedral /iP-RQ-M.
Art. 548.
But Z ABC is a right Z , Art. 505.
( for AB J. MN by hyp . ) .
/. PQ -L MN.
Art. 547.
Q. E. B.
556. COR. A plane perpendicular to the edge of a
dihedral angle is perpendicular to each of the two faces form-
ing the dihedral angle.
DIHEDEAL ANGLES 343
PROPOSITION XVIII. THEOREM
557. If two planes are perpendicular to each other, a
straight line drawn in one of them perpendicular to their
line of intersection is perpendicular to the other plane.
p
Given the plane PQ JL plane MN and intersecting it in
the line RQ-, and AB a line in PQ _L RQ.
To prove AB J_ plane MN.
Proof. In the plane MN draw BC JL RQ.
:. ZABCis the plane Z of the dihedral ^P-RQ-M.
Art. 548.
/. Z ABC is a rt. Z , Art. 554.
(for P-RQ-Mis a right dihedral Z ).
.*. AB JL BC and RQ at their intersection.
/. AB _L plane NN. (Why T)
Q. £. D.
558. COR. 1. If two planes are perpendicular to each
other, a perpendicular to one of them at any point of their
intersection will lie in the other plane.
For, in the above figure, a -L erected at the point B in
the plane MN must coincide with AB lying in the plane
PQ and _L MN, for at a given point in a plane only one JL
can be drawn to that plane (Art. 515).
559. COR. 2. If two planes are perpendicular to each
other, a perpendicular to one plane, from a point in the other
plane, will lie in tlie other plane.
344
BOOK VI. SOLID GEOMETRY
PROPOSITION XIX. THEOREM
560. If two intersecting planes are each perpendicular
to a third plane., their line of intersection is perpendicular
to the third plane.
Given the planes PQ and R8 JL plane MN, and inter-
secting in the line AB.
To prove
AB JL plane MN.
Proof. At the point B in which the three planes meet
erect a _L to the plane MN. This _L must lie in the plane PQ,
and also in the plane R8. Art- 558-
Hence this _L must coincide with AB, the intersection
of PQ and RS. Art. 508, 2.
/. AB J_ plane MN.
Q. £. D.
561. COR. If two planes, including a right dihedral
angle, are each perpendicular to a third plane, the intersec-
tion of any two of the planes is perpendicular to the third
plane, and each of the three lines of intersection is perpen'
dicular to the other two.
Ex. 1. Name all the dihedral angles on the above figure.
Ex. 2. If Z CBQ=3Q°, find the ratio of each pair of dihedral
DIHEDRAL ANGLES
345
PROPOSITION XX. THEOREM
562. Every point in the plane which bisects a given
dihedral angle is equidistant from the faces of the dihedral
angle.
Given plane CB bisecting the dihedral /.A-BR-D, P
any point in plane BC, PQ and PT JL faces BA and BD,
respectively.
To prove PQ = PT.
Proof. Through PQ and PT pass a plane intersecting
AB in QR, BD in JBT,and BG in PR.
Then plane PQT JL planes AB and £D. Art. 555.
/. plane PQT J_ line RB, the intersection of the planes
AB and BD. Art. 560.
/. I2£ J_ RQ, RP and -RT. Art. 505.
/. A QRP and PRT are the plane A of the dihedral
^ A-BR-P and P-BR-D. Art. 548.
But these dihedral ^ are equal. Hyp.
Art. 554.
(Why?)
(Why?)
Q. £. D.
rt.A PQR=rt. A PRT.
:. PQ=PT.
563. The locus of all points equidistant from the faces
of a dihedral angle is the plane bisecting the dihedral angle.
346
BOOK VI. SOLID GEOMETRY
PROPOSITION XXI. PROBLEM
564. Through any straight line not perpendicular to a
given plane, to pass a plane perpendicular to the given plane.
•if
Given the line AB not J_ plane MN.
To construct a plane passing through AB and _L MN.
Construction. From a point A in the line AB draw a _L
AC to the plane MN. Art. 516.
Through the intersecting lines AB and AC pass the
plane AD. Art. 503.
Then AD is the plane required.
Proof. The plane AD passes through AB. Constr.
Also plane AD _L plane MN, Art. 555.
(for it contains AC, which is _L MN).
Q. E. F.
565. COR. 1. Through a straight line not perpendicular
to a given plane only one plane can be passed perpendicu-
lar to that plane.
For, if two planes could be passed through AB _L plane
MN, this intersection AB would be J_ MN (Art. 560),
which is contrary to the hypothesis.
566. COR. 2. The projection upon a plane of a straight
line not perpendicular to that plane is a straight line.
For, if a plane be passed through the given line J_ to
the given plane, the foot of a _L from any point in the
line to the given plane will be in the intersection of the
two planes (Art. 559).
DIHEDRAL ANGLES 347
PROPOSITION XXII. THEOREM
567. The acute angle which a line makes with its pro-
jection on a plane is the least angle which it makes with any
line of the plane through its foot.
Given line AB meeting the plane MN in the point B,
BG the projection of AB on MN, and PB any other line
in the plane MN through B.
To prove that Z ABC is less than Z ABP.
Proof. Lay off PB equal to CB, and draw AC and AP.
Then, in the A ABC and ABP,
AB=AB. (Why?)
BC=BP. (Why!)
But AC < AP. Art. 521.
/. Z ABC is less than Z ABP Art. 108.
Q. £. D.
568. DEF. The inclination of a line to a plane is the
acute angle which the given line makes with its projection
upon the given plane.
Ex. 1. A plane has an inclination of 47° to each of the faces of a
dihedral angle and is parallel to the edge of the dihedral angle; how
many degrees are in the plane angle of the dihedral angle?
Ex. 2. In the figure on page 345, if PT=QT,how large is the
dihedral L A-BR-Dt if PT—RT. how large is it?
348
BOOK VI. SOLID GEOMETKY
PROPOSITION XXIII. PROBLEM
569. To draw a common perpendicular to any two lines
<ot in the same plane,
a L
Given the lines AB and CD not in the same plane. '
To construct a line perpendicular to both AB and CD.
Construction. Through AB pass a plane MN\\ line CD.
Art. 529.
Through CD pass a plane CF J_ plane MN (Kii. 564), and
intersecting plane MN in the line EF.
Then EF \\ CD (Art. 528), /. EF must intersect AB
(which is not || CD by hyp.) in some point K.
At K in the plane CF draw LK J_ EF. Art, 274.
Then LK is the perpendicular required.
Proof. LK _L EF. Constr.
.*. LK J_ CD. Art. 123.
Also LK _L plane MN. Art. 557.
/. LK _L line AS. (Why ?)
/. ££: -L both CD and AS.
<?. E. F.
570. Only one perpendicular can be drawn between two
lines not in the same plane.
For, if possible, in the above figure let another line BD
be drawn _L AB and CD. Then, if a line be drawn through
B || CD, BD _L this line (Art. 123), and .'. J_ plane MN
(Art. 509). Draw DF JL line EF; then DF J_ plane MN
(Art. 557) . Hence from the point D two _k , DB and DF,
are drawn to the plane MN, which is impossible (Art. 517).
POLYHEDRAL ANGLES
349
POLYHEDRAL ANGLES
571. A polyhedral angle is the amount
of opening between three or more planes
meeting at a point.
Such an angle may be regarded as a portion of
space cut out by the planes forming the angle.
572. The vertex of a polyhedral angle
is the point in which the planes forming the angle meet; the
edges are the lines in which the planes intersect; the faces
tore the portions of the planes forming the polyhedral angle
which are included between the edges ; the face angles are
the angles formed by the edges.
Each two adjacent faces of a polyhedral angle form a
dihedral angle.
The parts of a polyhedral angle are its face angles and
dihedral angles taken together.
573. Naming a polyhedral angle. A polyhedral angle
is named either by naming the vertex, as F; or by naming
the vertex and a point on each edge, as V-ABC.
In case two or more polyhedral angles have the same
vertex, the latter method is necessary.
In the above polyhedral angle, the vertex is V; the
edges are VA, VB, VC-, the face angles
are AVB, BVC, AVC.
574. A convex polyhedral angle is a
polyhedral angle in which a section
made by a plane cutting all the edges
is a convex polygon, as V-ABCDE.
575. A trihedral angle is a polyhe-
dral angle having three faces; a tetrahedral angle is one
having four faces, etc.
350
BOOK VI, SOLID GEOMETRY
576. A trihedral angle is rectangular, birectangular, or
trirectangular, according as it contains one, two, or three
right dihedral angles.
577. An isosceles trihedral angle is a trihedral angle
two of whose face angles are equal.
578. Vertical polyhedral angles are polyhedral angles
having the same vertex and the faces of one the faces of
the other produced.
579. Two equal polyhe-
dral angles are polyhedral an-
gles having their correspond-
ing parts equal and arranged
in the same order, as V-ABG
and V'-A'B'C1.
Two equal polyhedral angles
may be made to coincide. .
580. Two symmetrical poly-
hedral angles are polyhedral an-
gles having their corresponding
parts equal but arranged in
reverse order.
If the faces of a trihedral angle, F-ABC, be
produced, they will form a vertical trihedral angle,
V-A'B'C', which is symmetrical to V-ABC. For,
if V-A'B'C' be rotated forward about a horizontal
axis through V, the two trihedral angles are seen
to have their corresponding parts equal but ar-
ranged in reverse order.
Similarly, any two vertical polyhedral angles
are symmetrical.
CL-l
POLYHEDRAL ANGLES 351
581. Equivalence of symmetrical polyhedral angles. It
has been shown in Plane Geometry (Art. 488) that two
triangles (or polygons) symmetrical with respect to an
axis have their corresponding parts equal and arranged
in reverse order. By sliding
two such figures about in a plane
they cannot be made to coincide,
t
but by lifting one of them up
from the plane in which it lies and turning it over it may
be made to coincide with the other figure.
Symmetrical polyhedral angles, however, cannot be
made to coincide in any way; hence some indirect method
of showing their equivalence is necessary. See Ex. 29,
p. 358, and Arts. 789-792.
Ex. 1. Name the trihedral angles on the figure to Prop. XX. If
=>90°, what kind of trihedral angles are those on the figure ?
If ZPJ2$=30°, what kind are they?
Ex. 2. Are two trirectangular trihedral angles necessarily equal ?
Prove this.
Ex. 3. Are two lines which are perpendicular to the same plane
necessarily parallel ? Are two planes which are perpendicular to the
same plane necessarily parallel ? Are two planes which are perpen-
dicular to the same line necessarily parallel ?
Ex. 4. Let the pupil cut out three pieces of pasteboard of the form
indicated in the accompanying figures ; cut them half through where
the lines are dotted ; fold them and fasten the edges so as to form
three trihedral angles, two of which (Figs. 1 and 2) shall be equal
and two (Figs. 1 and 3) symmetrical. By experiment, let the pupil
find which pair may be made to coincide, and which not. |
Fig.
352
BOOK VI. SOLID GEOMETKY
PROPOSITION XXIV. THEOREM
582. The sum of any two face angles of a trihedral
angle is greater than the third face angle.
Given the trihedral angle S-ABC, with angle A8C its
greatest face angle.
To prove Z ASB + ^BSC greater than Z A8C.
Proof. In the face ASC draw SD, making ^ASD=
/.ASB.
Take SD = SB.
In the face ASC draw the line ADC in any convenient
direction, and draw AB and BC.
Then, in the A ASB and ASD, SA = SA.
SB = SD, and /.ASB= Z.ASD.
• .'. A ASB= A ASD.
/. AB = AD.
Also AB + BC > AC.
Hence, subtracting the equals AB and
BC > DC.
Hence, in the A D£C and DSC, SC=SC,
> DC.
/. Z BSC is greater than /.DSC.
(Why?)
(Why?)
(Why?)
(Why?)
(Why ?)
(Why?)
= SD, and
(Why?)
Art. 108.
To each of these unequals add the equals /.ASB and
:. /.ASB+ Z BSC is greater than /.ASC. (Why?)
Q. E. D.
Ex. In the above figure, if /.ASC equals one of the other face
angles at 5, as Z ASB, how is the theorem proved ?
POLYHEDRAL ANGLES 353
PROPOSITION XXV. THEOREM
i
583. The sum of the face angles of any convex polyhedral
angle is less than four right angles.
Given the polyhedral angle S-ABCDE.
To prove the sum of the face A at 8 less than 4 rt. A .
Proof. Pass a plane cutting the edges of the given poly-
hedral angle in the points A, B, C, D, E.
From any point O in the polygon ABCDE draw OA
OB, OC, OD, OE.
Denote the A having the common vertex S as the 8 A,
and those having the common vertex 0 as the 0 A .
Then the sum of the A of the 8 A = the sum of A of
the 0 A . Art. 134.
But Z SB A + Z SBC is greater than Z ABC,
Z£C£+Z SCD is greater than ^BCD, etc/Art
.*. the sum of the base A of the S A > the sum of the
base A of the 0 A. Ax. 9.
. •. the sum of the vertex A of the 8 A < the sum of
the vertex A of the 0 A, Ax. 11.
(if unequals be subtracted from equals, the remainders are unequal
in reverse order).
But the sum of the A at 0 = 4 rt. A . (Why?)
.% the sum of face A at S < 4 rt. A . Ax. 8.
Q. E. D.
W
354 BOOK VI. SOLID GEOMETRY
PROPOSITION XXVI. THEOREM
584. // two trihedral angles have the three face angles
of one equal to the three face angles of the other, the tri-
hedral angles have their corresponding dihedral angles equal,
and are either equal or symmetrical, according as their
corresponding face angles are arranged in the same or in
reverse order.
Given the trihedral A S-ABC and S'-A'B'C', having
the face A ASB, A SO and BSC equal to the face A A'S'B',
A'S'C' and B'S'C', respectively.
To prove that the corresponding dihedral A of 8- ABC
and S'-A'B'C' are equal, and that A S-ABC and S'-
A'B'C' are either equal or symmetrical.
Proof. On the edges of the trihedral A take 8A, SB,
SO, S'A', S'B', S'C' all equal.
Draw AB, AC, BC, A'B', A'C', B'C'.
Then, 1. In the A ASB and A'S'B', SA = S'Af, SB =
S'B', and /iASB = /.A'S'B'. (Why?)
/. A A8B = A A'8'B'. (Why?)
.'. AB=A'B'. (Why?)
2 In like manner AC=A'Cf, and BC=B'C'.
.'. A ABC=& A' B'C'. (Why?)
EXERCISES ON THE LINE AND PLANE 355
3. Take D a convenient point in SA, and draw DE in
the face ASB, and DF in the face ASC, each J_ 8A.
DE and DF meet AB and A G in points E and F,
T"P^T)Pf*f 1 VP 1 V
(/or /£ SAB and SAC are acute).
Similarly, take SfDf = SD and construct A D'E'F'.
Then, in the rt. A ADE and A'D'E', AD=A'D', and
/.DAE^D'A'E'. (Why ?)
.-. A ADE=& A'D'E'. (Why?)
/. AE=A'Ef, and DE=D'Ef. (Why?)
4. In like manner it maybe shown that AF=A'F'1 and
DF=D'F'.
:. A AEF=& A'E'F'. (Why?)
And ^7jP= JEJ'JF' (Why ?)
5. Hence, in the A DEF and D'^J^P, DE=D'E', DF
= DfFf and ^7^= E'F'. (Why ?)
/. A DEF=& D'E'F'. (Why?)
.*. Z EDF= Z ^D^. (Why ?)
But these A are the plane A of the dihedral A whose
edges are SA and S1 'A'.
:. dihedral Z B-A8~C= dihedral £B'-A'8'-C? Art. 552.
In like manner it may be shown that the dihedral A at
SB and S'B' are equal; and that those at 8C and S'G' are
equal.
.'. the trihedral A S and Sf are either equal or sym-
metrical. Arts. 579, 580.
Q. E. D.
EXERCISES. CROUP 64
THEOREMS CONCERNING THE LINE AND PLANE IN SPACE
Ex. 1. A segment of a line not parallel to a plane is longer than
its projection in the plane.
Ex. 2. Equal straight lines drawn from a point to a plane are
equally inclined to the plane,
356
BOOK VI. SOLID GEOMETRY
B
N
Ex. 3. A line and plane perpendicular to the same plane are
parallel.
Ex. 4. If three planes intersecting in three straight lines are
perpendicular to a plane, their lines of intersection are parallel.
Ex. 5. If a plane bisects any line at right
angles, any point in the plane is equidistant
from the ends of the line.
Ex. 6. Given AB _L plane MN,
and AC J_ plane BS;
prove BC _L NE.
Ex. 7. Given PQ J_ plane MN,
PR _L plane BL,
and BS J_ plane .MN;
prove QS I. AB. M
Ex. 8. If a line is perpendicular to one
of two intersecting planes, its projection
on the other plane is perpendicular to the
line of intersection of the two planes.
Ex. 9. Given CE _L DE,
AE J_ DE,
and / C-AD-E a rt. dihedral Z ;
prove CA J_ plane DAE.
Ex. 10. The projections of two parallel lines on
a plane are parallel. (Is the converse of this
theorem also true ?)
Ex. 11. If two parallel planes are cut by two non -parallel planes,
the two lines of intersection in each of the parallel planes will make
equal angles.
Ex. 12. If a line is perpendicular to a plane, any plane parallel
to the line is perpendicular to the plane. (Is the converse true ?)
Ex. 13. In the figure to Prop. VI, given AB J_ MN and AF J.
DC; prove BF J_ DC.
Ex. 14. Two planes parallel to a third plane are parallel to each
other.
[Sue. Draw a line -L third plane.]
EXERCISES ON THE LINE AND PLANE 357
Ex. 15. The projections upon a plane of two equal and parallel
straight lines are equal and parallel.
Ex. 16. A line parallel to two planes is parallel to their inter-
section.
Ex. 17. In the figure to Prop. XXII, if angle GBP is obtuse,
prove the angle ABP obtuse.
Ex. 18. In a quadrilateral in space (i. e., a
quadrilateral whose vertices are not all in the
same plane), show that the lines joining the
midpoints of the sides form a parallelogram.
Ex. 19. The lines joining the midpoints of
the opposite sides of a quadrilateral in space bisect each other.
Ex. 20. The planes bisecting the dihedral angles of a trihedral
angle meet in a line every point of which is equidistant from the
three faces.
[SuG. See Art. 562.]
Ex. 21. Given OQ bisecting
PQ JL plane ROS,
QR L OR,
and QS _L OS',
prove PR= PS, PR JL OR,
and PS _L OS.
Ex. 22. In a plane bisecting a given plane angle, and perpendicu-
lar to its plane, every point is equidistant from the sides of the angle.
[SuG. See Ex. 21; or through P any point in the bisecting plane
pass planes -L to the sides of the Z , etc.]
Ex. 23. In a trihedral angle, the three planes bisecting the three
face angles at right angles to their respective planes, intersect in a line
every point of which is equidistant from the three edges of the tri-
hedral angle.
Ex. 24. If two face angles of a trihedral angle are equal, the dihe-
dral angles opposite them are equal.
Ex. 25. In the figure to Prop. XXIV, prove that £ASC+ £BSC
is greater than Z ASD + Z BSD.
Ex. 26. The common perpendicular to two lines in space is the
shortest line between them.
358
BOOK VI. SOLID GEOMETKY
Ex. 27. Given MN \\ ES,
and PB = pi ;
prove /.ABC=/.abc,
and A ABC = A abc.
Ex. 28. Two isosceles
trihedral angles are equal.
symmetrical
r\
\
Ex. 29. Any two symmetrical trihe-
dral angles are equivalent.
[SUG. Take SA, SB, SC, S'A',
S'B', S'C', all equal. Pass planes ABC,
A'B'C'. Draw SO and S'O' ± these
planes. Then the trihedral A are
divided into three pairs of isosceles
symmetrical trihedral A, etc.]
EXERCISES. CROUP 65
LOCI IN SPACE
Find the locus of a point equidistant from
Ex. 1. Two parallel planes. Ex. 3. Three given points.
Ex. 2. Two given points. Ex. 4. Two intersecting lines.
Ex. 5. The three faces of a trihedral angle.
Ex. 6. The three edges of a trihedral angle.
Find the locus
Ex. 7. Of all lines passing through a given point and parallel to
a given plane.
Ex. 8. Of all lines perpendicular to a given line at a given "point
in the line.
Ex. 9. Of all points in a given plane equidistant from a given
point outside the plane.
Ex. 10. Of all points equidistant from two given points and from
two parallel planes.
Ex. 11. Of all points equidistant from two given points and from
two intersecting planes.
Ex. 12. Of all points at a given distance from a given plane and
equidistant from two intersecting lines.
EXERCISES ON THE LINE AND PLANE 359
EXERCISES. CROUP 66
PROBLEMS CONCERNING THE POINT, LINE AND PLANE
IN SPACE
Ex. 1. Through a given point pass a plane parallel to a given
plane.
Ex. 2. Through a given point pass a plane perpendicular to a
given plane.
Ex. 3. Through a given point to construct a plane parallel to two
given lines which are not in the same plane.
Prove that only one plane can be constructed fulfilling the given
conditions.
Ex. 4. Bisect a given dihedral angle.
Ex. 5. Draw a plane equally inclined to three lines which meet at
a point.
Ex. 6. Through a given point draw a line parallel to two given
intersecting planes.
Ex. 7. Find a point in a plane such that lines drawn to it from
two given points without the plane make equal angles with the plane.
[Sua. See Ex. 23, p. 176.]
Ex. 8. Find a point in a given line equidistant from two given
points.
Ex. 9. Find a point in a plane equidistant from three given points.
Ex. 10. Find a point equidistant from four given points not in a
plane.
Ex. 11. Through a given point draw a line which shall intersect
two given lines.
[SUG. Pass a plane through the given point and one of the given
lines, and pass another plane through the given point and the other
given line, etc.]
Ex. 12. Through a given point pass a plane cutting the edges of
a tetrahedral angle so that the section shall be a parallelogram.
[SuG. Produce each pair of opposite faces to intersect in a
Straight line, etc.]
BOOK VII
Polyhedron
POLYHEDRONS
585. A polyhedron is a solid bounded by
586. The faces of a polyhe-
dron are its bounding planes; the
edges of a polyhedron are the lines
of intersection of its faces.
A diagonal of a polyhedron is
a straight line joining two of its
vertices which are not in the
same face. The vertices of a poly-
hedron are the points in which its
edges meet or intersect.
587. A convex polyhedron is a polyhedron in which a
section made by any plane is a convex polygon.
Only convex polyhedrons are to be considered in this book.
588. Classification of polyhedrons. Polyhedrons are
sometimes classified according to the number of their
faces. Thus, a tetrahedron is a polyhedron of four faces;
a hexahedron is a polyhedron of six faces ; an octahedron
is one of eight, a dodecahedron one of twelve, and an
icosahedron one of twenty faces.
Tetrahedron Cube
Octahedron Dodecahedron icosahedron
(360)
POLYHEDEONS 361
The polyhedrons most important in practical life are those deter-
mined by their stability, the facility with which they can be made out
of common materials, as wood and iron, the readiness with which
they can be packed together, etc. Thus, prism means "something
sawed off."
PRISMS AND PARALLELOPIPEDS
589. A prism is a polyhedron bounded
by two parallel planes and a group of
planes whose lines of intersection are
parallel,
590. The bases o$ a prism are the
faces formed by the two parallel planes;
Prism
the lateral faces are the faces formed by
the group of planes whose lines of intersection are parallel.
The altitude of a prism is the perpendicular distance
between the planes of its bases.
The lateral area of a prism is the sum of the areas of the
lateral faces.
591. Properties of a prism inferred immediately.
1. The lateral edges of a prism are equal, for they are
parallel lines included between parallel planes (Art. 589)
and are therefore equal (Art. 532).
2. The lateral faces of a prism are parallelograms (Art.
160), for their sides formed by the lateral edges are equal
and parallel.
3. The bases of a prism are equal polygons, for their
homologous sides are equal and parallel, each to each,
(being opposite sides of a parallelogram), and their homol-
ogous angles are equal (Art. 538).
592. A right section of a prism is a section made by a
plane perpendicular "to the lateral edges.
362
BOOK VII. SOLID GEOMETRY
593. A triangular prism is a prism whose base is a
triangle ; a quadrangular prism is one whose base is
a quadrilateral, etc.
ObliQiie Prism?
Right Prism
Regular Prism
594. An oblique prism is a prism whose lateral edges
are oblique to the bases.
595. A right prism is a prism whose
lateral edges are perpendicular to the bases.
596. A regular prism is a right prism
whose bases are regular polygons.
597. A truncated prism is that part of
a prism included between a base and a
section made by a plane eliliqoo»tQ the
base and cutting all the lateral edges. Truncated Prism
598. A parallelepiped is a prism whose bases are paral-
lelograms.
Hence, all the faces of a parallelepiped are parallelograms.
Oblique
Parallelepiped
Right Rectangular Cube
Parallelepiped Parallelepiped
599. A right parallelepiped is a parallelepiped whose
lateral edges are perpendicular to the bases,
PRISMS AND PAKALLELOPIPEDS 363
600. A rectangular parallelepiped is a right parallele-
piped whose bases are rectangles.
Hence, all the faces of a rectangular parallelopiped are
rectangles.
601. A cube is a rectangular parallelopiped whose edges
are all equal.
Hence, all the faces of a cube are squares.
602. The unit of volume is a cube whose edge is equal
to some linear unit, as a cubic inch, a cubic foot, etc. o '*"*{'
603. The volume of a solid is the number of units of
volume which the solid contains.
Being a number, a volume may often be determined from other
numbers in certain expeditious ways, which it is one of the objects of
geometry to determine.
604. Equivalent solids are solids whose volumes are
equal.
Ex. 1. What is the least number of faces which a polyhedron can
have ?
Ex. 2. A square right prism is what kind of a parallelopiped ?
Ex. 3. Are there more right parallelepipeds or rectangular paral-
lelepipeds ? That is, which of these includes the other as a special
case ?
Ex. 4. Prove that if a given straight line is perpendicular to a
given plane, and another straight line is perpendicular to another
plane, and the two planes are parallel, then the two given lines are
^parallel.
364 BOOK vn. SOLID GEOMETRY
O PROPOSITION I. THEOREM
605. Sections of a prism made by parallel planes cutting
all the lateral edges are equal polygons.
Given the prism PQ cut by || planes forming the sections
AD and A'D' .
To prove section AD = section A'D' .
Proof. AB, EC, CD, etc., are || A'B', B'C', C'D', etc.,
respectively. Art. 531.
/. AB, BC, <7D? etc., are equal to A'B', B'C', C'D'r
etc., respectively. Art. 157.
Also A ABC, BCD, etc., are equal to A. A' B'C', E'UD',
etc., respectively. Art. 538.
/. ABCDE=A'B'C'D'E', Art. 47.
(for the polygons have all their parts equal, each to each, and .'. can be
made to coincide).
Q. E. D.
606. COR. 1. Every section of a prism made by a plane
parallel to the base is equal to the base.
607. COR. 2. All right sections of a prism are equal.
PRISMS 365
0 PROPOSITION II. THEOREM
608. The lateral area of a prism is equal to the product
of the perimeter of a right section by a lateral edge.
Given the prism RQ, with its lateral area denoted by 8
and lateral edge by E\ and AD a right section of the given
prism with its perimeter denoted by P.
To prove 8=PXE.
Proof. In the prism RQ, each lateral edge = ^7. Art. 591, 1.
Also AB JL OR, BG JL IJ, etc. Art. 505.
Hence area CU RH=AB X GH=AB X E^\
area C3 GJ= BC X E, \ Art. 385.
area EU IQ = CD X E, etc.
But S, the lateral area of the prism, equals the sum of
the areas of the ££7 forming the lateral surface.
/. adding, 8= (AB + BG + CD + etc. ) X E. Ax. 2.
Or S=PXE.
Q. £. D.
609. COR. The lateral area of a right prism equals the
product of the perimeter of the base by the altitude.
Ex. Find the lateral area of a right prism whose altitude is 12 in.,
and whose base is an equilateral triangle with a side of 6 in. Also
find the total area of this figure.
366 BOOK VII. SOLID GEOMETRY
O PROPOSITION III. THEOREM
i
610. If two prisms have the three faces including a
trihedral angle of one equal, respectively, to the three faces
including a trihedral angle of the other, and similarly
placed, the prisms are equal.
Given the prisms AJ and AfJf, having the faces AK,
AD, AG equal to the faces A'K', A'D', A'O', respectively,
and similarly placed.
To prove AJ=A'J'.
Proof. The face A EAF, EAB and EAF are equal,
respectively, to the face A E'A'F', EfA'Bf and B'A'F'. Hyp.
/. trihedral ZA = trihedral /.A' ' Art. 584.
Apply the prism A'J' to the prism AJ, making each of
the faces of the trihedral Z Af coincide with corresponding
equal face of the trihedral /.A. Geom. Ax. 2.
.*. the plane F'J' will coincide in position with the
plane FJ, Art. 500.
(for the points Gf, Ff, K' coincide with G, F, K, respectively).
Also the point Cr will coincide with the point C.
:. G'R' will take the direction of CH. Geom. Ax. 3.
.'. R' will coincide with H. Art. 508, 1.
In like manner J1 will coincide with J.
Hence the prisms AJ and A'J' coincide in all points.
.*. AJ=A'Jf. Art. 47.
611. COR. 1. Two truncated prisms are equal if tlie
three faces including a trihedral angle of one are equal to
the three faces including a trihedral angle of the other.
PRISIMS 367
612. COR. 2. Two right prisms are equal if they have
equal l)ases and equal altitudes.
0 PROPOSITION IV. THEOREM
613. An oblique prism is equivalent to a right prism
whose base is a right section of the oblique prism and whose
altitude is equal to a lateral edge of the oblique prism.
K
Given the oblique prism AD', with the right section FJ;
also the right prism FJf whose lateral edges are each equal
to a lateral edge of AD' .
To prove AD' =0= FJ'.
Proof. AA = FF'. Hyp.
Subtracting FA from each of these, AF=A'F'. (Why?)
Similarly BO = B'Gr.
Also AB = AB', and FG=F'G'. Art. 155.
And A of face AG = homologous A of face A'O'. Art. 130.
/. face A6r = face A'G', Art. 47.
(for they have all their parts equal, each to each, and .". can be made
to coincide).
In like manner face AK=facQ A'K' .
But face AD = face AD'. Art. 591, 3.
.*. truncated prism AJ= truncated prism AJ'. Art. 611.
To each of these equals add the solid FD' .
;.AD'^FJf. (why?)
Q. £. P.
368 BOOK VII. SOLID GEOMETRY
0 PROPOSITION V. THEOREM
614. The opposite lateral faces of a parallelopiped are
equal and parallel.
Given the parallelepiped AH with the base AC.
To prove AG = aud \\ EH, and AJ= and || EH.
Proof. The base AC is a ZZ7 . Art. 598.
/. AB=and \\ EG. (Why?)
Also the lateral face AJ is a. £17 , Art. 591, 2.
.'. JLF=and || EJ. . (Why ?)
/. Z J3J^= Z 0^7. Art. 538.
And £~7AG=£JEH. Art.162.
Also plane AG \\ plane J£#. Art. 538.
In like manner it may be shown that AJ and BH are
equal and parallel.
Q. E. D.
615. COR. Any two opposite -faces of a parallelopiped
may be taken as the bases.
Ex. 1. How many edges has a parallelopiped ? How many faces ?
How many dihedral angles ? How many trihedral angles ?
Ex. 2. Find the lateral area of a prism whose lateral edge is 10
and whose right section is a triangle whose sides are 6, 7, 8 in.
Ex. 3. Find the lateral area of a right prism whose lateral edge is
16 and whose ba§e is a rhombus with diagonals of 6 arid 8 in.
PRISMS 369
0 PROPOSITION VI. THEOREM
616. A plane passed through two diagonally opposite
edges of a parallelepiped divides the parallelepiped into two
equivalent triangular prisms.
Given the parallelepiped AH with a plane passed through
the diagonally opposite edges AF and CH, forming the tri-
angular prisms ABC-G and ADC-K.
To prove ABC-Go ADC-K.
Proof. Construct a plane J_ to one of the edges of the
prism forming the right section PQRS, having the diagonal
PR formed by the intersection of the plane FHCA.
Then PQ \\ SR, and QR \\ PS. (Why ?)
/. PQRS is a ZZ7 . (Why ?)
' .*. A PQR= A PSR. (Why ?)
But the triangular prism ABC-G =c= a prism whose base
is the right section PQR and whose altitude is AF. Art. 613.
Also the triangular prism ADC-K o a prism whose base
is the right section PSR and whose altitude is AF. (Why?)
But the prisms having the equal bases, PQR and PSR,
and the same altitude, AF, are equal. Art. 612.
/. ABC-G =0= ADC-K. Ax l.
Q. £. D.
370 BOOK VII. SOLID GEOMETRY
0 PROPOSITION VII. THEOREM
617. If two rectangular parallelepipeds have equal bases,
they are to each other as their altitudes.
Given the rectangular parallelepipeds Pf and P having
equal bases and the altitudes A'B' and AB.
To prove Pf : P=A'B' : AB.
CASE I. When the altitudes A'B' and AB are com-
mensurable.
Proof. Find a common measure of A'B' and AB, as
AK, and let it be contained in A'B' n times and in AB m
times.
Then A'B' : AB = n : m.
Through the points of division of A'B' and AB pass
planes parallel to the bases.
These planes will divide P' into n, and P into m small
rectangular parallelepipeds, all equal. Art. 612.
/. P : P=n : m.
:. P' : P=A'B' : AB. (Why ?)
CASE II. When the altitudes A'B' and AB are incom-
mensurable.
Let the pupil supply the proof, using the method of
limits. (See Art. 554).
618. DEF. The dimensions of a rectangular parallele-
piped are the three edges which meet at one vertex.
PAKALLELOPIPEDS
371
619. COR. If two rectangular parallelopipeds have two
dimensions in common, they are to each other as their third,
dimensions.
PROPOSITION VIII. THEOREM
620. Two rectangular parallelopipeds having equal alti-
tudes are to each other as their bases.
Given the rectangular parallelopipeds P and Pf having
the common altitude a, and the dimensions of their bases
b, c and bf, c', respectively.
P bXc
To prove = '
Then
Art. 619.
Proof. Construct the rectangular parallelepiped, Q, whose
altitude is a and the dimensions of whose base are b and c'.
P=c_
Q c''
AISO J = |7. (Why?)
Multiplying the corresponding members of these
equalities,
£ = -£*£.. Ax. 4.
f b'Xc'
Q. E. D.
621. COR. Two rectangular parallelopipeds having one
dimension in common are to each other as the products of the
other two dimensions.
372
BOOK VII. SOLID GEOMETRY
PROPOSITION IX. THEOREM
622. Any two rectangular parallelepipeds are to each
other as the products of their three dimensions.
Given the rectangular parallelepipeds P and Pf having
the dimensions a, 5, c and af, ~b' ', c', respectively.
P aXbXc
Proof. Construct the rectangular parallelepiped Q hav-
ing the dimensions a, &, cf '.
P=c_
Q c''
Q = aX b
P1 a' X I''
Multiplying the corresponding members of these
equalities,
P n . V 7) V r.
Ax. 4.
Q. E. B.
Then
Also
Art. 619,
Art. 621.
P ^ a XbX c
P' o'X&'Xc'
Ex. 1. Find the ratio of the volumes of two rectangular parallelo-
pipeds whose edges are 5, 6, 7 in. and 7, 8, 9 in.
Ex.2. Which will hold more, a bin 10x2x7 ft., or one 8x
4 x 5 ft. ?
Ex. 8. How many bricks 8 x 4 x 2 in. are necessary to build a wall
80x6 ft, x8 in. f
PAKALLELOPIPEDS 373
PROPOSITION X. THEOREM
623. The volume of a rectangular parallelopiped is equal
to the product of its three dimensions.
u
Given the rectangular parallelopiped P having the three
dimensions a, 6, c.
To prove volume of P=a Xb Xc.
Proof. Take as the unit of volume the cube Z7, whose
edge is a linear unit.
P aX bXc
rhen rTxTxV Art'622'
The volume of P is the number of times P contains the
p
unit of volume U, or — • Art. 603.
/. volume of P=a X b X c.
Q. E. D.
For significance of this result, see Art. 2.
624. COR. 1. The volume of a cube is the cube of its
edge.
625. COR. 2. The volume of a rectangular parallelo-
piped is equal to the product of its base by its altitude.
Ex. 1. Find the number of cubic inches in the volume of a cube
whose edge is 1 ft. 3 in. How many bushels does this box contain, if
lbushel = 2150.42cu. in. ?
Ex. 2. The measurement of the volume < f a cube reduces to the
measurement of the length of what single straight line ?
374
BOOK VII. SOLID GEOMETRY
PROPOSITION XI. THEOREM
626. The volume of any parallelopiped is equal to the
product of its base ly its altitude.
Given the oblique parallelopiped P, with its base denoted
by B, and its altitude by H.
To prove volume of P=B X H.
Proof. In P produce the edge CD and all the edges
parallel to CD.
On CD produced take FG=CD.
Pass planes through F and G J_ the produced edges,
forming the parallelopiped Q, with the rectangular base
denoted by Bf.
Similarly produce the edge GI and all the edges || GI.
Take IK= GI, and pass planes through 7 and K J_ the
edges last produced, forming the rectangular parallelopiped
R, with its base denoted by B'f.
Then
Also
But
Or
[Outline Proof.
volume of R = B" X H.
volume of P=B"XH.
volume of P=BX H.
^R=B" XH=BX H.]
Art. 613.
Art. 386.
Art. 625.
Ax. 1.
Ax. 8.
Q. £. P.
'PRISMS 375
PROPOSITION XII. THEOREM
627. The volume of a triangular prism is equal to the
product of its base by its altitude.
Given the triangular prism PQR-M, with its volume
denoted by F, area of base by J5, and altitude by H.
To prove V=BXH.
Proof. Upon the edges PQ, QR, QM, construct the
parallelepiped QK.
Hence QK= twice PQR-M. Art. 616.
But volume of QK= area PQRT X H. Art. 626.
= 2BXH. Ax. 8.
.'. twice volume PQR-M=2B X H. Ax. i.
.*. volume PQR-M=B X H. Ax. 5.
Q. E. D.
Ex. 1. If the altitude of a triangular prism is 18 in., and the base
is a right triangle whose legs are 6 and 8 in., find the volume.
Ex. 2 Find the volume of a triangular prism whose altitude is 24,
and the edges of whose base are 7, 8, 9. Also find the total surface.
tfc $- -c •
376 BOOK VII. SOLID GEOMETRY
o PROPOSITION XIII. THEOREM
628. The volume of any prism is equal to the product of
its base by its altitude.
L
Given the prism AK, with its volume denoted by F,
area of base by B, and altitude by H.
To prove V=BXH.
Proof. Through any lateral edge, as AF, and the diag-
onals of the base, AC and AD, drawn from its foot, pass
'planes.
These planes will divide the prism into triangular
prisms.
Then F, the volume of the prism AK, equals the sum
of the volumes of the triangular prisms. Ax. 6.
But the volume of each triangular prism = its base X H.
Hence the sum of the volumes of the A prisms = the
sum of the bases of the A prisms X H.
= BXH. Ax. 8.
/. V=BXH. AX. l.
Q. E. D.
629. COR. 1. Two prisms are to each other as the pro-
ducts of their bases by their altitudes; prisms having equiva-
lent bases and equal altitudes are equivalent.
630. COR. 2. Prisms having equivalent bases are to
each other as their altitudes; prisms having equal altitudes
are to each other as their bases.
PYKAMIDS
377
Pyramid
0 PYRAMIDS
631. A pyramid is a polyhedron
bounded by a group of planes passing
through a common point, and by another
plane cutting all the planes of the group.
632. The base of a pyramid is the
face formed by the cutting plane; the €^~
lateral faces are the faces formed by the
group of planes passing through a com-
mon point; the vertex is the common point through which
the group of planes passes ; the lateral edges are the inter-
sections of the lateral faces.
The altitude of a pyramid is the perpendicular from the
vertex to the plane of the base.
The lateral area is the sum of the areas of the lateral
faces.
633. Properties of pyramids inferred immediately.
1. The lateral faces of a pyramid are triangles (Art.
508, 2).
2. The base of a pyramid is a polygon (Art. 508, 2).
634. A triangular pyramid is a pyramid whose base is
a triangle; a quadrangular pyramid is a pyramid whose
base is a quadrilateral, etc.
A triangular pyramid is also called a tetrahedron, for it has four
faces. All these faces are triangles, and any one of them may be
taken as the base.
635. A regular pyramid
is a pyramid whose base is a
regular polygon, and the foot
of whose altitude coincides
with the center of the base.
378 BOOK VII. SOLID GEOMETRY
636. Properties of a regular pyramid inferred immedi-
ately.
1. The lateral edges of a regular pyramid are equal, for
they are oblique lines drawn from a point to a plane
cutting off equal distances from the foot of the per-
pendicular from the point to the plane (Art. 518).
2. The lateral faces of a regular pyramid are equal isos-
celes triangles.
637. The slant height of a regular pyramid is the alti-
tude of any one of its lateral faces.
The axis of a regular pyramid is its altitude.
638. A truncated pyramid is the portion of a pyramid
included between the base and a section cutting all the
lateral edges.
639. A frustum of a pyramid is the
part of a pyramid included between the
base and a plane parallel to the base.
The altitude of a frustum of a pyramid is
the perpendicular distance between the planes of its bases.
640. Properties of a frustum of a pyramid inferred im-
mediately.
1. The lateral faces of a frustum of a pyramid are
trapezoids.
2. The lateral faces of a frustum of a regular pyramid
are equal isosceles trapezoids.
641. The slant height of the frustum of a regular pyra-
mid is the altitude of one of its lateral faces.
Ex. 1. Show that the foot of the altitude of a regular pyramid
coincides with the center of the circle circumscribed about the base.
Ex. 2. The perimeter of the midsection of the frustum of a pyra-
mid equals one -half the sum of the perimeters of the bases,
PYKAMIDS 379
0 PROPOSITION XIV. THEOREM
642. The lateral area of a regular pyramid is equal to
half the product of the slant height by the perimeter of the
base.
o
Given 0-ABCDF a regular pyramid with its lateral
&rea denoted by 8, slant height by L, and perimeter of its
base by P.
To prove S=%LXP.
Proof. The lateral faces GAB, OBC, etc., are equal
isosceles A. Art. 636, 2.
Hence each lateral face has the same slant height, L.
.'. the area of each lateral face = J L X its base.
.'. the sum of all the lateral faces = J L X sum of bases.
. 643. COR. The lateral area of
Hie frustum of a regular pyramid is
equal to one-half the sum of the
perimeters of its bases multiplied
by its slant height.
Ex. Find the lateral area of a regular square pyramid whose slant
height is 32, and an edge of whose base is 16. Find the total
area also,
5
380
BOOK VII. SOLID GEOMETRY
PROPOSITION XV. THEOREM
644. If a pyramid is cut by a plane parallel to the lase,
I. The lateral edges and the altitude are divided pro-
portionally;
II. The section is a polygon similar to the base.
Given the pyramid S-ABCDF, with the altitude SO cut
by a plane MN, which is parallel to the base and intersects
the lateral edges in a, b, c, d, /and the altitude in o.
Sa Sb Sc So
To prove I. — =— =-^= = ^Q'
II. The section abcdf similar to the base
ABCDF.
Proof. I. Pass a plane through the vertex 8 II MN.
Then SA, SB, SC . . . SO, are lines intersected by three
|| planes.
• -^=_^=_^L= . . . =^L. Art. 539.
" SA SB SC SO
II. ab || AB. (Why?)
.*. A Sab and SAB are similar. Art. 328.
In like manner the A Sbc, Scd, etc., are similar to the
A SBC, SCD, etc., respectively.
ab _ / Sb \ _ be _ / Sc \ _ cd _
' AE
SB
\
)
BC \SC CD
PVKAMIDS 381
That is, the homologous sides of abcdf and ABCDF are
proportional.
Also Z.dbc = Z ABC, £bcd = Z BCD, etc. Art. 538.
.*. section obcdf is similar to the base ABCDF. Art. 321.
Q. E. D.
645. COR. 1. A section of a pyramid parallel to the
base is to the base as the square of its distance from the
vertex is to the square of the altitude of the pyramid.
abcdf ab
For (Why ?>
ABCDF
ab Sa So ab2 So
But - = - = - /. - (Why!)
AB 8A SO AB2 SO2
(Whyf)
ABCDF
646. COR. 2. // two pyramids having equal altitudes
are cut by a plane parallel to their bases at equal distances
from the vertices, the sections have the same ratio as the
bases.
Let S-ABCDF and V-PQR be two pyramids cut as
described.
Then = (Art. 645); also = . ( }
ABCDF so2 PQR ~VTL
But VT=SO, and Vt = So. (Why?)
pqr abcdf ABCDF
= °r - - -- (Why ?)
647. COR. 3. If two pyramids have equal altitudes and
equivalent bases, sections made by planes parallel to the bases
at equal distances from the vertices are equivalent.
382
BOOK VII. SOLID GEOMETRY
PROPOSITION XVI. THEOREM
648. The volume of a triangular pyramid is the limit of
the sum of the volumes of a series of inscribed, or of a series
of circumscribed prisms of equal altitude, if the number of
prisms be indefinitely increased.
r o
Given the triangular prism O-ABC with a series of in-
scribed, and also a series of circumscribed prisms, formed by
passing planes which divide the altitude into equal parts, and
by making the sections so formed first upper bases, then
lower bases, of prisms limited by the next parallel plane.
To prove O-ABC the limit of the sum of each series, if
the number of prisms in each be indefinitely increased.
Proof. Each inscribed prism equals the circumscribed
prism immediately above it. Art. 629.
.*. (sum of circumscribed prisms) — (sum of inscribed
prisms) = lowest circumscribed prism, or ABC-K.
If the number of prisms be indefinitely increased, the
altitude of each approaches zero as a limit.
Hence volume ABC-K=0, Art. 253, 2.
(for its base, ABC, is constant while its altitude = 0).
.*. (sum of circumscribed prisms) — (sum of inscribed
prisms) = 0.
.*. volume O-ABC — (either series of prisms) = 0.
(for this difference < difference between the two series, which
last difference =0).
/. O-ABC is the limit of the sum of the volumes of
either series of prisms. Q. E. D.
PYRAMIDS
PROPOSITION XVII. THEOREM
383
649. If two triangular pyramids have equal altitudes
and equivalent bases, they are equivalent.
o
B
Given the triangular pyramids O-ABC and O'-A'B'C'
having equivalent bases ABC and A'B'O ', and equal
altitudes.
To prove O-ABC- O'-A'B'C'.
Proof. Place the pyramids so that they have the com-
mon altitude H, and divide H into any convenient num-
ber of equal parts.
Through the points of division and parallel to the plane of
the bases of the pyramids, pass planes cutting the pyramids.
Using the sections so formed as upper bases, inscribe a
series of prisms in each pyramid, and denote the volumes
of the two series of prisms by V and V.
The sections formed by each plane, as KLM and K'L'M',
are equivalent. Art. 647.
.". each prism in O-ABC =e= corresponding prism in
Of-A'B'Cf (Art. 629). /. V= Vf. Ax. 2.
Let the number of parts into which the altitude is
divided be increased indefinitely.
Then V and V become variables with O-ABC and
O'-A'B'C' as their respective limits. Art. 648.
But V^V always. (Why?)
.-. O-ABC^O'-A'B'C', (Why?)
Q. E. D.
384
BOOK VII. SOLID GEOMETRY
PROPOSITION XVIII. THEOREM
650. The volume of a triangular pyramid is equal tc
one-third the product of its base ly its altitude.
D
Given the triangular pyramid 0-ABC, having its volume
denoted by V, the area of its base by B, and its altitude by 27".
To prove V=$BXH.
Proof. On ABC as a base, with OB as a lateral edge,
construct the prism ABC-DOF.
Then this prism will be composed of the original pyra-
mid O-ABC and the quadrangular pyramid O-ADFC.
Through the edges OD and OC pass a plane intersecting
the face ADFC in the line DC, and dividing the quadrangular
pyramid into the triangular pyramids 0-ADC and 0-DFC.
Then 0-ADC^ 0-DFC. Art. 649.
(for they have the common vertex O, and the equal bases ADC and DFC}.
But 0-DFC may be regarded as having C as its vertex
and DOF as its base. Art. 634.
/. 0-DFC^O-ABC. Art. 649.*
.*. the prism is made up of three equivalent pyramids.
.*. 0-ABC= J the prism. ^ Ax. 5.
But volume of prism = B X H. Art. 627.
.-. 0-ABC, or F=J B X H. Ax. 5.
0. E. D.
Ex. Find the volume of a triangular pyramid whose altitude is 12
ft., and whose base is an equilateral triangle with a side of 15 ft.
PYRAMIDS 385
PROPOSITION XIX. THEOREM
651. The volume of any pyramid is equal to one-third
the product of its base by its altitude.
Given the pyramid 0-ABCDF, having its volume denoted
by V, the area of its base by B, and its altitude by H.
To prove V=$BXH.
Proof. Through any lateral edge, as OD, and the diago-
nals of the base drawn from its foot, as AD and BD, pass
planes dividing the pyramid into triangular pyramids.
Then V, the volume of the pyramid 0-ABCDF, will
equal the sura of the volumes of the triangular pyramids.
But the volume of each A pyramid = & its base X H.
Art. 650.
Hence the sum of the volumes of A pyramids = J sum of
their bases X H. Ax. 2.
= %BXH. Ax. 8.
.-. V=$ BX H. Ax. 1.
Q. E. D.
652. COR. 1. The volumes of two pyramids are to each
other as the products of their bases and altitudes; pyramids
having equivalent bases and equal altitudes are equivalent.
653. COR. 2. Pyramids having equivalent bases are to
each other as their altitudes; pyramids having equal alti-
tudes are to each other as their bases.
654. SCHOLIUM. The volume of any polyhedron may be
found by dividing the polyhedron into pyramids, finding the
volume of each pyramid separately, and taking their sum.
386
BOOK VII. SOLID GEOMETRY
0 PROPOSITION XX. THEOREM
655. The frustum of a triangular pyramid is equivalent
to the sum of three pyramids whose common altitude is the
altitude of the frustum, and whose bases are the lower base,
the upper base, and a mean proportional between the two
bases of the frustum .
Given ABC-DEF the frustum of a triangular pyramid,
having the area of its lower base denoted by B, the area
of its upper base by b, and its altitude by H.
To prove ABC-DEF =c= three pyramids whose bases are
B, b and vBb, and whose common altitude is H.
Proof. Through E and A C, E and DC, pass planes divid-
ing the frustum into three triangular pyramids. Then
1. E-ABC has the base B and the altitude H.
2. E-DFC, that is, C-DEF, has the base b and the
altitude H.
3. It remains to show that E-ADC is equivalent to a
pyramid having an altitude H, and a base that is a mean
proportional between B and b.
Denoting the three pyramids by I, II, III,
I=A ABE=AB = AC=&ADC=H
II A ADE DE DF A DFC III"
(Arts. 653, 391, 644, 321. Let the pupil supply the reason for each
I II
step in detail).
or
=VlXIII. Art. 303.
b.
Hence, ABC-DEFosum of three pyramids, as described.
Q. £. D.
PYRAMIDS
387
656. Formula for volume of frustum of a triangular
pyramid. y= J H (B + b +
PROPOSITION XXI. THEOREM
657. The volume of the frustum of any pyramid is
equivalent to the sum of the volumes of three pyramids, whose
common altitude is the altitude of the frustum, and ivhose
bases are the lower base, the upper base, and a mean propor-
tional between the two bases <f the frustum.
K T
A
T\
^
yt
Given the frustum of a pyramid Ad, having its volume
denoted by V, the area of its lower base by B, of its upper
base by 5, and its altitude by H.
To prove V= J H (B + b -
Proof. Produce the lateral faces of Ad to meet in K.
Also construct a triangular pyramid with base PQR
equivalent to ABCDF, and in the same plane with it, and
with an altitude equal to the altitude of K-ABCDF. Pro-
duce the plane of ad to cut the second pyramid in pqr.
Then pqr^abcdf. Art. 647. \
:. pyramid K-ABCDF ^ pyramid T-PQR. Art. 652.
Also pyramid K-abcdf o pyramid T-pqr. (Why ?)
Subtracting, frustum Ad =c= frustum Pr. Ax. 3.
But volume Pr = $ H (B + b + V Eb) .
;. volume Ad = J H (B + b + V ' Bb) . (Why ?)
Q. £. D.
388 BOOK VII. SOLID GEOMETRY
PROPOSITION XXII. THEOREM
658. A truncated triangular prism is equivalent to the
sum of three pyramids, of which the base of the prism is the
common base, and ivhose vertices are the three vertices of the
inclined section.
,R
Fig. 1 Fig. 2
Given the truncated triangular prism ABC-PQR,
To prove ABC-PQR = the sura of the three pyramids
P-ABC, Q-ABC and R-ABC.
Proof. Pass planes through Q and AC, Q and PC, divid-
ing the given figure into the three pyramids Q-ABC,
Q-APC and Q-PRC.
1. Q-ABC has the required base and the required
vertex Q.
2. Q-APC =0 B-APC, Art. 652.
(for they have the same base, A PC, and the same altitude, their ver-
tices being in a line \\ base APC).
But B-APC may be regarded as having P for its ver-
tex, and ABC for its base, as desired. Art. 634.
3. Q-PRC =0 B-ARC (see Fig. 2) . Art. 652.
{for the base ARC ^ base PRC (Art. :WO); and the altitudes of the two
pyramids are equal, the vertices Q ami />' hcimj in line \\ plane
PACE, in 'which the bases tie).
But B-ARC may be regarded as having R for its ver-
tex, and ABO for its base, as desired. Art, 634,
PKISMATOIDS
389
.'. ABC-PQR =0= sum of three pyramids whose common
base is ABC, and whose vertices are P, Q, JR.
Q. E. D.
Pig. 3
Fig. 4
659. COR. 1. The volume of a truncated right triangu-
lar prism (Fig. 3) is equal to the product of its base by
one- third the sum of its lateral edges.
660. COR. 2. The volume of any truncated triangular
prism (Fig. 4) is equal to the product of the area of its
right section by one- third the sum of its lateral edges.
PRISMATOIDS
661. A prismatoid is a polyhedron
bounded by two polygons in parallel planes,
called bases, and by lateral faces which are
either triangles, trapezoids or parallelograms.
662. A prismoid is a prismatoid in
which the bases have the same number of
sides and have their corresponding sides
parallel.
Erlamoid
Ex. The volume of a truncated right parallelepiped equals the
area of the lower base multiplied by oue- fourth the sum of the lateral
edges (or by a perpendicular from the center of the upper base to tiie
Ipwer base),
390
BOOK VII. SOLID GEOMETKY
PROPOSITION XXIII. THEOREM
663. The volume of a prismatoid is equal to one-sixth
the product of its altitude ~by the sum of its bases and of
four times the area of its midsection.
G
Given the prismatoid ABCD-FGK, with bases B and b,
midsection M, volume F, and altitude H.
To prove F=i H (B + I + 4 M).
Proof. Take any point 0 in the midsection, and through
it and each edge of the prismatoid let planes be passed.
These planes will divide the figure into parts as follows:
1. A pyramid with vertex 0, base ABCD and altitude
J _5T, and whose volume /. =i SX B. Art. 651.
2. A pyramid with vertex 0, base FGK, and altitude J
H, and whose volume .*. =i J3"X &. (Why?)
3. Tetrahedrons like 0-ABG whose volume may be
determined as follows (see Fig. 2) :
AB = 2 PQ. (Why?)
.-. A AGB = ± A PGQ. Art. 398.
/. 0-AGB = 4 0-PGQ. Art. 653.
But 0-PGQ (or G-PQO) = i PQO X J fi=i HXPQO.
:. 0-AGB = i HX4A PQO. (Why ?)
.'. the sum of all tetrahedrons like 0-A GB = i jff X 4 M.
Or
KEGULAK POLYHEDKONS 391
REGULAR POLYHEDRONS
664. DEF. A regular polyhedron is a polyhedron all of
whose faces are equal regular polygons, and all of whose
polyhedral angles are equal. Thus, the cube is a regular
polyhedron.
PROPOSITION XXIV. THEOREM
665. But five regular polyhedrons are possible.
Given regular polygons of 3, 4, 5, etc., sides.
To prove that regular polygons of the same number of
sides can be joined to form polyhedral A of a regular
polyhedron in but five different ways, and that, conse-
quently, but five regular polyhedrons are possible.
Proof. The sum of the face A of any polyhedral angle
< 360°. Art. 583.
1. Each Z of an equilateral triangle is 60°. Art. 134.
3 X 60°, 4 X 60° and 5 X 60° are each less than 360°;
but any larger multiple of 60° = or > 360°.
.*. but three regular polyhedrons can be formed with
equilateral A as faces.
2. Each Z of a square contains 90°. Art. 151.
3 X 90° is less than 360°, but any larger multiple of
90° = or > 360°.
.*. but one regular polyhedron can be formed with
squares as faces.
3. Each Z of a regular pentagon is 108°. Art. 174.
3 X 108° is less than 360°, but any larger multiple of
108° > 360°.
392
BOOK VII. SOLID GEOMETRY
/. but one regular polyhedron can be formed with regu*
lar pentagons as faces.
4. Each Z of a regular hexagon is 120°, and 3 X 120°
-360°.
.*. no regular polyhedron can be formed with hexagons,
or with polygons with a greater number of sides as faces.
/. but five regular polyhedrons are possible.
Q. E. D.
666. The construction of the regular polyhedrons, by the
use of cardboard, may be effected as follows:
Draw on a piece of cardboard the diagrams given below.
Cut the cardboard half through at the dotted lines and en-
tirely through at the full lines. Bring the free edges
together and keep them in their respective positions by
some means, such as pasting strips of paper over them.
Octahedron
Tetrahedron
Hexahedron
Dodecahedron
Icosahedron
POLYHEDRONS
393
POLYHEDRONS IN GENERAL
PROPOSITION XXV. THEOREM
667. In any polyhedron, the number of edges increased
by two equals the number of vertices increased by the number
of faces. p
Given the polyhedron AT, with the number of its ver-
tices, edges and faces denoted by F, E and F, respectively.
To prove E + 2=V+F.
Proof. Taking the single face ABCD, the number of
edges equals the number of vertices, or E= V.
If another face, CRTD, be annexed (Fig. 2), three new
edges, CR, RT, TD, are added and two new vertices, R and T.
:. the number of edges gains one on the number of ver-
tices, or E=V+ 1.
If still another face, BQRC, be annexed, two new edges,
BQ and QR, are added, and one new vertex, Q. .*. E= F+2.
With each new face that is annexed, the number of
edges gains one on the number of vertices, till but one
face is lacking.
The last face increases neither the number of edges nor
of vertices.
Hence number of edges gains one on number of vertices,
for every face except two, the first and the last, or gains
F—2 in all.
.*. for the entire figure, E= F + F— 2.
That is E + 2= V + F. Ax. 2.
Q. B. D.
394 BOOK VII. SOLID GEOMETRY
PROPOSITION XXVI. THEOREM
668. The sum of the face angles of any polyhedron
equals four right angles taken as many times, less two, as
the polyhedron has vertices.
Given any polyhedron, with the sum of its face angles
denoted by S, and the number of its vertices, edges and
faces denoted by F, E, F, respectively.
To prove S=(V— 2) 4 rt. A.
Proof. Each edge of the polyhedron is the intersection
of two faces, .'. the number of sides of the faces = 2 E. ^
:. the sum of the interior and exterior A of the faces =
2 E X 2 rt. A , or E X 4 rt. A . Art. 73.
But the sum of the exterior A of each face = 4 rt. A .
Art. 175 „
/. the sum of exterior A of the FfsLces = FX 4 rt. A .
Ax. 4.
Subtracting the sum of the exterior A from the sum of
all the A , the sum of the interior A of the F faces =
(EXlrt. A)—(FX4rt. A).
Or S=(E—F) 4rt. A.
But E + 2=V+F. Art. 667.
Hence E—F= F— 2. Ax. 3.
Substituting for E— F, S= ( F— 2) 4 rt. A . Ax. 8.
Q. E. D.
Ex. Verify the last two theorems in the case of the cube.
COMPARISON OF POLYHEDRONS
395
COMPARISON OF POLYHEDRONS. SIMILAR POLYHEDRONS
PROPOSITION XXVII. THEOREM
669. // two tetrahedrons have a trihedral angle of one
equal to a trihedral angle of the other, they are to each other
as the products of the edges including the equal trihedral
angles.
c
c'
Given the tetrahedrons O-ABC and 0'-A'B'Cr, with
their volumes denoted by Fand V ', respectively, and having
the trihedral A 0 and Of equal.
V_= OAXOBX PC
V 0'A'XO'B'Y.O'C''
Proof. Apply the tetrahedron O'-A'B'C' to O-ABC so
that the trihedral Z 0' shall coincide with its equal, the
trihedral Z 0.
Draw CP and G'P' JL plane QAB, and draw OP the
projection of OC in the plane OAB.
Taking OAB and OA'B' as the bases, and CP and C'P'
as the altitudes of the pyramids C-OAB and C'-OA'B',
respectively.
V A OAB X OP
&OAB ^
CP
Art. 652.
Art. 397.
(Why ?)
But
In the
V A OA
A
'B' X O'P'
OA£ OA
X OB
C'P'
OC
~00'*
X 00
A
similar rt. /
OA X 6
OA'B' OA' X 0£'
^7P
& OOPandOO'P', 7^7 =
0 X
U3 X 00 OA X 05
"7
' OA' X 0
£' X 00X C
VA> X O'B'
X 0'(
^ . • AX . o .
Q. £. D.
396
BOOK vii. SOLID GEOMETRY
670. DEP. Similar polyhedrons are polyhedrons having:
the same number of faces, similar, each to each, and simi-
larly placed, and having their corresponding polyhedral
angles equal.
PROPOSITION XXVIII. THEOREM
671. Any two similar polyhedrons may be decomposed
into the same number of tetrahedrons, similar, each to each,
and similarly placed.
c
Given P and Pf, two similar polyhedrons.
To prove that P and Pf may be decomposed into the
same number of tetrahedrons, similar, each to each.
Proof. Take H and Hf any two homologous vertices of
P and P'. Draw homologous diagonals in all the faces of
P and P' except those faces which meet at H and Hf, sepa-
rating the faces into corresponding similar triangles.
Through H and each face diagonal thus formed in P,
and through R' and each face diagonal in P', pass planes.
Each corresponding pair of tetrahedrons thus formed
may be proved similar.
Thus, in H-ABC and R'-A'B'Cf, the A HBA and
R'E'A! are similar. Art. 329.
In like manner A REG and R'E'G' are similar; and A
ABC and A'B'C' are similar.
SIMILAR POLYGONS 397
AC
' Art'321'
/. A ARC and A'H'Cf are similar. Art. 326.
Hence the corresponding faces of H- ABC and Hf-A'BfCf
are similar.
Also their homologous trihedral A are equal. Art. 584.
•. tetrahedron H-ABC is similar to H'-AfB'Cf. Art. 670.
After removing H-ABC from P, and H'-A'B'C' from
P', the remaining polyhedrons are similar, for their faces
are similar, and the remaining polyhedral A are equal.
Ax. 3.
By continuing this process, P and Pf may be decom-
posed into the same number of tetrahedrons, similar, each
to each, and similarly placed.
Q. £. D.
672. COR. 1. The homologous edges of similar polyhe-
drons are proportional;
Any two homologous lines in two similar polyhedrons
have the same ratio as any other two homologous lines.
673. COB. 2. Any two homologous faces of two similar
polyhedrons are to each other as the squares of any two
homologous edges or lines;
The total areas of any two similar polyhedrons are to
each other as the squares of any two homologous edges.
Ex. 1. In the figure, p. 395, if the edges meeting at OareS, 9, 12 in.,
and those meeting at 0' are 4, 6, 8 in., find the ratio of the volumes
of the tetrahedrons.
Ex. 2. If the linear dimensions of one room are twice as great as
the corresponding dimensions of another room, how will their surfaces
(and .'. cost of papering) compare ? How will their volumes compare ?
Ex. 3. How many 2 in. cubes can be cut from a 10 in. cube ?
Ex. 4. If the bases of a prismoid are rectangles whose dimensions
are a, b and b, a, and altitude is //, find the formula for the volume.
Vnfjy
398 BOOK VII. SOLID GEOMETRY
PROPOSITION XXIX. THEOREM
674. The volumes of two similar tetrahedrons are to
each other as the cubes of any pair of homologous edges.
Given the similar tetrahedrons 0-ABC and O'-A'B'C'.
V OA
To prove — ^=
F
_ , V OA X OB X OC
Proof. - = —————. (Why!)
OA x OS ^ 00
O'B' 0>0
OA OB OG
But ==' Art'672'
y 0.4 ^ OA ^ OA ~OA3
— = - X - X - = Ax. 8.
F
Ex. 1. In the above figures, if AB=2 A'B', find the ratio of V
to F'. Find the same, if AB=H A'B'.
Ex. 2 The measurement of the volume of a regular triangular
prism reduces to the measurement of the lengths of how many straight
lines ? of a frustum of a regular square pyramid ?
Ex. 3. Show how to construct out of pasteboard a regular prism,
a parallelepiped, and a truncated square prism.
SIMILAK POLYGONS 399
PROPOSITION XXX. THEOREM
675. The volumes of any two similar polyhedrons are
to 'each other as the cubes of any two homologous edges, or
of any other two homologous lines.
BO B1 O1
Given the polyhedrons AK and A'K' having their vol-
umes denoted by V and Vft and HB and H'B' any pair of
homologous edges.
V _HB3
To prove — —
y H'B'*
Proof. Let the polyhedrons be decomposed into tetra-
hedrons, similar, each to each, and similarly placed. Art. 671.
Denote the volumes of the tetrahedrons in P by vi, V2,
vs . . . and of those in Pf by v'i, 1/2, v's • • •
Then
Also
HfBfi
!_!_
1/1
Art. 674, Ax. 1.
(for each of these ratios^.
Vl
j±at=:-2Lj that is> *1=£L. Art. 312.
HB
Ax. 1.
Q. E. D-
400 BOOK VII. SOLID GEOMETRY
EXERCISES. CROUP 67
THEOREMS CONCERNING POLYHEDRONS
Ex. 1. The lateral faces of a right prism are rectangles.
Ex. 2. A diagonal plane of a prism is parallel to every lateral edge
of the prism not contained in the plane.
Ex. 3. The diagonals of a parallelepiped bisect each other.
Ex. 4. The square of a diagonal of a rectangular parallelepiped
equals the sum of the squares of the three edges meeting at a vertex.
Ex. 5. Each lateral face of a prism is parallel to every lateral edge
not contained in the face.
Ex. 6. Every section of a prism made by a plane parallel to a
lateral edge is a parallelogram.
Ex. 7. If any two diagonal planes of a prism which are not par-
allel to each other are. perpendicular to the base of the prism, the
prism is a right prism.
Ex. 8. What part of the volume of a cube is the pyramid whose
"base is a face of the cube and whose vertex is the center of the cube ?
Ex. 9. Any section of a regular square pyramid made by a plane
through the axis is an isosceles triangle.
Ex. 10. In any regular tetrahedron, an altitude equals three times
the perpendicular from its foot to any face.
Ex. 11. In any regular tetrahedron, an altitude equals the sum of
the perpendiculars to the faces from any point within the tetrahedron.
Ex. 12. Find the simplest formula for the lateral area of a trun-
cated regular prism of n sides.
Ex. 13. The sum of the squares of the four diagonals of a paral-
lelopiped is equal to the sum of the squares of the twelve edges.
[Sue. Use Art. 352.]
Ex. 14. A parallelepiped is symmetrical with respect to what point?
Ex. 15. A rectanglar parallelepiped is symmetrical with respect to
how many planes ? (Let the pupil make a definition of a figure sym
metrical with respect to a plane. See Arts. 486, 487.)
EXERCISES ON POLYHEDRONS 401
Ex. 16. The volume of a pyramid whose lateral edges are the
three edges of the parallelepiped meeting at a point is what part of
the volume of the parallelepiped ?
Ex. 17. If a plane be passed through a vertex of a cube and the
diagonal of a face not adjacent to the vertex, what part of the volume
of the cube is contained by the pyramid so formed ?
Ex. 18. If the angles at the vertex of a triangular pyramid are
right angles and each lateral edge equals a, show that the volume of
the pyramid is — *
b
Ex. 19. How large is a dihedral angle at the base of a regular
pyramid, if the apothem of the base equals the altitude of the pyramid ?
Ex. 20. The area of the base of a pyramid is less than the area of
the lateral surface.
Ex. 21. The section of a triangular pyramid by a plane parallel
to two opposite edges is a parallelogram.
If the pyramid is regular, what kind of a parallelogram does the
section become ?
Ex. 22. The altitude of a regular tetrahedron divides an altitude
of the base into segments which are as 2 : 1.
Ex. 23. If the edge of a regular tetrahedron is a, show that the
slant height is -~- ; and hence that the altitude is ^— , and the vol-
2t o
Ex. 24. If the midpoints of all the edges of
a tetrahedron except two opposite edges be
joined, a parallelogram is formed.
Ex. 25. Straight lines joining the midpoints
of the opposite edges of a tetrahedron meet in
a point and bisect each other.
Ex. 26. The midpoints of the edges of a regular tetrahedron are,
the vertices of a regular octahedron.
402 BOOK VII. SOLID GEOMETRY
EXERCISES. GROUP 68
PROBLEMS CONCERNING POLYHEDRONS
Ex. 1 . Bisect the volume of a given prism by a plane parallel to
the base.
Ex. 2. Bisect the lateral surface of a given pyramid by a plane par-
allel to the base.
Ex. 3. Through a given point pass a plane which shall bisect the
volume of a given parallelepiped.
Ex. 4. Given an edge, construct a regular tetrahedron.
Ex. 5. Given an edge, construct a regular octahedron.
Ex. 6. Pass a plane through the axis of a regular tetrahedron so
that the section shall be an isosceles triangle.
Ex. 7. Pass a plane through a cube so that the section shall be a
regular hexagon.
Ex. 8. Through three given lines no two of which are parallel
pass planes which shall form a parallelepiped.
Ex. 9. From cardboard construct a regular square pyramid each
of whose faces is an equilateral triangle.
EXERCISES. CROUP 69
REVIEW EXERCISES
Make a list of the properties of
Ex. 1. Straight lines in space. Ex. 9. Right prisms.
Ex. 2. One line and one plane. Ex. 10. Parallelopipeds in gen-
Ex. 3. Two or more lines and
one plane. Ex. 11. Rectangular parallelo-
Ex. 4. Two planes and one line. pipeds.
Ex.5. Two planes and two lines. Ex« 12« Pyramids in general.
Ex. 6. Polyhedrons in general. Ex- 13- Regular pyramids.
Ex. 7. Similar polyhedrons. Ex. 14 Frusta of pyramids.
Ex.8. Prisms in general. Ex.15. Truncated prisms.
BOOK VIII
CYLINDERS AND CONES
CYLINDERS
676. A cylindrical surface is
a curved surface generated .by a
straight line which moves so as
constantly to touch a given fixed
curve and constantly be parallel
to a given fixed straight line.
Thus, every shadow cast by a point
of light at a great distance, as by a
star or the sun, approximates the
cylindrical form, that is, is bounded Cylindrical surface
by a cylindrical surface of light. Hence, in all radiations (as of light,
heat, magnetism, etc.) from a point at a great distance, we are
concerned with cylindrical surfaces and solids.
677. The generatrix of a cylindrical surface is the mov-
ing straight line; the directrix is the given curve, as CDE\
an element of the cylindrical surface is the moving straight
line in any one of its positions, as DF.
678. A cylinder is a solid bounded by
a cylindrical surface and by two parallel
planes.
The bases of a cylinder are its parallel
plane faces; the lateral surface is the
cylindrical surface included between the
parallel planes forming its bases; the alti- cylinder
tude of a cylinder is the distance between the bases.
The elements of a cylinder are the elements of the
cylindrical surface bounding it.
(403)
404
BOOK VIII. SOLID GEOMETRY
679. Property of a cylinder inferred immediately. All
the elements of a cylinder are equal, for they are parallel lines
included between parallel planes (Arts. 532, 676).
The cylinders most important in practical life are those determined
by their stability, the ease with which they can be made from com-
mon materials, etc.
680. A right cylinder is a cylinder
whose elements are perpendicular to the
Oblique circular
cylinder
681. An oblique cylinder is one whose
elements are oblique to the bases.
682. A circular cylinder is a cylinder
whose bases are circles.
683. A cylinder of revolution is a cylin-
der generated by the revolution of a rect-
angle about one of its sides as an axis.
Hence, a cylinder of revolution is a right
circular cylinder.
Some of the properties of this solid are derived
most readily by considering it as generated by a re-
volving rectangle ; and others, by regarding it as a
particular kind of cylinder derived from the general
definition.
684. Similar cylinders of revolution are cylinders gen-
erated by similar rectangles revolving about homologous
sides.
685. A tangent plane to a cylinder is a plane which
contains one element of the cylinder, and which does not cut
the cylinder on being produced.
Ex. 1. A plane passing through a tangent to the base of a circu-
lar cylinder and the element drawn through the point of contact is
tangent to the cylinder. (For if it is not, etc.)
Ex. 2. If a plane is tangent to a circular cylinder, its intersection
the itlane of the base is tangent to the base.
CYLINDEKS 405
686. A prism inscribed in a cylinder is a prism whose
lateral edges are elements of the cylinder, and whose bases
are polygons inscribed in the bases of the cylinder.
inscribed prism Circumscribed prism
687. A prism circumscribed about a cylinder is a prism
whose lateral faces are tangent to the cylinder, and whose
bases are polygons circumscribed about the bases of the
cylinder.
688. A section of a cylinder is the figure formed by the
intersection* of the cylinder by a plane.
A right section of a cylinder is a section formed by a
plane perpendicular to the elements of the cylinder.
689. Properties of circular cylinders. By Art. 441 the
area of a circle is the limit of the area of an inscribed or
circumscribed polygon, and the circumference is the limit
of the perimeters of these polygons; hence
1. The volume of a circular cylinder is the limit of the
volume of an inscribed or circumscribed prism.
2. The lateral area of a circular cylinder is the limit of
the lateral area of an inscribed or circumscribed prism.
Also, 3. By methods loo advanced for this book, it may be proved that
the perimeter of a right section is the limit of the perimeter of a right sec-
tion of an inscribed or circumscribed jpr ism.
406 BOOK VIII. SOLID GEOMETKY
PROPOSITION I. THEOREM
690. Every section of a cylinder made by a plane pass-
ing through an element is a parallelogram.
Given the cylinder AQ cut by a plane passing through
the element AB and forming the section ABQP.
To prove ABQP a ZZ7 .
Proof. AP\\BQ. Art. 531.
It remains to prove thatvP$ is a straight line || AB.
Through P draw a line in the cutting plane || AB.
This line will also lie in the cylindrical surface. Art. 676.
.*. this line must coincide with PQ,
(for the line drawn lies in both the cutting plane and the cylindrical
surface, hence, it must be their intersection).
:. PQ is a --straight line 1 1 AB.
:. ABQP is a ZZ7 . (Why ?)
Q. E. D.
691. COR. Every section of a right cylinder made by a
plane passing through an element is a rectangle.
Ex. 1. A door swinging on its hinges generates what kind of a
solid f
Ex. 2. Every section of a parallelepiped made by a plane inter-
eectiug all its lateral edges is a parallelogram,
CYLINDEKS 407
PROPOSITION II. THEOREM
692. The bases of a cylinder are equal.
Given the cylinder AQ with the bases APB and CQD.
To prove base APJ5 = base CQD.
Proof. Let AC and BD be any two fixed elements in
the surface of the cylinder AQ.
Take P, any point except A and B in the perimeter of
the base, and through it draw the element PQ.
Draw AB, AP, PB, CD, CQ, QD.
Then AC and BD are = and ||. (Why?)
.-. AD is a ZZ7. (Why?)
Similarly AQ and BQ are CU .
:. AB=CD, AP=CQ, and BP=DQ. (Why?)
/. A APB = A CQD. (Why ?)
Apply the base APB to the base CQD so that AB coin-
cides with CD. Then P will coincide with Q,
(for A APB = & CQD).
But P is any point in the perimeter of the base APB.
.*. every point in the perimeter of the lower base will
coincide with a corresponding point of the perimeter of
the upper base.
.*. the bases will coincide and are equal. Art. 47.
9. £. D.
408 BOOK VIII. SOLID GEOMETRY
693. COR. 1. The sections of a cylinder made by two
parallel planes cutting all the elements arc equal.
For the sections thus formed are the bases of the cylinder
included between the cutting planes.
694. COR. 2. Any section of a cylinder parallel to the
base is equal to the base.
PROPOSITION III. THEOREM
695. The lateral area of a circular cylinder is equal to
the product of the perimeter of a right section of the cylin-
der by an element.
Given the circular cylinder AJ, having its lateral area
denoted by 8, an element by E, and 'the perimeter of a
right section by P.
To prove S=PXE.
Proof. Let a prism with a regular polygon for its base
be inscribed in the cylinder.
Denote the lateral area of the inscribed prism by $',
and the perimeter of its right section by Pf.
Then the lateral edge of the inscribed prism is an ele-
ment of the cylinder. Constr
CYLINDERS 409
Art. 608.
If the number of lateral faces of the inscribed prism be
indefinitely increased,
8' will approach 8 as a limit. Art. 689, 2.
P' will approach P as a limit. Art. 689, 3.
And P'XE will approach PXE as a limit. Art. 253, 2.
But S' = P'XE always. (Why?)
/. S=PXE. (Why?)
Q. £. D.
696. COR. 1. The lateral area of a cylinder of revolu-
tion is equal to the product of the circumference of its base
by its altitude.
697. Formulas for lateral area and total area of a cylin-
der of revolution. Denoting the lateral area of a cylinder of
revolution by 8, the total area by T, the radius by R, and
the altitude by H.
8=2 nRH.
T=2 nRH + 2 nR2 :. T=2 nR (H+ R).
Ex. 1. If, in a cylinder of revolution, H=W in. and R=7 in.,
find 8 and T.
Ex. 2. If the altitude of a cylinder of revolution equals the radius
of the base (H=R), what do the formulas for 8 and T become in
terms of -K ? also, in terms of H ?
Ex. 3. What do they become, if the altitude equals the diameter
of the base ?
Ex. 4. In a cylinder of revolution, what is the ratio of the lateral
area to the area of the base ? to the total area ?
410 BOOK VIII. SOLID GEOMETRY
PROPOSITION IV. THEOREM
698. The volume of a circular cylinder is equal to the
product of its base by its altitude.
Given the circular cylinder AJ, having its volume
denoted by F, its base by B, and its altitude by H.
To prove V=BXH.
Proof. Let a prism having a regular polygon for its
base be inscribed in the cylinder, and denote the volume of
the inscribed prism by T7, and its base by B' '.
The prism will have the same altitude, H, as the
cylinder.
.-. V' = B'XH. (Why?)
If the number of lateral faces of the inscribed prism
be indefinitely increased,
V will approach V as a limit. Art. 689, 1.
B' will approach B as a limit. (Why?)
And B' X H will approach B X H as a limit (Why ?)
But V' = B'XH always, (Why ?)
/. V=BXH. (Why?)
Q. E. D.
699. Formula for the volume of a circular cylinder,
By use of Art. 450,
CYLINDERS
PROPOSITION V. THEOREM
411
700. The lateral areas, or the total areas, of two simi-
lar cylinders of revolution are to each other as the squares
of their radii, or as the squares of their altitudes; and
their volumes are to each other as the cubes of their radii,
or as the cubes of their altitudes.
H'
Given two similar cylinders of revolution having their
lateral areas denoted by S and $', their total areas by T
and T', their volumes by Fand V, their radii by R and
R', and their altitudes by H and R' , respectively.
To prove 8 : S'=T : T' = R2 : R'2 = R2 : H'2-,
and V: Vf = R* : Rf3 = H3 : H'3.
g_^_ R+R
R R' H' + R
8' 2
RXH R ^R
(H+B) = B H+B
(R' + Rf) R' R'
Also
R'
=R2 R =R* ^H3
nR'2Rf ~ R'2 H! Rf3 ~ R'*
Arts. 321, 3fflL
R2
W2
R'2
(Why?)
(Why ?)
Q. E. D.
Ex. If a cylindrical cistern is 12 ft. deep, how much more cement
is required to line it than to line a similar cistern 6 ft. deep ? How
much more water will the former cistern hold ?
412
BOOK VIII. SOLID GEOMETRY
B
Conical surface
CONES
701. A conical surface is a sur-
face generated by a straight line
which moves so as constantly to touch
a given fixed curve, and constantly
pass through a given fixed point.
Thus every shadow cast by a near point
of light is conical in form, that is, is! bounded
by a conical surface of light. Hence, the
study of conical surfaces and solids is im-
portant from the fact that it concerns all
cases of forces radiating from a near point.
702. The generatrix of a conical c\
surface is the moving straight line, as
AAf; the directrix is the given fixed
curve, as ABC-, the vertex is the fixed point, as 0; an
element is the generatrix in any one of its positions, as BB' '.
703. The upper and lower nappes of a conical surface
are the portions above and below the vertex, respectively,
as 0-ABC and 0-A'B'C''.
Usually it is convenient to limit a conical surface to a single nappe.
704. A cone is a solid bounded by a
conical surface and a plane cutting all the
elements.
705. The base of a cone is the face
formed by the cutting plane; the lateral
surface is the bounding conical surface;
the vertex of the cone is the vertex of the
conical surface; the elements of the cone
are the elements of the conical surface;
the altitude of a cone is the perpendicular distance from the
Vertex to the plane of the base.
Oblique circular
cone
CONES 413
706. A circular cone is a cone whose base is a circle.
The axis of & circular cone is the line drawn from the
vertex to the center of the base.
707. A right circular cone is a circular cone whose axis
is perpendicular to the plane of the base.
An oblique circular cone is a circular cone
whose axis is oblique to the base.
708. A cone of revolution is a cone gene-
rated by the revolution of a right triangle
about one of its legs as an axis.
Hence a cone of revolution and a right
circular cone are the same solid. Cone of revolution
709. Properties of a cone of revolution inferred im-
mediately.
1. The altitude, of a cone of revolution is the axis of the
cone.
2. All the elements of a cone of revolution are equal.
710. The slant height of a cone of revolution is any one
of its elements.
711. Similar cones of revolution are cones generated by
similar right triangles revolving about homologous sides.
712. A plane tangent to a cone is a plane which con-
tains one element of the cone, but which does not cut the
conical surface on being produced.
Ex. 1. A plane passing through a tangent to the base of a circular .
cone and the element drawn through the point of contact is tangent to
the cone.
Ex. 2. If a plane is tangent to a circular cone, its intersection with
the plane of the base is tangent to the cone.
414
BOOK VTTT. SOLID GEOMETRY
713. A pyramid inscribed in
a cone is a pyramid whose lateral
edges are elements of the cone
and whose base is a polygon in-
scribed in the base of the cone.
714. A pyramid circum-
scribed about a cone is a pyra-
mid whose lateral faces are tan
gent to the cone and whose base is a polygon circumscribed
about the base of the cone.
715. Properties of circular cones. By Art. 441 the area
of a circle is the limit of the area of an inscribed, or of a
circumscribed polygon, and the circumference is the limit
of the perimeters of these polygons; hence
1. The volume of a circular cone is the limit of the vol-
ume of an inscribed or circumscribed pyramid.
2. The lateral area of a circular cone is the limit of the
lateral area of an inscribed or circumscribed pyramid.
716. A frustum of a cone is the por-
tion of the cone included between the base
of the cone and a plane parallel to the
base.
The lower base of the frustum is the
base of the cone, and the upper base of
the frustum is the section made by the
plane parallel to the base of the cone.
Frustum of a cone.
What must be the altitude and the lateral surface of a
frustum of a cone; also the slant height of the frustum of
a cone of revolution ?
CONES
415
X PROPOSITION VI. THEOREM
717. Every section of a cone made by a plane passing
through its vertex is a triangle ^^ &£ Y^vv-vr^ *»*•*• <<J2
B
Given the cone S-APBQ with a plane passing through
the vertex S9 and making the section SPQ.
To prove SPQ a triangle.
Proof. PQ, the intersection of the base and the cutting
plane, is a straight line. ,
r/°v<i ^*t*^~ffrf" &**- *- 6-~— r»-~
Draw the straight lines SP and SQ.
Then SP and /SQ must be in the cutting plane ; Art. 498.
And be elements of the conical surface. Art. 701.
.'. the straight lines 8P and SQ are the intersections of
the conical surface and the cutting plane.
/. the section SPQ is a triangle,
(for it is bounded by three straight lines).
Art. 81.
Q. £. D.
Ex. What kind of triangle is a section of a right circular cone
made by a plane through the vertex ?
416
BOOK VIII. SOLID GEOMETKY
PROPOSITION VII. THEOREM
718. Every section of a circular cone made by a plane
parallel to the base is a circle.
Given the circular cone SAB with apb a section made
by a plane parallel to the base.
To prove apb a circle.
Proof. Denote the center of the base by 0, and draw
the axis, SO, piercing the plane of the section in o.
Through SO and any element, SP, of the conical sur-
face, pass a plane cutting the plane of the base in the
radius OP, and the plane of the section in op.
In like manner, pass a plane through SO and SB form-
ing the intersections OB and ob.
:. OP\\op, and OB \\ ob. (Why?)
.'. A SPO and SBO are similar to A Spo and Sbo,
respectively.
But
op _ So_ _ °b
OP~ \soj ~ OB'
OP= OB.
.*. op = ob.
:. opb is a circle.
Art. 328.
(Why?)
(Why?)
(Why?)
(Why?)
Q. E. D.
719. COR. The axis of a circular cone passes through
the center of every section parallel to the base.
CONES 417
PROPOSITION VIII. THEOREM
720. The lateral area of a cone of revolution is equal to
half the product of the slant height by the circumference of
the base.
Given a cone of revolution having its lateral area de-
noted by 8, its slant height by L, and the circumference
of its base by (7.
To prove #=J G X L.
Proof. Let a regular pyramid be circumscribed about
the cone.
Denote the lateral area of the pyramid by $', and the
perimeter of its base by P.
Then 8' = J P X L. Art. 642.
If the number of lateral faces of the circumscribed
pyramid be indefinitely increased, .
8' will approach 8 as a limit. Art. 715, 2.
P will approach C as a limit. Art. 441.
And J P XL will approach J CXL as a limit. Art. 253, 2.
But 8' = % P X L always. (Why ?)
/. $=J CXL. (Why?)
Q. E. D.
721. Formulas for lateral area and total area of a cone
of revolution. Denoting the radius of the base by R,
8= i (2 TiR X L) :. S=nRL.
Also T= nRL + nR2 .'. T=nR(L + R).
AA
418
BOOK VIII. SOLID GEOMETRY
PROPOSITION IX. THEOREM
722. The volume of a circular cone is equal to one-third
of the product of its base by its altitude.
Given a circular cone having its volume denoted by V,
its base by B, and its altitude by H.
To prove V= J B X H.
Proof. Let a pyramid with a regular polygon for its
base be inscribed in the given cone.
Denote the volume of the inscribed pyramid by V ', and
its base by B' .
Hence V = J B' X H. Art. esi.
If the number of lateral faces of the inscribed pyramid
be indefinitely increased,
V will approach V as a limit. (Why?)
B' will approach B as a limit. (Why?)
And i Bf X H will approach J B X H as a limit. (Why ?)
But V = $ B'X H always. (Why ?)
/. F=i£ X If. (Why?)
Q. E. D.
723. Formula for the volume of a circular cone.
Ex. 1. If, in a cone of revolution, H=3 and .R=4, find £, T and F.
Ex. 2. If the altitude of a cone of revolution equals the radius of
the base, what do the formulas for S, T and V become ?
CONES 419
PROPOSITION X. THEOREM
724. The lateral areas, or the total areas, of two simi-
lar cones of revolution are to each other as the squares of
their radii, or as the squares of their altitudes, or as the
squares of their slant heights; and their volumes are to each
other as the cubes of these lines.
Given two similar cones of revolution having their
lateral areas denoted by 8 and $', their total areas by T
and T' , their volumes by V and V, their radii by R and
Rf, their altitudes by R and H', and their slant heights by
L and U ', respectively.
To prove 8:8'=T: T' = ft2: R'2 = H2 : H'2 = L2 : £'2;
and V: V' =
T? T. T. 4- 7?
(Why?)
= ^ = =
~ ~ ~
#' TiE'l/ J? J7 £/2 L12 H'2
==TT,X'
T' TCE'dX + .R') R1 L' + Rf R'2 L'2 H'2
(Why?)
^ F=4 n&H! = &**H' = W = H* = lfi' (Why?)
Q. E. D.
725. DEP. An equilateral cone is a cone of revolution
such that a section through the axis is an equilateral
triangle.
420 BOOK VIII. SOLID GEOMETRY
y PROPOSITION XL THEOREM
726. The lateral area of a frustum of a cone of revolu-
tion is equal to one-half the sum of the circumferences of its
bases multiplied by its slant height.
Given a frustum of a cone of revolution having its
lateral area denoted by 8, its slant height by L, the radii
of its bases by R and r, and the circumferences of its bases
by C and c.
To prove fl=J (C+c)XL.
Proof. Let the frustum of a regular pyramid be circum-
scribed about the given frustum. Denote the lateral area
of the circumscribed frustum by S', the perimeter of the
lower base by P, and the perimeter of the upper base by p.
The slant height of the circumscribed frustum is L.
Hence 8' = % (P + p) X L. Art. 643.
Let the pupil complete the proof.
Q. E. D.
727. Formula for the lateral area of a frustum of a cone
of revolution. £=J (2 nR-\- 2 nr) L.
:. S=n (B + r)L.
728. COR. The lateral area of a frustum of a cone of
revolution is equal to the product of the circumference of its
rtiidsection by its slant heiyht.
CONES 421
>^ PROPOSITION XII. THEOREM
729. The volume of Hie frustum of a circular cone is
equivalent to the volume of three cones, whose common alti-
tude is the altitude of the frustum, and whose bases are the
lower base, the upper base, and a mean proportional between
the two bases.
Given a frustum of a circular cone having its volume
denoted by F, its altitude by H, the area of its lower base
by B, and that of its upper base by b.
To prove F = 4 H (B + b + l/BXb).
Proof. Let the frustum of a pyramid with regular poly-
gons for its bases be inscribed in the given frustum.
Denote the volume of the inscribed frustum by F', and the
areas of its bases by B' and bf.
':. F = J H (Bf + b' + VB* X b') . (Why?)
If the number of lateral faces of the inscribed frustum
be indefinitely increased, V will approach F, Br and bf
approach B and b respectively, and Bf X b' approach B X
5, as limits. Art. 715.
Hence, also, B' + b'+T/B' X b' will approach B + b +
V B X 6 as a limit. Art. 253.
But 7/= J H (Br + V + *JB' X V) always. (Why t)
/. V= J H (B + b + l/BXb). (Why f )
Q. E. J>.
422 BOOK VIII. SOLID GEOMETKY
730. Formula for the volume of the frustum of a circu-
lar cone.
F=4 H (nR2 + nr2 + n& X nr2) .
Ex. 1. The measurement of the volume of a frustum of a cone of
revolution reduces to the measurement of the lengths of what straight
lines?
Ex. 2. If a conical oil -can is 12 in. high, how much more tin is
required to make it than to make a similar oil -can 6 in. high ? How
much more oil will it hold ?
Ex. 3. The linear dimensions of a conical funnel are three times
those of a similar funnel. How much more tin is required to make
the first ? How much more liquid will it hold ?
Ex. 4. Make a similar comparison of cylindrical oil -tanks. Of
conical canvas tents.
EXERCISES. CROUP 7O
THEOREMS CONCERNING CYLINDERS AND CONES
Ex. 1. Any section of a cylinder of revolution through its axis is
a rectangle.
Ex. 2. On a cylindrical surface only one straight line can be
drawn through a given point.
[SuG. For if two straight lines could be drawn, etc.]
Ex. 3. The intersection of two planes tangent to a cone is a
straight line through the vertex.
Ex. 4. If two planes are tangent to a cylinder, their line of inter-
section is parallel to an element of the cylinder.
[Sue. Pass a plane J_ to the elements of the cylinder.]
Ex. 5. If tangent planes be passed through two diametrically
opposite elements of a circular cone, these planes intersect in a
straight line through the vertex and parallel to the plane of the base.
Ex. 6. In a cylinder of revolution the diameter of whose base
equals the altitude, the volume equals one-third the product of the
total surface by the radius of the base.
EXEECISES ON THE CYLINDER AND CONE 423
Ex. 7. A cylinder and a cone of revolution have the same base
and the same altitude. Find the ratio of their lateral surfaces, and also
of their volumes.
Ex. 8. If an equilateral triangle whose side is a be revolved about
one of its sides as an axis, find the area generated in terms of a.
Ex. 9. If a rectangle whose sides are a and & be revolved first
about the side a as an axis, and then about the side &, find the ratio of
the lateral areas generated, and also of the volumes.
Ex. 10. The bases of a cylinder and of a cone of revolution are
concentric. The two solids have the same altitude, and the diameter
of the base of the cone is twice the diameter of the base of the cylin-
der. What kind of line is the intersection of their lateral surfaces,
and how far is it from the base ?
Ex. 11. Determine the same when the radius of the cone is three
times the radius of the cylinder. Also when r times.
Ex. 12. Obtain a formula in terms of r for the volume of the
frustum of an equilateral cone, in which the radius of the upper base
is r and that of the lower base is 3r.
Ex. 13. A regular hexagon whose side is a revolves about a diag-
onal through the center as axis. Find, in terms of a, the surface
and volume generated.
Ex. 14. Find the locus of a point at a given distance from a given
straight line.
Ex. 15. Find the locus of a point whose distance from a given
line is in a given ratio to its distance from a fixed plane perpen-
dicular to the line.
Ex. 16. Find the locus of all straight lines which make a given
angle with a given line at a given point.
Ex. 17. Find the locus of all straight lines which make a given
angle with a given plane at a given point.
Ex. 18. Find the locus of all points at a given dis-
tance from the surface of a given cylinder of revolution.
Ex. 19. Find the locus of all points at a given dis-
tance from the surface of a given cone of revolution.
424 BOOK VIII. SOLID GEOMETRY
EXERCISES. CROUP 71
PROBLEMS CONCERNING THE CYLINDER AND CONE
Ex. 1. Through a given element of a circular cylinder, pass a
plane tangent to the cylinder.
Ex. 2. Through a given element of a circular cone, pass a plane tangc pt
to the cone.
Ex. 3. About a given circular cylinder circumscribe a prism, with
a regular polygon for its base.
Ex. 4. Through a given point outside a circular cylinder, pass a
plane tangent to the cylinder.
Ex. 5. Through a given point outside a given circular cone, pass
a plane tangent to the cone.
[Suo. Through the vertex of the cone and the given point pass a
line, and produce it to meet the plane of the base.]
Ex. 6. Into what segments must the altitude of a cone of revolution
be divided by a plane parallel to the base, in order that the volume of
the cone be bisected ?
Ex. 7. Divide the lateral surface of a given cone of revolution into
two equivalent parts by a plane parallel to the base.
Ex. 8. If the lateral surface of a cylinder of revolution be cut
along one element and unrolled, what sort of a plane figure is formed ?
Hence, out of cardboard construct a cylinder of revolution with
given altitude and given circumference.
Ex. 9. If the lateral surface of a cone of . revolution be cut along
one element and unrolled, what sort of a plane figure is formed ?
Hence, out of cardboard construct a cone of revolution of given
slant height.
Ex. 10. Construct an equilateral cone out of pasteboard.
Ex. 11. Construct a frustum of a cone of revolution out of paste-
board.
BOOK IX
THE SPHERE
731. A sphere is a solid bounded by a surface all
points of which are equally distant from a point within
called the center.
732. A sphere may also be defined as a solid generated
by the revolution of a semicircle about its diameter as an
axis.
. Some of the properties of a sphere may be obtained more readily
from one of the two definitions given, and some from the other.
A sphere is named by naming the point at its center, or by naming
three or more points on its surface.
733. A radius of a sphere is a line drawn from the
center to any point on the surface.
A diameter of a sphere is a line drawn through the
center and terminated at each end by the surface of the
sphere.
734. A line tangent to a sphere is a line having but one
point in common with the surface of the sphere, however
far the line be produced.
(425)
426
BObK IX. SOLID GEOMETKY
735. A plane tangent to a sphere is a plane having but
one point in common with the surface of the sphere, how-
ever far the plane be produced.
736. Two spheres tangent to each other are spheres
whose surfaces have one point, and only one, in common.
737. Properties of a sphere inferred immediately.
1. All radii of a sphere, or of equal spheres, are equal.
2. All diameters of a sphere, or of equal spheres, are
equal.
3. Two spheres are equal if their radii or their diame-
ters are equal.
PROPOSITION I. THEOREM
738. A section of a sphere made by a plane is a .circle.
Given the sphere 0, and PCD a section made by a plane
cutting the sphere.
To prove that PCD is a circle.
Proof. From the center 0, draw OA J_ the plane of
the section.
THE SPHERE 427
Let 0 be a fixed point on the perimeter of the section,
and P any other point on this perimeter.
Draw A C, AP, OC, OP.
Then the A GAP and OAC are rt. A. Art. 505.
OP= OC. (Why ?)
OA=OA. (Why?)
A0AP=AOJLO. (Why?)
.'. AP=AC. (Why?)
But P is any point on the perimeter of the section PCD.
.'. every point on this perimeter is at the distance AC
from A.
:. PCD is a circle with center A. Art. 197.
Q. E. D.
739. COR. 1. Circles which are sections of a sphere
made by planes equidistant from the center are equal; and
conversely.
740. COR. 2. Of two circles on a sphere, the one made
by a plane more remote from the center is smaller; and
conversely.
741. DEF. A great circle of a sphere is a circle whose
plane passes through the center of the sphere.
742. DEF. A small circle of a sphere is a circle whose
plane does not pass through the center of the sphere.
743. DEF. The axis of a circle of a sphere is the
diameter of a sphere which is perpendicular to the plane of
the circle. Thus, on figure p. 426, BB' is the axis of PCD.
744. DEF. The poles of a circle of a sphere are the
extremities of the axis of the circle. Thus, B and B', of
figure p. 426, are poles of the circle PCD.
428 BOOK IX. SOLID GEOMETRY
745. Properties of circles of a sphere inferred imme-
diately.
1. The axis of a circle of a sphere passes through the
center of the circle; and conversely.
2. Parallel circles have the same axis and the same poles.
3. All great circles of a sphere are equal.
1 great circle on a sphere bisects the sphere and
its surface.
5. Two great circles on a sphere bisect each other.
For the line of intersection of the two planes of the
circles passes through the center, and hence is a diameter
of each circle.
6. Through two points (not the extremities of a diameter)
on the surface of a sphere, one, and only one, great circle can
be passed.
For the plane of the great circle must also pass through
the center of the sphere (Art. 741), and through three
points not in a straight line only one plane can be passed
(Art. 500).
7. Through any three points on the surface of a sphere,
not in the same plane with the center, one small circle, and
only one, can be passed.
746. DEF. The distance between two points on the sur-
face of a sphere is the length of the minor arc of a great
circle joining the points.
Ex. I. If the radius of a sphere is 13 in., find the radius of a
circle on the sphere made by a plane at a distance of 1 ft. from the
center.
Ex. 2. What geographical circles on the earth's surface are great,
and what small circles ?
Ex. 3. What is the largest number of points in which two circles
on the surface of a sphere can intersect ? Why ?
THE SPHERE
429
PROPOSITION II. THEOREM
747. All points in the circumference of a circle of a
sphere are equally distant from each pole of the circle.
Given ABC a circle of a sphere, and P and Pf its poles.
To prove the arcs PA, PB, PC equal, and arcs P'A,
P'B, P'G equal.
Proof. Draw the chords PA, PB, PC.
The chords PA, PB and PC are equal. Art. 518.
/. arcs PA, PB and PC are equal. Art. 218.
In like manner, the arcs P'A, P'B and P'C may be
proved equal.
Q. £. D.
748. DEF. The polar distance of a small circle on a
sphere is the distance of any point on the circumference of
the circle from the nearer pole.
The polar distance of a great circle on a sphere is the dis-
tance of any point on the circumference of the great circle
from either pole.
749. COR. The polar distance of a great circle is the
quadrant of a great circle.
430 BOOK IX. SOLID GEOMETRY
PROPOSITION III. THEOREM
750. If a point on the surface of a sphere is at a quad-
rant's distance from two other points on the surface, it is
the pole of the great circle through those points.
Given PB and PC quadrants on the surface of the
sphere 0, and ABC a great circle through B and C.
To prove that P is the pole of ABC.
Proof. From the center 0 draw the radii OB, OC, OP.
The arcs PB and PC are quadrants. (Why ?)
.'. A POB and POC are rt. A . (Why ?)
/. PO I. plane ABC. (Why?)
.'. P is the pole of the great circle ABC. (Why ?)
Q. £. D.
751. COR. Through two given points on the surface of
a sphere to describe a great circle.
Let A and B be the given points.
From A and B as centers, with a
quadrant as radius, describe arcs
on the surface of the sphere inter-
secting at P. With P as a center
and a quadrant as a radius, describe a great circle.
THE SPHERE 431
•
PROPOSITION IV. THEOREM
752. A plane perpendicular to a radius at its extremity
is tangent to the sphere.
Given the sphere 0, and the plane MN J. the radius OA
of the sphere at its extremity A.
To prove MN tangent to the sphere.
Proof. Take P any point in plane MN except A. Draw
OP. Then OP > OA. (Why?)
.*. the point P is outside the surface of the sphere.
But P is any point in the plane MN except A.
.'.plane MN is tangent to the sphere at the point A,
(for every paint in the plane, except A, is outside the surface of the sphere).
Art. 735. Q. E. D.
753. COR. 1. A plane, or a line, which is tangent to a
sphere, is perpendicular to the radius drawn to the point of
contact. Also, if a plane is tangent to a sphere, a perpendicu-
lar to the plane at its point of contact passes through the center
of the sphere.
754. COR. 2. A straight line perpendicular to a radius
of a sphere at its extremity is tangent to the sphere.
755. COR. 3. A straight line tangent to a circle of a
sphere lies in the plane tangent to the sphere at the point of
contact.
432 BOOK IX. SOLID GEOMETKY
756. COR. 4. A straight line drawn in a tangent plane,
and through the point of contact is tangent to the sphere at
that point.
757. COR. 5. Two straight lines tangent to a sphere at
a given point determine the tangent plane at that point.
758. DEP. A sphere circumscribed about a polyhedron
is a sphere in whose surface lie all the vertices of the
polyhedron.
759. DEF. A sphere inscribed in a polyhedron is a
sphere to which all the faces of the polyhedron are tangent.
PROPOSITION V. PROBLEM
760. To circumscribe a sphere about a given tetra-
hedron.
Given the tetrahedron ABCD.
To circumscribe a sphere about ABCD.
Construction and Proof. Construct E and F the centers
of circles circumscribed about the A ABC and BCD, re-
spectively. Art. 286.
Draw EH _L plane ABC and FK _L plane BCD. Art, 514.
Draw EQ and FG to <? the midpoint of
THE SPHERE 433'
Then EG and FG are JL EG. Art. 113.
/. plane EGF _L BC. Art. 509.
/. plane EGF JL plane JU30. Art. 555.
/. EH lies in the plane FGE. Art. 558.
In like manner FK lies in the plane FGE.
The lines EG and .F& are not ||,
(/or Mey TweeZ in the point G).
:. the lines EH and JPfiT are not j|. Art. 122.
Hence EH must meet FK in some point 0.
But EH is the locus of all points equidistant from A,
B and (7; and -FZT is the locus of all points equidistant
from B, G and D. Art. 520.
/. 0, which is in both EH and FK, is equidistant from
A, B, CandD. (Why?)
Hence a spherical surface constructed with 0 as a center
and OA as a radius will pass through A, B, C and D, and
form the sphere required. Q. E. r.
761. COB. 1. Four points not in the same plane deter-
mine a sphere.
762. COR. 2. The four perpendiculars erected at the
centers of the faces of a tetrahedron meet in a point.
763. COR. 3. The six planes perpendicular to the edges
of a tetrahedron at their midpoints intersect in a point.
764. DEF. An angle formed by two curves is the angle
formed by a tangent to each curve at the point of inter-
section.
765. DEF. A spherical angle is an angle formed by
two intersecting arcs of great circles on a sphere, and
hence by tangents to these arcs at the point of intersection.
BB
434 BOOK IX. SOLID GEOMETRY
PROPOSITION VI. PROBLEM
766. To inscribe a sphere in a given tetrahedron.
Given the tetrahedron ABGD.
To inscribe a sphere in ABCD.
Construction and Proof. Bisect the dihedral angle D-
A B- C by the plane GAB; similarly bisect the dihedral A
whose edges are BG and AG by the planes GBG and GAG,
respectively.
Denote the point in which the three bisecting planes
intersect by 0.
Every point in the plane GAB is equidistant from the
faces DAB and GAB. . Art. 562.
Similarly, every point in GBG is equidistant from the
two faces intersecting in BG, and every point in GAG is
equidistant from the two faces intersecting in AG.
:. 0 is equidistant from all four faces of the tetra-
hedron. Ax. 1.
Hence, from 0 as a center, with the JL from 0 to any
one face as a radius, describe a sphere.
This sphere will be tangent to the four faces of the
tetrahedron and .'. inscribed in the tetrahedron. Art. 759.
Q. E. F.
767. COR. The planes bisecting the six dihedral angles
of a tetrahedron meet in one point.
THE SPHEKE 435
PROPOSITION VII. PROBLEM
768. To find the radius of a given material sphere.
p
{\
:^B I \ \ *j ~V
' *--:>- jjs
\ s^\ | /
2*~ ''' . P'
Fig-. 2 ^8' 3
Given the material sphere 0.
To construct the radius of the sphere.
Construction. With any point P (Fig. 1) of the surface
of the sphere as a pole, describe any convenient circum-
ference on the surface.
On this circumference take any three points A, B and C.
Construct the A ABC (Fig. 2) having as sides the three
chords AB, BC, AC, obtained from Fig. 1, by use of the
compasses. Art. 283.
Circumscribe a circle about the A ABC. Art. 286.
Let KB be the radius of this circle.
Construct (Fig. 3) the right A Jcpb, having for hypot-
enuse the chord pb (Fig. 1) and the base kb. Art. 284.
Draw bp' J_ bp and meeting pk produced at p' .
Bisect ppf at 0.
Then op is the radius of the given sphere.
Proof. Let the pupil supply the proof,
Q. E. F.
436 BOOK IX. SOLID GEOMETRY
PROPOSITION VIII. THEOREM
769. The intersection of two spherical surfaces is the
circumference of a circle whose plane is perpendicular to the
line joining the centers of the spheres, and whose center is w
that line.
Given two intersecting © 0 and 0' which, by rotation
about the line 00' as an axis, generate two intersecting
spherical surfaces.
To prove that the intersection of the spherical surfaces
is a O, whose plane J_ 00', and whose center lies in 00'.
Proof. Let the two circles intersect in the points P and
Q, and draw the common chord PQ,
Then, as the two given © rotate about 00' as an axis,
the point P will generate the line of intersection of the two,
spherical surfaces that are formed.
But PR is constantly J_ 00'. Art. 241,
/. PR generates a plane A. 00f Art. 510.
Also PR remains constant in length.
.*. P describes a circumference in that plane. Art. 197.
Hence the intersection of two spherical surfaces is a O ,
whose plane J_ the line of centers, and whose center is in
the line of centers.
Q. E. D.
The above demonstration is an illustration of the use of the second
definition of a sphere (Art. 732).
THE SPHERE 437
PROPOSITION IX. THEOREM
770. A spherical angle is measured by the arc of a great
circle described from the vertex of the angle as a pole, and
included between its sides, produced, if necessary.
Given /.BAG a spherical angle formed by the intersec-
tion of the arcs of the great circles BA and CA, and BO
an arc of a great circle whose pole is A.
To prove /.BAG measured by arc BC.
Proof. Draw AD tangent to AB, and AF tangent to
AC. Also draw the radii OB and OG.
Then AD _L AO. Art. 230.
Also OB _L AO ( for AB is a quadrant) .
:. OB \\AD. (WhyT)
Similarly OG\\AF.
:. Z BOG= Z DAP Art. 538.
But /.BOG is measured by arc BG. Art. 257.
/. /.DAF, that is, /.BAG, is measured by arc BG.
Us- (Why?)
Q. E. D.
771. COR. A spherical angle is equal to the plane angle
of the dihedral angle formed by the planes of its sides.
438 BOOK IX. SOLID GEOMETRY
SPHERICAL TRIANGLES AND POLYGONS
772. A spherical polygon is a portion of the surface of
a sphere bounded by three or more
arcs of great circles, as ABCD.
The sides of the spherical polygon
are the bounding arcs; the vertices
are the points in which the sides in-
tersect; the angles are the spherical
angles formed by the sides.
The sides of a spherical polygon are usually limited to arcs less
than a semicircumference.
773. A spherical triangle is a spherical polygon of three
sides.
Spherical triangles are classified in the same way as
plane triangles; viz., as isosceles, equilateral, scalene,
right, obtuse and acute.
774. Relation of spherical polygons to polyhedral angles.
If radii be drawn from the center of a sphere to the ver-
tices of a spherical polygon on its surface (as OA, OB,
etc., in the above figure), a polyhedral angle is formed at
O, which has an important relation to the spherical poly-
gon ABCD
Each face angle of the polyhedral angle equals (in num-
ber of degrees contained) the corresponding side of the spheri-
cal polygon;
Each dihedral angle of the polyhedral angle equals the
corresponding angle of the spherical polygon.
Hence, corresponding to each property of a polyhedral
angle, there exists a property of a spherical polygon, and
conversely.
THE SPHERE 439
Hence, also, a trihedral angle and its parts correspond
to a spherical triangle and its parts.
Of the common properties of a polyhedral angle and a spherical
polygon, some are discovered more readily from the one figure and
some from the other. In general, the spherical polygon is simpler
to deal with than a polyhedral angle. For instance, if a trihedral
angle were drawn with the plane angles of its dihedral angles, nine
lines would be used, forming a complicated figure in solid space;
whereas, the same magnitudes are represented in a spherical triangle
by three lines in an approximately plane figure.
On the other hand, the spherical polygon, because of its lack of
detailed parts, is often not so suggestive of properties as the poly-
hedral angle.
PROPOSITION X. THEOREM
775. The sum of two sides of a spherical triangle is
greater than the third side.
Given the spherical triangle ABC, of which no side is
larger than AB.
To prove AC+ BC > AB.
Proof. From the center of the sphere, 0, draw the radii
OA, OB, OC.
Then, in the trihedral angle 0-ABC,
£AOC+£BOC > AOB. Art. 582.
/. AC-}-BG > AB. Art. 774.
Q. £. D.
440 BOOK IX. SOLID GEOMETRY
776. COR. 1. Any side of a spherical triangle is greater
than the difference between the other two sides.
111. COR. 2. The shortest path between two points on
the surface of a sphere is the arc of the great circle joining
those points.
For any other path between the two points may be
made the limit of a series of arcs of great circles connect-
ing successive points on the path, and the sum of this
series of arcs of great circles connecting the two points is
greater than the single arc of a great circle connecting
them.
PROPOSITION XI. THEOREM
778. The sum of the sides of a spherical polygon is less
than 360°.
Given the spherical polygon ABCD.
To prove the sum of the sides of ABCD < 360°.
Proof. From 0, the center of the sphere, draw the radii
OA, OB, OC, OD.
Then Z AOB + Z.BOC + Z COD-}- ^DOA < 300°.
(Why ?)
/. AB + BG + GD + DA < 360° . Art. 774.
Q. £. D.
SPHEKICAL TKIANGLES
441
D
A'
779. DEF. The polar p' D'
triangle of a given triangle
is the triangle formed by
taking the vertices of the
given triangle as poles, and
describing arcs of great cir-
cles. (Hence, if each pole
be regarded as a center, the radius used in describing each
arc is a quadrant.) Thus A'B'C 'is the polar triangle of
ABC-, also D'E'F' is the polar triangle of DEF.
PROPOSITION XII. THEOREM
780. // one spherical triangle is the polar of another,
then the second triangle is the polar of the first.
Given A'B'C' the polar triangle of ABC.
To prove ABC the polar triangle of A'B'C'.
Proof. B is the pole of the arc A'C'. Art. 779.
.*. arc A'B is a quadrant. (Why ?)
Also C is the pole of the arc A'B'. (Why ?)
.*. arc A'C is a quadrant. (Why ?)
/. A' is at a quadrant's distance from both B and C.
/. Af is the pole of the arc BC. Art. 750.
In like manner it may be shown that B' is the pole of
AC, and C' the pole of AB. q. E. j>.
442 BOOK IX. SOLID GEOMETRY
PROPOSITION XIII. THEOREM
781. In a spherical triangle and its polar, each angle of
one triangle is the supplement of the side opposite in the
other triangle.
A'
Given the polar A ABC and A'B'C' with the sides of
ABC denoted by a, &, c, and the sides of A'B'C' denoted
by a', &', c', respectively.
To prove A + a'= 180°, B + &' = 180°, C + c' = 180°f
A' + a = 180°, JB' + 6 = 180°, (7 + c = 180°.
Proof. Produce the sides AB and A G till they meet
B'O in the points D and .F, respectively.
Then B' is the pole of AF :. arc 5^=90°. Art. 780.
Also C' is the pole of AD :. arc 0/D = 90°. (Why?)
Adding, £'^ + (71)= 180°. (Why?)
Or J3'.F+ ^O7 + D^-1800. Ax. 6.
Or JB/C/ + D^T-180°.
But B'C' = a', and D^is the measure of the Z A. Art. 770.
/. A + a' = 180°.
In like manner the other supplemental relations may be
proved as specified.
Q. E. D.
782. DEF. Supplemental triangles are two spherical
triangles each of which is the polar triangle of the other.
This new name for two polar triangles is due to the
property proved in Art. 781.
SPHERICAL TRIANGLES 443
PROPOSITION XIV. THEOREM
783. The sum of the angles of a spherical triangle is
greater than 180°, and less than 540°.
Given the spherical triangle ABO.
To prove A + B -f G > 180° and < 540°.
Proof. Draw A'B'C', the polar triangle of ABC, and
denote its sides by «', b' ', c' .
Then ^ + a' = 180° )
J? _j- £' — 180° f Art. 781.
(7 _j_ c' = 180° )
.-. A + J5 + (7 + «' + ^ + c' = 540°. . . (1) AX. 2.
But C a' + £' -f cr < 360° Art. 778.
{ a' + J' + cr > 0°
Subtracting each of these .in turn from (1),
^ _j_ B + C > 180° and < 540°. Ax. 11.
Q. E. D.
784. COR. A spherical triangle may have one, two or
three right angles; or it may have one, two or three obtuse
angles.
785. 'DEF. A birectangular spherical triangle is a spher-
ical triangle containing two right angles.
786. DEF. A trirectangular spherical triangle is a
spherical triangle containing three right angles.
444
BOOK IX. SOLID GEOMETRY
787. COR. The surface of a sphere may be divided into
eight trirectangular spherical triangles. For let three planes
J_ to each other be passed through the center of a sphere, etc.
788. DBF. The spherical excess of a spherical triangle
is the excess of the sum of its angles over 180°.
789. DEF. Symmetrical spherical triangles are triangles
which have their parts equal, but arranged in reverse order.
B'
Fig. 2
Fig-. 3
Fig.l
' Three planes passing through the center of a sphere
form a pair of symmetrical spherical triangles on opposite
sides of the sphere (see Art. 580), as A ABC and A'B'C*
of Fig. 1.
790. Equivalence of symmetrical spherical triangles.
Two plane triangles which have their parts equal, but ar-
ranged in reverse
order, may be
made to coincide
by lifting up one
triangle, turning
it over in space, and placing it upon the other triangle.
But two symmetrical spherical triangles cannot-be made
to coincide in this way, because of the curvature of a
spherical surface. Hence the equivalence of two sym-
metrical spherical triangles must be demonstrated in some
indirect way.
SPHE1UCAL TRIANGLES
445
791. Property of symmetrical spherical triangles. Two
isosceles symmetrical spherical triangles are equal, for they
can be made to coincide.
PROPOSITION XV. THEOREM
792. Two symmetrical spherical triangles are equivalent.
Given the symmetrical spherical A ABC and A'B'C',
formed by planes passing through 0, the center of a sphere.
(See Art. 789.)
To prove A ABC=a= A A'B'C'.
Proof. Let Pbe the pole of a small circle passing through
the points A, J3, C. Draw the diameter POP'.
Also draw PA, PB, PC, P'A', P'B', P'C', all arcs of
great ©.
Then PA = PB = PC. Art. 747.
Also P'A' = PA, P'B' = PB, P'C' = PC.
Arts. 78, 215.
; .% P'A' = P'B' =P'C'. Ax. 1.
Hence PAB and P'A'B' are symmetrical isosceles A.
Similarly A PA C= A P'A'G'. Art. 791.
And A PBC= A P'E'G'}
Adding A PAB+ A PAC+ A PEG
^ A P*A'B'+ A PA'O+ A P'B' a. Ax. 2.
Or A ABC- A A'B'C'. Ax. 6.
In case the poles P and P' fall outside the A A.B<7 and
A'B'C1 ', let the puuil suoply the demonstration. Q. E. D.
446 BOOK IX. SOLID GEOMETRY
PROPOSITION XVI. THEOREM
793. On the same sphere, or on equal spheres, two tri-
angles are equal,
I. // two sides and the included angle of one are equal to
two sides and the included angle of the other; or
II. If two angles and the included side of one are equal
to two angles and the included side of the other,
the corresponding equal parts being arranged in the same
order in each case.
I. Given the spherical A ABC and DEF, in which
AC=DF, CB=FE, and Z<7= /.F.
To prove A ABC= A DEF.
Proof. Let the pupil supply the proof (see Book I,
Prop. VI).
II. Given the spherical A ABC and DEF, in which
To prove A ABC= A DEF.
Proof. Let the pupil supply the proof (see Book I,
Prop. VII) . _
Ex. 1. If the line of centers of two spheres is 10 in., and the radii
are 12 in. and 3 in., how are the spheres situated with reference to
each other ?
Ex. 2. The tank on a motor car is a cylinder 35 inches long and
15 inches in diameter. How many gallons of gasolene will it hold ?
Ex. 3. In an equilateral cone, find the ratio of the lateral area to
the area of the base.
SPHERICAL TRIANGLES 447
PROPOSITION XVII. THEOREM
794. On the same sphere, or on equal spheres, two tri-
angles are symmetrical and equivalent,
I. If two sides and the included angle of one are equal to
two sides and the included angle of the other; or
II. // two angles and the included side of one are equal
to two angles and the included side of the other,
the corresponding equal parts being arranged in reverse
order.
A
I. Given the spherical A ABC and DEF, in which AB
= DE, AC = DF, and /.A = /.D, the corresponding parts
being arranged in reverse order.
To prove A ABC symmetrical with A DEF.
Proof. Construct the &D'E'F' symmetrical with ADEF.
>t
Then A ABC may be made to coincide with A D'E'F',
Art. 793.
(having two sides and the included Z equal and arranged in the
same order).
But A D'E'F' is symmetrical with the A DEF.
:. A ABC, which coincides with A D'E'F', is symmet-
rical with A DEF.
II. The second part of the theorem is proved in the
same way.
Q. £. D.
448 BOOK IX. SOLID GEOMETRY
PROPOSITION XVIII. THEOREM
795. If two triangles on the same sphere, or equal
spheres, are mutually equilateral, they are also mutually
equiangular, and therefore equal or symmetrical.
Given two mutually equilateral spherical A ABC and
A'B'C' on the same or on equal spheres.
To prove A ABC and A'B'C' equal or symmetrical.
Proof. From 0 and 0', the centers of the spheres to
which the given triangles belong, draw the radii OA, OB,
OC, O'A', 0'Bf, O'C'.
Then the face A at 0 = corresponding face A at 0' '.
Art. 774.
Hence dihedral A at 0= corresponding dihedral A at Of.
Art. 584.
.*. A of spherical A A BC= homologous A of spherical
& A'B'C'. Art. 774.
.*. the A ABC and A'B'C' are equal or symmetrical,
according as their homologous parts are arranged in the
same or in reverse order. Art. 789.
796. NOTE. The conditions in Props. XVI and XVIII which make
two spherical triangles equal are the same as those which make
two plane triangles equal. Hence many other propositions occur in
spherical geometry which are identical with corresponding proposi-
tions in plane geometry. Thus, many of the construction problems of
spherical geometry are solved in the same way as the corresponding
construction problems in plane geometry j as, to bisect a given
angle, etc,
SPHERICAL TRIANGLES 449
PROPOSITION XIX. THEOREM
797. // two triangles on the same sphere are mutually
equiangular, they are also mutually equilateral, and there-
fore equal or symmetrical.
Given the mutually equiangular spherical A Q and Q*
on the same sphere or on equal spheres.
To prove that Q and Qf are mutually equilateral, and
therefore equal or symmetrical.
Proof. Construct P and Pf the polar A of Q and #',
respectively.
Then A P and Pf are mutually equilateral. Art. 781.
/. A P and Pf are mutually equiangular. Art. 795.
But Q is the polar A of P, and Q' of P' '. Art. 780.
/. A Q and Q' are mutually equilateral. Art. 781.
Hence Q and Q' are equal or symmetrical, according as
their homologous parts are arranged in the same or in
reverse order. Art. 789.
Q. E. D.
798. COR. If two mutually equiangular triangles are
on unequal spheres, their corresponding sides have the same
ratio as the radii of their respective spheres.
CC
450
BOOK IX. SOLID GEOMETRY
PROPOSITION XX. THEOREM
799. In an isosceles spherical triangle the angles oppo<
site the equal sides are equal.
Given the spherical A ABC in which AB=AC.
To prove Z£=ZO.
Proof. Draw an arc from the vertex A to D, the mid-
point of the base.
Let the pupil supply the remainder of the proof.
PROPOSITION XXI. THEOREM (CONY. OF PROP. XX)
800. If two angles of a spherical triangle are equal, the
sides opposite these angles are equal, and the triangle is
isosceles.
A.
Given the spherical A ABC in which £B= Z G.
To prove AB = AC.
Proof. Construct A A'B'C' the polar A of ABC.
Then A'C' = A'B'. Art. 781.
/. Z C'= Z-B'. Art. 799.
.'. AB = AG. Art. 781.
Q. £. D.
SPHEKICAL TRIANGLES 451
PROPOSITION XXII. THEOREM
801. In any spherical triangle, if two angles are un-
equal, the sides opposite these angles are unequal, and the
greater side is opposite the greater angle, and CONVERSELY^,
Given the spherical A ABC in which Z BAG is greater
than Z.C.
To prove BC > BA.
Proof. Draw the arc AD making /.D AC equal to Z C.
Then DA = DC. Art. 800.
To each of these equals add the arc BD.
:. BD + DA = BD + DC, or EG. (Why f)
But in A BDA, BD + DA > BA. (Why ?)
/. BC > BA. . Ax. 8.
Let the pupil prove the converse by the indirect method
(see Art. 106).
Q. E. D.
Ex. 1. Bisect a given spherical angle.
• Ex. 2. Bisect a given arc of a great circle on a sphere.
Ex. 3. At a given point in an arc on a sphere, construct an angle
equal to a given spherical angle on the same sphere.
Ex. 4. Find the locus of the centers of the circles of a sphere
formed by planes perpendicular to a given diameter of the given
sphere.
452
BOOK IX. SOLID GEOMETRY
SPHERICAL AREAS
802. Units of spherical surface. A spherical surface
may be measured in terms of, either
1. The customary units of area, as a square inch, a
square foot, etc., or
2. Spherical degrees, or spherids.
803. A spherical degree, or spherid, is one -ninetieth
part of one of the eight trirectangular triangles into which
the surface of a sphere may be divided (Art. 787), or TTO
part of the surface of the entire sphere.
A solid degree is one-ninetieth part of a trirectangular angle (see
Art. 774).
804. A lune is a portion of the surface of a sphere
bounded by two semicircumferences
of great circles, as PBP'C of Fig. 1,
The angle of a lune is the angle
formed by the semicircumferences
which bound it, as the angle BPC.
805. A zone is the portion of the
surface of the sphere bounded by
two parallel planes.
A zone may also be defined as the sur-
face generated by an arc of a revolving
semicircumference. Thus, if QFQ' (Fig. 2)
generates a sphere by rotating about QQf,
its diameter, any arc of QFQ', as EF,
generates a zone.
806. A zone of one base is a zone
one of whose bounding planes is
tangent to the sphere, as the zone
generated by the arc QE of Fig. 2.
807. The altitude of a zone is the perpendicular dis-
tance between the bounding planes of the zone.
The bases of a zone are the circumferences of the circles
of the sphere formed by the bounding planes of the zone.
SPHERICAL AREAS
453
PROPOSITION XXIII. THEOREM
0
808. The area generated ~by a straight line revolving
about an axis in its plane is equal to the projection of the
line upon the axis, multiplied by the circumference of a circle
ivhose radius is the perpendicular erected at the midpoint of
the line and terminated by the axis.
Given AB and XYin the same plane, CD the projection
of AB on JTY, P# the -L bisector of A£; and a surface
generated by the revolution of AB about XT, denoted as
To prove area AB=CDX2 nPQ.
Proof. 1. In general, the surface generated by AB is
the surface of a frustum of a cone (Fig. 1).
/. area AB = AB X 2 nPR.
Draw AF LED, then
A ABF and PQR are similar.
/. A£: AF=PQ: PR.
:. AB X PR = AFX PQ, or CD X PQ.
Substituting, area AB= CD X 2 nPQ.
2. If AB\\XY (Fig. 2), the surface generated by AB
is the lateral surface of a cylinder.
/. area AB=CDX2 nPQ. Art. 697.
3. If the point A lies in the axis XY (Fig. 3), let the
pupil show that the same result is obtained. p. E. p,
Art. 728.
Art. 328.
(Why?)
(Why?)
Ax. 8.
454
BOOK IX. SOLID GEOMETKY
PROPOSITION XXIV. THEOREM
809. The area of the surface of a sphere is equal to the
product of the diameter of me sphere by the circumference
of a great circle.
A
Given a sphere generated by the revolution of the semi-
circle ACE about the diameter AE, with the surface of the
sphere denoted by S, and its radius by R.
To prove 8=AE X 2 nR.
Proof. Inscribe in the given semicircle the half of a
regular polygon of an even number of sides, as ABCDE.
Draw the apothem to each side of the semipolygon, and
denote it by a.
From the vertices B, C, D draw Ja to AE.
Then area AB = AF X 2 na.
area BC= FOX 2 na.
area CD=OKX2 na.
area DE=KEX 2 na.
Adding, area ABCDE=AE X 2 na.
If, now, the number of sides of the polygon be indefi-
nitely increased,
area ABCDE approaches 8 as a limit. Art. 441.
And a approaches R as a limit. (Why?)
.*. AE X 2 na approaches AE X 2 nR as a limit, f Why ?)
J3ut area ABCDE '=AE X 2 na always.
,-, $=AEX2 nil. (WhyT)
. E. 9.
Art. 808.
SPHERICAL AREAS 455
810. Formulas for area of surface of a sphere.
Substituting for AE its equal 2 R, #=4 nR2.
Also denoting the diameter of the sphere by D, R= J D.
811. Con. 1. T/ie surface of a sphere is equivalent to four
times the area of a great circle of the sphere.
812. COR. 2. The areas of the surfaces of tivo spheres
are to each other as the squares of their radii, or of their
diameters.
For, if S and $' denote the surfaces, R and R' the radii,
and D and Df the diameters of two spheres,
8 = 4KR2 =R2 8 = KD2 = D2
S' InR'2 R'2' !° 8' TiD'2 D'2'
813. Property of the sphere. The following property of
the sphere is used in the proof of Art. 809 : //, in the generat-
ing arc of any zone, a broken line be inscribed, whose vertices
divide the arc into equal parts, then, as the number of these
parts is increased indefinitely, the area generated by the broken
line approaches the area of the zone as a limit. Hence
COR. 3. The area of a zone is equal to the circumference
of a great circle multiplied by the altitude of the zone.
Thus the area generated by the arc BC = FOX2 rtR.
814. COR. 4. On the same sphere, or on equal spheres, the
areas of two zones are to each other as the altitudes of the
zones. _
Ex. 1. Find the area of a sphere whose diameter is 10 in.
Ex. 2. Find the area of a zone of altitude 3 in., on a sphere whose
radius is 10 in.
456
BOOK IX. SOLID GEOMETKY
, PROPOSITION XXV. THEOREM
815. The area of a lune is to the area of the sur-
face of the sphere as the angle of the lune is to four right
is.
Given a sphere having its area denoted by 8, and on the
sphere the lune ABCD of Z A with its area denoted by L.
To prove
L:S=A° : 360°.
Proof. Draw FBH, the great O whose pole is A, inter-
secting the bounding arcs of the lune in B and D.
CASE I. When the arc BD and the circumference FBH
are commensurable.
Find a common measure of BD and .FBH, and let it be
contained in the arc BD m times, and in the circumference
FBHn times.
Then arc BD : circumference FBH=m : n.
Through the diameter AC, and the points of division of
the circumference FBH pass planes of great © .
The arcs of these great © will divide the surface of the
sphere in n small equal lunes, m of them being contained
in the lune ABCD.
:. L : S=m : n.
:. L : $=arc BD : circumference FBH. (Why ?)
Or L : S = A° : 360°. Art. 257.
CASE II. When the arc BD and the circumference FBH
are incommensurable.
Let the pupil supply the proof. Q. E. D.
SPHERICAL AREAS 457
816. Formula for the area of a lune in spherical degrees,
or spherids. The surface of a sphere contains 720 spherids
(Art. 803). Hence, by Art. 815,
L L spherids A m L_
'' fl°r720 spherids" 360' * 2 "
that is, the area of a lune in spherical degrees is equal to
twice the number of angular degrees in the angle of the lune.
817. Formula for area of a lune in square units o
S=4 TtR* (Art. 810) /.r^2-^, or(L = ^g-
818. COE. 1. On the same sphere, or on equal spheres,
two lunes are to each other as their angles.
819. COR. 2. Two lunes ivith equal angles, but, on un-
equal spheres, are to each other as the squares of the radii of
their spheres.
' Ex. 1. Find the area in spherical degrees of a lune of 27°.
"Ex. 2. Find the number of square inches in the area of a lune oft
27°, on a sphere whose radius is 10 in.
A solid symmetrical with respect to a plane is a solid in which
a line drawn from any point in its surface JL the given plane and
produced its own length ends in a point on the surface ; hence
Ex. 3. How many planes of symmetry has a circular cylinder ? A
cylinder of revolution ?
Ex. 4. Has either of these solids a center of symmetry f
Ex. 5. Answer the same questions for a circular cone.
Ex. 6. For a cone of revolution. For a sphere.
^ Ex. 7. For a regular square pyramid. For a regular pentagonal
pyramid.
0
458 BOOK IX. SOLID GEOMETKY
» PROPOSITION XXVI. THEOREM
820. If two great circles intersect on a hemisphere, the
sum of two vertical triangles thus formed is equivalent to a
lune ivhose angle is that angle in the triangles which is
formed by the intersection of the two great circles.
Given the hemisphere ADBF, and on it the great circles
AFB and DFC, intersecting at F.
To prove A AFC+ A BFD o lune whose Z is BFD. '
Proof. Complete the sphere and produce the given arcs
of the great circles to intersect at F' on the other hemisphere.
Then, in the A AFG and BF'D,
arc AF=arc BF' ,
(each being the supplement of the arc BF).
In like manner arc CF= 'arc DF'.
And arc AC = arc DB.
. •• A AFC^ A BF'D. Art. 795.
Add the A BFD to each of these equals ;
/. A AFC + A BFD<* A BF'D + A BFD. Ax. 3.
Or .-. A AFC + A ^D^lune FBF'D, Ax. 6
Q. E. D,
SPHERICAL AREAS 459
PROPOSITION XXVII. THEOREM
821. The number of spherical degrees, or spherids, in
the area of a spherical triangle is equal to the number of
angular degrees in the spherical excess of the triangle.
Given the spherical A ABO whose A are denoted by A ,
B, C, and whose spherical excess is denoted by E.
To prove area of A ABC —E spherids.
Proof. Produce the sides AC and BC to meet AB pro-
duced in the points D and F, respectively.
A ABC + & CDB = luuQ ABDC=2A spherids. )
V Art. 816.
A ABC + A ACF=\unQ BCFA = 2 B spherids. j
A ABC + & GFD=lune of /_BCA = 2 C spherids. Art. 820-
Adding, and observing that A ABC+A CDB+&ACF
-f A CFD = hemisphere ABDFC,
2 A ABC-}- hemisphere = 2 (A + B + C) spherids. Or,
2 A ABC+ 360 spherids = 2 (A + B + C) spherids. Ax. 8.
/. A ABC + 180 spherids = (A +B'+ C) spherids. Ax. 5.
/. A ABC= ( A + B + C—180) spherids. Ax. 3.
Or area A ABC=E spherids. Art. 788.
Q. £. D.
460 BOOK IX. SOLID GEOMETRY
822, Formula for area of a spherical triangle in square
units of area.
Comparing the area of a spherical A with the area of
the entire sphere .
area A : 4 nR2 = E spherids : 720 spherids.
4 KR2 X E KR2E
:. area A = -705—, or area A--—
823. The spherical excess of a spherical polygon is the
sum of the angles of the polygon diminished by (n — 2) 180° ;
that is, it is the sum of the spherical excesses of the tri-
angles into which the polygon may be divided.
PROPOSITION XXVIII. THEOREM
824. The area of a spherical polygon, in spherical de-
grees or spherids^ is equal to the spherical excess of the
polygon. A
Given a spherical polygon ABCDF of n sides, with its
spherical excess denoted by E.
To prove area of ABCDF=E spherical degrees.
Proof. Draw diagonals from A, any vertex of the poly-
gon, and thus divide the polygon into n — 2 spherical A.
The area of each A = (sum of its A — 180) spherids.
Art. 821.
/. sum of the areas of the A = [sum of A of the A —
(n— 2) 180] spherids. Ax. 2.
/. area of polygon = E spherids, Art. 823.
(for the sum of 4 of the A— (n— 2) 180°=^°).
SPHERICAL VOLUMES
461
SPHERICAL VOLUMES
825. A spherical pyramid is a por-
tion of a sphere bounded by a spheri-
cal polygon and the planes of the
great circles forming the sides of the
polygon. The base of a spherical
pyramid is the spherical polygon
bounding it, and the vertex of the
spherical pyramid is the center of the sphere.
Thus, in the spherical pyramid O-ABCD, the base is
A BCD and the vertex is 0.
826. A spherical wedge (or ungula) is the portion of a
sphere bounded by a lune and the planes of the sides of
the lune.
827. A spherical sector is the portion of a sphere gene-
rated by a sector of that semicircle whose rotation generates
the given sphere.
828. The base of a spherical sector is the zone gene-
rated by the revolution of the arc of the plane sector which
generates the spherical sector.
Let the pupil draw a spherical sector in which the base
is a zone of one base.
462 BOOK IX. SOLID GEOMETRY
829. A spherical segment is a portion of a sphere
included between two parallel planes.
The bases of a spherical segment are the sections of the
sphere made by the parallel planes which bound the given
segment ; the altitude is the perpendicular distance between
the bases.
830. A spherical segment of one base is a spherical seg-
ment one of whose bounding planes is tangent to the
sphere.
PROPOSITION XXIX. THEOREM
831. The volume of a sphere is equal to one -third the
product of the area of its surface by its radius.
Given a sphere having its volume denoted by V, sur-
face by $, and radius by R.
To prove V=%$XR.
Proof. Let any polyhedron be circumscribed about the
sphere.
Pass a plane through each edge of the polyhedron and
the center of the sphere.
These planes will divide the polyhedron into as many
pyramids as the polyhedron has faces, each pyramid hav-
ing a face of the polyhedron for its base, the center of the
SPHERICAL VOLUMES 4G3
sphere for its vertex, and the radius of the sphere for its
altitude.
.*. volume of each pyramid = J base X R. (Why ?)
.*. volume of polyhedron = J (surface of polyhedron) XR.
If the number of faces of the polyhedron be increased
indefinitely, the volume of the polyhedron approaches the
, volume of the sphere as a limit, and the surface of the
polyhedron approaches the surface of the sphere as a
limit.
Hence the volume of the polyhedron and J (surface of
the polyhedron )X R, are two variables always equal.
Hence their limits are equal,
Or V=bSXR. (Why?)
Q. £. D.
832. Formulas for volume of a sphere. Substituting
8=4: TtR2, or &=7tD2 in the result of Art. 831,
also F=
D
833. COR. 1. The volumes of two spheres are to each
other as the cubes of their radii, or as the cubes of their
diameters.
834. COR. 2. The volume of a spherical pyramid is
equal to one -third the product of its base by the radius of
the sphere.
835. COR. 3. The volume of spherical sector is equal to
one -third the product of its base (the bounding zone) by the
radius of the sphere,
464
BOOK IX. SOLID GEOMETRY
836. Formula for the volume of a spherical sector. De-
noting the altitude of the sector by H and the volume by V,
F=4 (area of zone) X J2,
= j(2 nEH) E. Art. 813.
PROPOSITION XXX. THEOREM
837. The volume of a spherical segment is equal to one-
half the product of its altitude by the sum of the areas of
its bases, plus the volume of a sphere whose diameter is the
altitude of the segment.
Given the semicircle ABC A' which generates a sphere
by its rotation about the diameter A A '; ED and CF semi-
chords _L AA', and denoted by r and r'; and DF denoted
by tf.
To prove volume of spherical segment generated by
BCFD, or F=J(7ir2 + 7ir/2) H+i nH8.
Proof. Draw the radii OB and 0(7.
Denote OF by h, and OD by k.
Then F=vol. OBC + vol. OCF— vol. OBD.
.'. F=§ 7l#2#-f 4 7tr'2h— J 7tr2A'. Arts. 836, 723.
But H=h-k, h*=lP-r*9 and /t2=^2-r2, (Why?)
SPHERICAL VOLUMES 465
.-. F=4 n [2 R2 (h-k)-i (R2-h2) h—(R2-k2) k] . Ax. 8.
= 4 7t [2 R2'(h-k) + R2 'U-AO — (A3-*3)].
= i 7t JT [3 £2— U2 + M + k2)] . x
But 7i2— 2 M -{- k2 = H2. s As. 4.
,<*
Subtract each member from 3 ft2 -f 3 &2 and divide by 2.
£*s. 3, 5.
Then A2 + M + ^2 = l (/i2 + k2)—-
/. F= i 7t# [f (r2 + r'2) + Y ] • Ax. 8.
Q. E. D.
838. Formula for volume of a spherical segment of one
base. In a spherical segment of one base rf = o, and r2=
(2 R-H) H (Art. 343).
Substituting for r and r' these values in the result of
Art. 837,
839. Advantage of measurement formulas. The student
should observe carefully that, by the results obtained in
Book IX, the measurement of the areas of certain curved
surfaces is reduced to the far simpler work of the measure-
ment of the lengths of one or more straight lines; in like
manner the measurement of certain volumes bounded by a
curved surface is reduced to the simpler work of linear
measurements. A similar remark applies to the results of
Book VIII.
Ex. 1. Find the volume of a sphere whose radius is 7 in.
Ex. 2. Find the volume of a sphere whose diameter is 7 in.
Ex. 3. In a sphere whose radius is 8 in., find the volume of a
spherical segment of one base whose altitude is 3.
DD
466 BOOK IX. SOLID GEOMETRY
EXERCISES. GROUP. 72
THEOREMS CONCERNING THE SPHERE
Ex. 1. Of circles of a sphere whose planes pass through a given
point within a sphere, the smallest is that circle whose plaiie is
perpendicular to the diameter through the given point.
Ex. 2. If a point on the surface of a given sphere is equidistant
from three points on a given small circle of the sphere, it is the pole
of the small circle.
Ex. 3. If two sides of a spherical triangle are quadrants, the third
side measures the angle opposite that side in the triangle.
Ex. 4. If a spherical triangle has one right angle, the sum of its
other two angles is greater than one right angle.
Ex. 5. If a spherical triangle is isosceles, its polar triangle is
isosceles.
Ex. 6. The polar triangle of a birectangular triangle is birectan-
gular.
Ex. 7. The polar triangle of a trirectangular triangle is identical
with the original triangle.
Ex. 8. Prove that the sum of the angles of a spherical quadri-
lateral is greater than 4 right angles, and less than 8 right angles.
What, also, are the limits of the sum of the angles of a spherical
hexagon ? Of the sum of the angles of a spherical w-gon ?
Ex. 9. On the same sphere, or on equal spheres, two birectan-
gular triangles are equal if their oblique angles are equal.
Ex. 10. Two zones on the same sphere, or on equal spheres, are
to each other as their altitudes.
Ex. 11. If one of the legs of a right spherical triangle is greater^
than a quadrant, another side is also greater than a quadrant.
[SuG. Of the leg which is greater than a quadrant, take the end
remote from the right angle as a pole, and describe an arc.]
Ex. 12. If ABC and A'B'C' are polar triangles, the radius OA is
perpendicular to the plane OB'C'.
EXEKCISES ON THE SPHEKE 467
Ex. 13. On the same sphere, or on equal spheres, spherical tri-
angles whose polar triangles have equal perimeters are equivalent.
Ex. 14. Given OAO' ', OBO> ', and AB arcs of great
circles, intersecting so that Z OAB= Z CfBA; prove that
AOA£=A(yAB. A
[Sue. Show that /.OBA = /.WAS.]
Ex. 15. Find the ratio of the volume of a sphere to
the volume of a circumscribed cube.
Ex. 16. Find the ratio of the surface of a sphere to
the lateral surface of a circumscribed cylinder of revolution; also
find the ratio of their volumes.
Ex. 17. If the edge of a regular tetrahedron is denoted by a, find
the ratio of the volumes of the inscribed and circumscribed sphez'js.
Ex. 18. Find the ratio of the two segments into which a hemi-
sphere is divided by a plane parallel to the base of the hemisphere and
at the distance $ R from the base.
EXERCISES. GROUP 73
SPHERICAL LOCI
Ex. 1. Find the locus of a point at a given distance a from the
surface of a given sphere.
Ex. 2. Find the locus of a point on the surface of a sphere that; is
equidistant from two given points on the surface.
Ex. 3. If, through a given point outside a given sphere, tangent
planes to the sphere are passed, find the locus of the points of
tangency.
Ex. 4. If straight lines be passed through a given fixed point in
space, and through another given point other straight lines be passed
perpendicular to the first set, njicl the locus of the feet of the
perpendiculars.
468 BOOK IX. SOLID GEOMETRY
EXERCISES. GROUP 74
PROBLEMS CONCERNING THE SPHERE
Ex. 1. At a given point on a sphere, construct a plane tangent to
the sphere.
Ex. 2. Through a given point on the surface of a sphere, draw an
arc of a great circle perpendicular to a given arc.
Ex. 3. Inscribe a circle in a given spherical triangle.
Ex. 4. Construct a spherical triangle, given its polar triangle.
Given the radius, r, construct a spherical surface which shall pass
through
Ex. 5. - Three given points.
Ex. 6. Two given points and be tangent to a given plane.
Ex. 7. Two given points and be tangent to a given sphere
Ex. 8. One given point and be tangent to two given planes.
Ex. 9. One given point and be tangent to two given spheres.
Given the radius, r, construct a spherical surface which shall be
tangent to
Ex. 10. Three given planes.
Ex. 11. Two given planes and one given sphere.
Ex. 12. Construet a spherical surface which shall pass through
three given points and be tangent to a given plane.
Ex. 13. Through a given straight line pass a plane tangent to a
given sphere.
[SuG. Through the center of the sphere pass a plane X given
line, etc.]
When is the solution impossible ?
Ex. 14. Through a given point on a sphere, construct an arc of
a great circle tangent to a given small circle of the sphere.
[Sua. Draw a straight line from the center of the sphere to the
given point, and produce it to intei-sect the plane of the small circle,
NUMERICAL EXERCISES IN SOLID
GEOMETRY
For methods of facilitating numerical computations,
see Arts. 493-6.
EXERCISES. CROUP 75
LINES AND SURFACES OF POLYHEDRONS
Find the lateral area and total area of a right prism whose
/ Ex. 1. Base is an equilateral triangle of edge 4 in., and whose
altitude is 15 in.
Ex. 2. Base is a triangle of sides 17, 12, 25, and whose altitude
is 20.
f Ex. 3. Base is an isosceles trapezoid, the parallel bases being 10
and 15 and leg 8, and whose altitude is 24.
* Ex. 4. Base is a rhombus whose diagonals are 12 and 16, and
whose altitude is 12.
Ex. 5. Base is a regular hexagon with side 8 ft., and whose alti-
tude is 20 ft.
Ex. 6. Find the entire surface of a rectangular parallelepiped
8 X 12 X 16 in. ; of one p X q X r ft.
Ex. 7. Of a cube whose edge is 1 ft. 3 in.
Ex. 8. The lateral area of a regular hexagonal prism is 120 sq. ft.
and an edge of the base is 10 ft. Find the altitude.
Ex. 9. How many square feet of tin are necessary to line a box
20X6X4 in. ?
/ Ex. 10. If the surface of a cube is 1 sq. yd., find an edge in inches,
Ex. 11. Find the diagonal of a cube whose edge is 5 in.
Ex. 13. If the diagonal of a cube is 12 ft., find the surface,
470 SOLID GEOMETKY
Ex. 13. If the surface of a rectangular parallelepiped is 208 sq. in.,"
and the edges are as 2 : 3 : 4, find the edges.
In a regular square pyramid
Ex. 14. If an edge of the base is 16 and slant height is 17, find
the altitude.
j Ex. 15. If the altitude is 15 and a lateral edge is 17, find an edge
of the base.
Ex. 16. If a lateral edge is 25 and an edge of the base is 14, find
the altitude.
In a regular triangular pyramid
Ex. 17. If an edge of the base is 8 and the altitude is 10, find the
slant height.
Ex. 18. Find the altitude of a regular tetrahedron whose edge is 6.
Find the lateral surface and the total surface of
Ex. 19. A regular square pyramid an edge of whose base is 16,
and whose altitude is 15.
Ex. 20. A regular triangular pyramid an edge of whose base is 10,
and whose altitude is 12.
^ Ex. 21. A regular hexagonal pyramid an edge of whose base is 4,
and whose altitude is 21.
Ex. 22. A regular square pyramid whose slant height is 24, and
•whose lateral edge is 25.
/ Ex. 23. A regular tetrahedron whose edge is 4.
Ex. 24«. A regular tetrahedron whose altitude is 9.
Ex. 25. A regular hexagonal pyramid each edge of whose base is
a, and whose altitude is 6.
Ex. 26. In the frustum of a regular square pyramid the edges of
the bases are 6 and 18, and the altitude is 8. Find the slant height.
Hence find the lateral area.
Ex. 27. In the frustum of a regular triangular pyramid the edges
of the bases are 4 and 6, and the altitudi is 5, Find the slant height.
Hence and the lateral area.
NUMERICAL EXERCISES IN SOLID GEOMETRY 471
Ex. 28. In the frustum of a regular tetrahedron, if the edge of the
lower base is &i, the edge of the upper base is 62, and the altitude is a,
show that £=iT/H&i — W -f-4a2.
Ex. 29. In the frustum of a regular square pyramid the edges of
the bases are 20 and 60, and a lateral edge is 101. Find the lateral
surface.
EXERCISES. GROUP 76
LINES AND SURFACES OP CONES AND CYLINDERS
Ex. 1 . How many square feet of lateral surface has a tunnel 100
yds. long and 7 ft. in diameter.
Ex. 2. The lateral area of a cylinder of revolution is 1 sq. yd.,
and the altitude is 1 ft. Find the radius of the base:
Ex. 3. The entire surface of a cylinder of revolution is 900 sq. ft.
and the radius of the base is 10 ft. Find the altitude.
In a cylinder of revolution
Ex. 4. Find R in terms of 8 and H.
Ex. 5. Find H in terms of E and T.
Ex. 6. Find T in terms of S and H.
Ex. 7. How many sq. yds- of canvas are required to make a coni-
cal tent 20 ft. in diameter and 12 ft. high ?
Jt Ex. 8. A man has 400 sq. yds. of canvas and wants to make a
conical tent 20 yds. in diameter. What will be its altitude ?
^ Ex. 9. The altitude of a cone of revolution is 10 ft. and the lat-
eral area is 11 times the area of the base. Find the radius of the base.
In a cone of revolution
Ex. 10. Find T in terms of S and L.
Ex. 11. Find 12 in terms of T and L.
J* Ex. 12. How many square feet of tin are necessary to make a
funnel the diameters of whose ends are 2 in. and 8 in., and whose
altitude is 7 in. ?
It Ex. 13. If the slant height of a frustum of a cone of revolution
makes an angle of 45° with the base, show that the lateral area of the
frustum is (n2 — r22) TT i/2.
472 SOLID GEOMETRY
EXERCISES. CROUP 77
SPHERICAL LINES AND SURFACES
Ex. 1. Find in square feet the area of the surface of a sphere
whose radius is 1 ft. 2 in.
f> Ex. 2. How many square inches of leather will it take to cover a
baseball whose diameter is 3i in.?
Ex. 3. How many sq. ft. of tin are required to cover a dome in
the shape of a hemisphere 6 yds. in diameter ?
K Ex. 4. What is the radius of a sphere whose surface is 616 sq. in. ?
Ex. 5. Find the diameter of a globe whose surface is 1 sq. yd.
Ex^jS. If the circumference of a great circle on a sphere is 1 ft.,
find the area of the surface of the sphere.
Ex. 7. If a hemispherical dome is to contain 100 sq. yds. of sur-
face, what must its diameter be ?
* Ex. 8. Find the radius of a sphere in* which the area of the sur-
face equals the number of linear units in the circumference of a great
circle.
Find the area of a lune in which
X Ex. 9. The angle of the lune is 36°, and the radius of the sphere
is 14 in.
Ex. 10. The angle of the lune is 18° 20', and the diameter of the
sphere is 20 in.
ft Ex. 11. The angle of the lune is 24°, and the surface of the
sphere is 4 sq. ft.
Find the area of a spherical triangle in which
* Ex. 12. The angles are 80°, 90°, 120°, and the diameter of the
sphere is 14 ft.
+ Ex. 13. The angles are 74° 24', 83° 16', 92° 20', and the radius of
the sphere is 10.
% Ex. 14. The angles are 85°, 95°, 135°, and the surface of the
sphere is 10 sq. ft.
NUMERICAL EXERCISES IN SOLID GEOMETRY 473
Ex. 15. If the sides of a spherical triangle are 100°, 110°, 120°
and the radius of the sphere is 16, find the area of the polar triangle.
Ex. 16. If the angles of a spherical triangle are 90°, 100°, 120*
and its area is 3900, find the radius of the sphere.
Ex. 17. If the area of an equilateral spherical triangle is one-
third the surface of the sphere, find an angle of the triangle.
Ex. 18. In a trihedral angle the plane angles of the dihedral
angles are 80°, 90°, 100°; find the number of solid degrees in the
trihedral angle.
S Ex. 19. Find the area of a spherical hexagon each of whose
angles is 150°, on a sphere whose radius is 20 in.
Ex. 20. If each dihedral angle of a given pentahedral angle is
120°, how many solid degrees does the pentahedral angle contain ?
Ex. 21. In a sphere whose radius is 14 in., find the area of a zone
3 in. high.
Ex. 22. What is the area of the north temperate zone, if the
earth is taken to be a sphere with a radius of 4,000 miles, and the
distance between the plane of the arctic circle and that of the tropic
of Cancer is 1,800 miles ?
Ex. 23. If Cairo, Egypt, is in latitude 30°, show that its parallel
of latitude bisects the surface of the northern hemisphere.
Ex. 24. How high must a person be above the earth's surface to
see one -third of the surface ?
S/
Ex. 25. How much of the earth's surface will a man see who is
2,000 miles above the surface, if the diameter is taken as 8,000 miles?
Ex. 26. If the area of a zone equals the area of a great circle,
find the altitude of the zone in terms of the radius of the sphere.
Ex. 27. If sounds from the Krakatoa explosion were heard at a
distance of 3,000 miles (taken as a chord) on the surface of the earth,
over what fraction of the earth's surface were they heard ?
Ex. 28. The radii of two spheres are 5 and 12 in. and their cen-
s are 13 in. apart. Find the area of the circle of intersection and
also of that part of the surface of each sphere not included by the
other sphere.
474 SOLID GEOMETKY
EXERCISES. CROUP 78
VOLUMES OF POLYHEDRONS
Find the volume of a prism
XEx. 1. Whose base is an equilateral triangle with side 5 in., and
rhose altitude is 16 in.
)/ Ex. 2. Whose base is a triangle with sides 12, 13, 15, and whose
Altitude is 20.
Ex. 3. Whose base is an isosceles right triangle with a leg equal
to 2 yds., and whose altitude is 25 ft.
Ex. 4. Whose base is a regular hexagon with a side of 8 ft., and
hose altitude is 10 yds.
S Ex. 5. Whose base is a rhombus one of whose sides is 25, and one
of whose diagonals is 14, and whose altitude is 11.
Ex. 6. Whose base contains 84 sq. yds., and whose lateral faces
are three rectangles with areas of 100, 170, 210 sq. yds., respectively.
XEx. 7. How many bushels of wheat are held by a bin 30 x 10 x 6 ft. ,
if a bushel is taken as li cu. ft.?
Ex. 8. How many cart-loads of earth are in a cellar 30 x 20x6 ft.,
if a cart-load is a cubic yard ?
/ Ex. 9. 'If a cubical block of marble costs $2, what is the cost of
a cube whose edge is a diagonal of the first block ?
Ex. 10. Find the edge of a cube whose volume equals the sum of
the volumes of two cubes whose edges are 3 and 5 ft.
Ex. 11. Find the edge of a cube whose volume equals the area of
its surface.
»
Ex. 12. If the top of a cistern is a rectangle 12x8 ft., how deep
ust the cistern be to hold 10,000 gallons ?
Ex. 13. Find the inner edge of a peck measure which is in the
shape of a cube.
Ex. 14. A peck measure is to be a rectangular parallelepiped with
square base and altitude equal to twice the edge of the base, Find
its dimensions,
NUMERICAL EXERCISES IN SOLID GEOMETRY 475
Ex. 15. Find the volume of a cube whose diagonal is a.
Find the volume of a pyramid
Ex. 16. Whose base is an equilateral triangle with side 8 in.r
and whose altitude is 12 in.
Ex. 17. Whose base is a right triangle with hypotenuse 29 and
one leg 21, and whose altitude is 20.
Ex. 18. Whose base is a square with side 6, and each of whose
lateral edges is 5.
Ex. 19. Whose base is a square with side 10, and each of whose
lateral faces makes an angle of 45° with the base.
Ex. 20. If the pyramid of Memphis has an altitude of 146 yds.
and a square base of side 232 yds., how many cubic yards of stone
does it contain ? What is this worth at $1 a cu. yd.?
Ex. 21. A church spire 150 ft. high is hexagonal in shape and
each side of the base is 10 ft. The spire has a hollow hexagonal
interior, each side of whose base is 6 ft., and whose altitude is 45 ft.
How many cubic yards of stone does the spire contain ?
Ex. 22. If a pyramid contains 4 cu. yds. and its base is a square
with one side 2 ft., find the altitude.
Ex. 23. A heap of candy in the shape of a frustum of a regular
square pyramid has the edges of its bases 25 and 9 in. and its altitude
12 in. Find the number of pounds in the heap if a pound is a rectan-
gular parallelepiped 4x3x2 in. in size.
Ex. 24. Find the volume of a frustum of a regular triangular
pyramid, the edges of the bases being 2 and 8, and the slant height 12.
Ex. 25. The edges of the bases of the frustum of a regular square
pyramid are 24 and 6, and each lateral edge is 15; find the volume.
Ex. 26. If a stick of timber is in the shape of a frustum of a
regular square pyramid with the edges of its ends 9 and 15 in., and
with a length of 14 ft., find the number of feet of lumber in the stick.
What is the difference between this volume and that of a stick of
the same length having the shape of a prism with a base equal to the
area of a midsection of the first stick ?
476 SOLID GEOMETRY
Ex. 27. Find the volume of a prismoid whose bases are rectangles
5 x 2 ft. and 7 x 4 ft., and whose altitude is 12 ft.
Ex. 28. How many cart-loads of earth are there in a railroad cut
12 ft. deep, whose base is a rectangle 100x8 ft., and whose top is a
rectangle 30x50 ft.?
Ex. 29. Find the volume of a prismatoid whose base is an equi-
lateral triangle with side 12 ft., and whose top is a line 12 ft. long
parallel to one side of the base, and whose altitude is 1& ft.
Ex. 30. If the base of a prismatoid is a rectangle with dimensions
a and &, the top is a line c parallel to the side b of the base, and the
altitude is h, find the volume.
EXERCISES. CROUP 79
VOLUMES OF CONES AND CYLINDERS
Ex. 1. How many barrels of oil are contained in a cylindrical
tank 20 ft. long and 6 ft. in diameter, if a barrel contains 4 cu. ft. f
)^ Ex. 2. How many cu. yds. of earth must be removed in making a
tunnel 450 ft. long, if a cross-section of the tunnel is a semicircle of
15 ft. radius ?
V Ex. 3. A cylindrical glass 3 in. in diameter holds half a pint.
Find its height in inches.
y Ex. 4. If a cubic foot of brass be drawn out into wire ^ inch in
^ diameter, how long will the wire be ?
Ex. 5. A gallon measure is a cylinder whose altitude equals the
diameter of the base. Find the altitude. *
Ex. 6. Show that the volumes of two cylinders, having the altitude
of each equal to the radius of the other, are to each other as E : E' .
Ex. 7. In a cylinder, find E in terms of V and H ; also V in terms
of 8 and E.
A Ex. 8. A conical heap of potatoes is 44 ft. in circumference and
6 ft. high. How many bushels does it contain, if a bushel is lieu, ft.f
y Ex. 9. What fraction of a pint will a conical wine-glass hold, if
its altitude is 3 in. and the diameter of the top ia 2 in. ?
NUMERICAL EXERCISES IN SOLID GEOMETRY 477
Ex. 10. Find the ratio of the volumes of the two cones inscribed
in, and circumscribed about, a regular tetrahedron.
Ex. 11. If an equilateral cone contains 1 quart, find its dimen-
sions in inches.
Ex. 12. In a cone of revolution find V in terms of R and L ; also
find V in terms of R and S.
Ex. 13. Find the volume of a frustum of a cone of revolution,
whose radii are 14 and 7 ft., and whose altitude is 2 yds.
' Ex. 14. What is the cost, at 50 cts. a cu. ft., of a piece of marble
in the shape of a frustum of a cone of revolution, whose radii are 6
and 9 ft., and whose slant height is 5 ft.?
Ex. 15. In a frustum of a cone of revolution, the volume is 88
cu. ft., the altitude is 9 ft., and R = 2r. Find r.
EXERCISES. CROUP 8O
SPHERICAL VOLUMES
Ex. 1. Find the volume of a sphere whose radius is 1 ft. 9 in.
Ex. 2. If the earth is a sphere 7,920 miles in diameter, find its
***+***.. volume in cubic miles.
T'
/ Ex. 3. Find the diameter of a sphere whose volume is 1 cu. ft.
Ex. 4. What is the volume of a sphere whose surface is 616 sq. in. ?
ML
//^rty
Ex. 5. Find the radius of a sphere equivalent to the sum of two
spheres, whose radii are 2 and 4 in.
Ex. 6. Find the radius of a sphere whose volume equals the area
of its surface.
Ex. 7. Find the volume of a sphere circumscribed about a cube
whose edge is 6.
•" Ex. 8. Find the volume of a spherical shell whose inner and
outer diameters are 14 and 21 in.
Ex. 9. Find the volume of a spherical shell whose inner and outer
Surfaces are 20 TT and 12 TT.
478 SOLID GEOMETRY
Find the volume of
Ex. 10. A spherical wedge whose angle is 24°, the radius of the
sphere being 10 in.
- Ex. 11. A spherical sector whose base is a zone 2 in. high, the
radius of the sphere being 10 in.
V
"icallsc
Ex. 12. A spherical ^segment of two bases whose radii are 4 and
7 and altitude 5 in.
' Ex. 13. A wash-basin in the shape of a segment of a sphere is
6 in. deep and 24 in. in diameter. How many quarts of water will the
basin hold ?
/ Ex. 14. A plane parallel to the base of a hemisphere and bisect-
ing the altitude divides its volume in what ratio ?
/ Ex. 15. A spherical segment 4 in. high contains 200 cu. in. ; find
the radius of the sphere.
f Ex. 16. If a heavy sphere whose diameter is 4 in. be placed in a
conical wine-glass full of water, whose diameter is 5 in. and altitude
6 in., find how much water will run over.
EXERCISES. GROUP 81
EQUIVALENT SOLIDS
I Ex. 1. If a cubical block of putty, each edge of which is 8 inches,
be molded into a cylinder of revolution whose radius is 3 inches, find
the altitude of the cylinder.
Ex. 2. Find the radius of a sphere equivalent to a cube whose
edge is 10 in.
-*,
Ex. 3. Find the radius of a sphere equivalent to a cone of revolu-
tion, whose radius is 3 in. and altitude 6 in.
Ex. 4. Find the edge of a cube equivalent to a frustum of a cone
of revolution, whose radii are 4 and 9 ft. and altitude 2 yds.
Ex. 5. Find the altitude of a rectangular parallelepiped, whose
base is 3 x5 in. and whose volume is equivalent to a sphere of radius
7 in.
NUMEKICAL EXERCISES IN SOLID GEOMETEY 479
V Ex. 6. Find the base of a square rectangular parallelepiped,
whose altitude is 8 in. and whose volume equals the volume of a cone
of revolution with a radius of 6 and an altitude of 12 in.
Ex. 7. Find the radius of a cone of revolution, whose altitude is
15 and whose volume is equal to that of a cylinder of revolution with
radius 6 and altitude 20.
Ex. 8. Find the altitude of a cone of revolution, whose radius is
15 and whose volume equals the volume of a cone of revolution with
radius 9 and altitude 24.
X.
Ex. 9. On a sphere whose diameter is 14 the altitude of a zone of
one base is 2. Find the altitude of a cylinder of revolution, whose
base equals the base of the zone and whose lateral surface equals the
surface of the zone.
EXERCISES. GROUP 82
SIMILAR SOLIDS
Ex. 1. If on two similar solids L, L' and I, V are pairs of homol-
ogous lines; Af A1 and a, a' pairs of homologous areas, V, V and
v, vf pairs of homologous volumes,
CL :L' = l : l' = VA : i/A'=&v : tfv'.
show that A : A' = LZ : Ln = a : a'= vl : F'i.
Ex. 2. If the edge of a cube is 10 in., find the edge of a cube
having 5 times the surface.
Ex. 3. If the radius of a sphere is 10 in., find the radius of a
sphere having 5 times the surface.
Ex. 4. If the altitude of a cone of revolution is 10 in., find the
altitude of a similar cone of revolution having 5 times the surface.
Ex. 5. In the last three exercises, find the required dimension if
the volume is to be 5 times the volume of the original solid.
Ex. 6. The linear dimensions of one trunk are twice as great as
those of another trunk. How much greater is the faunae ? Tbe surface?
480 SOLID GEOMETRY
Ex. 7. How far from the vertex is the cross -section which bisects
the volume of a cone of revolution ? Which bisects the lateral
surface ?
Ex. 8. If the altitude of a pyramid is bisected by a plane parallel
to the base, how does the area of the cross-section compare with the
area of the base ? How does the volume cut off compare with the
volume of the entire pyramid ?
Ex. 9. Planes parallel to the base of a cone divide the altitude
into three equal parts ; compare the lateral surfaces cut off. Also the
volumes.
Ex. 10. A sphere 10 in. in diameter is divided into three equiva-
lent parts by concentric spherical surfaces. Find the diameters of
these surfaces.
Ex. 11. If the strength of a muscle is as the area of its cross-
section, and Goliath of Gath was three times as large in each linear
dimension as Tom Thumb, how much greater was his strength ? His
weight ? How, then, does the activity of the one man compare with
that of the other ?
Ex. 12. If the rate at which heat radiates, from a body is in pro-
portion to the amount of surface, and the planet Jupiter has a diame-
ter 11 times that of the earth, how many times longer will Jupiter be
in cooling off ?
[Suo. How many times greater is the volume, and therefore .the
original amount of heat in Jupiter ? How many times greater is its
surface ? What will be the combined effect of these factors ?]
CROUP 83
MISCELLANEOUS NUMERICAL EXERCISES IN SOLID
GEOMETRY
Find S, T and V of
Ex. 1. A right triangular prism whose altitude is 1 ft., and the
sides of whose base are 26, 28, 30 in.
Ex. 2. A cone .of revolution the radius of whose base is 1 ft. 2 in.,
and whose altitude is 35 in.
Ex. 3. A frustum of a square pyramid the areas of whose bases
are 1 sq, ft. and 36 sq, in., and whose altitude is 9 in.
NUMEKICAL EXEECISES IN SOLID GEOMETEY 481
Ex. 4. A pyramid whose slant height is 10 in., and whose base
is an equilateral triangle whose side is 8 in.
k Ex. 5. A cube whose diagonal is 1 yd.
% Ex. 6. A frustum of a cone of revolution whose radii are 6 and
11 in. and slant height 13 in.
/^ Ex. 7. A rectangular parallelepiped whose diagonal is 2^29, and
whose dimensions are in the ratio 2 : 3 : 4.
* Ex. 8. Find the volume of a sphere inscribed in a cube whose
'edge is 6 ; also find the area of a triangle on that sphere whose angles
are 80°, 90°, 150°.
X Ex. 9. Find the volume of the spherical pyramid whose base is
the above triangle.
X Ex. 10. Find the angle of a lune on the same sphere, equivalent
to that triangle.
£ Ex. 11. On a cube whose edge is 4, planes through the midpoints
of the edges cut off the corners. Find the volume of the solid re-
maining.
Ex. 12. How is V changed if H of a cone of revolution is doubled
and R remains unchanged ? If R is doubled and H remains unchanged?
If both H and B are doubled ?
jL Ex, 13. In an equilateral cone, find S and V in terms of E.
Ex. 14. A piece of lead 20x8 x2 in. will make how many spher-
ical bullets, each f in. in diameter ?
Ex. 15. How many bricks are necessary to make a chimney in
the shape of a frustum of a cone, whose altitude is 90 ft., whose outer
diameters are 3 and 8 ft., and whose inner diameters are 2 and 4 ft.,
counting 12 bricks to the cubic ft.?
Ex. 16. If the area of a zone is 300 and its altitude 6, find the
radius of the sphere.
Ex. 17. If the section of a cylinder of revolution through its axis
is a square, find S, T, V in terms of E.
Ex. 18. If every edge of a square pyramid is fy find b in terms of T.
DD
452 SOLID GEOMETRY
^ Ex.19. A regular square pyramid has a for its altitude and also for
each side of the base. Find the area of a section made by a plane parallel
to the base and bisecting the altitude. Find also the volumes of the
two parts into which the pyramid is divided.
/ Ex. 20. If the earth is a sphere of 8,000 miles diameter and its
atmosphere extends 50 miles from the earth, find the volume of the
atmosphere.
Ex. 21. On a sphere, find the ratio of the area of an equilateral
spherical triangle, each of whose angles is 95°, to the area of a lune
whose angle is 80°.
Ex. 22. A square right prism has an altitude 6a and an edge of
the base 2a. Find the volume of the largest cylinder, sphere, pyramid
and cone which can be cut from it.
I/ Ex. 23. Obtain a formula for the area of that part of a sphere
'illuminated by a point of light at a distance a from tne sphere
whose radius is E.
*/ Ex. 24. On a sphere whose radius is 6 in., find an angle of an
/ equilateral triangle whose area is 12 sq. in.
Ex. 25. Find the volume of a prismatoid, whose altitude is 24 and
whose bases are equilateral triangles, each side 10, so placed that the
mid-section of the prismatoid is a regular hexagon.
/ Ex. 26. On a sphere whose radius is 16, the bases of a zone are
equal and are together equal to the area of the zone. Find the alti-
tude of the zone.
/•" Ex. 27. Find the volume of a square pyramid, the edge of whose
base is 10 and each of whose lateral edges is inclined 60° to the base.
^ Ex. 28. An irregular piece of ore, if placed in a cylinder partly
filled with water, causes the water to rise 6 in. If the radius of the
cylinder is 8 in., what is the volume of the ore ?
Ex. 29. Find the volume of a truncated right triangular prism, if
the edges of the base are 8, 9, 11, and the lateral edges are 12, 13, 14.
-„
Ex. 30. In a sphere whose radius is 5, a section is taken at tho
distance 3 from the center. On this section as a base a cone is formed
whose lateral elements are tangent to the sphere, Find tho lateral
surface and volume of the cone.
NUMEKICAL EXERCISES IN SOLID GEOMETRY 483
Ex. 31 . The volume of a sphere is l,437i cu. in. Find the surface.
Ex. 32. A square whose side is 6 is revolved about a diagonal as
an axis; find the surface and volume generated.
Ex. 33. Find the edge of a cubical cistern that will hold 10 tons
of water, if 1 cu. ft. of water weighs 62.28 Ibs.
Ex. 34. A water trough has equilateral triangles, each side 3 ft.,
for ends, and is 18 ft. long. How many buckets of water will it hold,
if a bucket is a cylinder 1 ft. in diameter and li ft. high ?
Ex. 35. The lateral area of a cylinder of revolution is 440 sq. in.,
and the volume is 1,540 cu. in. Find the radius and altitude.
Ex. 36. The angles of a spherical quadrilateral are 80°, 100°,
120°, 120°. Find the angle of an equivalent equilateral triangle.
if Ex. 37. A cone and a cylinder have equal lateral surfaces, and
their axis sections are equilateral. Find the ratio of their volumes.
Ex. 38. A water-pipe £ in. in diameter rises 13 f.t. from the ground.
How many quarts of water must be drawn from it before the water
from under the ground comes out ? If a quart runs out in 5 seconds,
how long must the water run ?
Ex. 39. A cube immersed in a cylinder partly filled with water
causes the water to rise 4 in. If the radius of the cylinder is 6 in.,
what is an edge of the cube ?
Ex. 40. An auger hole whose diameter is 3 in. is bored through
the center of a sphere whose diameter is 8 inches. Find the volume
remaining.
Ex. 41. Show that the volumes of a cone, hemisphere, and cylin-
der of the same base and altitude are as 1 '. 2 : 3.
Ex. 42. The volumes of two similar cylinders of revolution are
as 8 : 125 ; find the ratio of their radii. If the radius of the smaller is
10 in., what is the radius of the larger ?
Ex. 43. An iron shell is 2 in. thick and the diameter of its outer
surface is 28 in. Find its volume.
Ex. 44. The legs of an isosceles spherical triangle each make an
angle of 75° with the base. The legs produced form a lune whose
area is four times the area of the triangle. Find the angle of the lune.
484 SOLID GEGMETEY
CROUP 84
EXERCISES INVOLVING THE METRIC SYSTEM
Find S, T, V of
Ex. 1. A right prism the edges of whose base are 6m., 70 dm.,
900 cm., and whose altitude is 90 dm.
Ex. 2. A regular square pyramid an edge of whose base is 30 dm.,
and whose altitude is 1.7 mP
Ex. 3. A sphere whose radius is 0.02 m.
Ex. 4. A frustum of a cone of revolution whose radii are 10 dm.
and 6 dm., and whose slant height is 50 cm.
Ex. 5. A cube whose diagonal is 12 cm.
Ex. 6. A cylinder of revolution whose radius equals 2 dm., and
whose altitude equals the diameter of the base.
Ex. 7. Find the area of a spherical triangle on a sphere whose
radius is 0.02 m., if its angles are 110°, 120°, 130°.
Ex. 8. Find the number of square meters in the surface of a
sphere, a great circle of which is 50 dm. long.
Ex. 9. How many liters will a cylindrical vessel hold that is
10 dm. in diameter and 0.25 m. high ? How many liquid quarts ?
Ex. 10. A liter measure is a cylinder whose diameter is half the
altitude. Find its dimensions in centimeters.
Ex. 11. Find the surface of a sphere whose volume is 1 cu. m,,
APPENDIX
J. MODERN GEOMETRIC CONCEPTS
840. Modern Geometry. In recent times many new
geometric ideas have been invented, and some of them
developed into important new branches of geometry.
Thus, the idea of symmetry (see Art. 484, etc.) is a
modern geometric concept. A few other of these modern
concepts and methods will be briefly mentioned, but
their thorough consideration lies beyond the scope of
this book.
841. Projective Geometry. The idea of projections
(see Art. 345) has been developed in comparatively
recent times into an important branch of mathematics with
many practical applications, as in engineering, architec-
ture, construction of maps, etc.
842. Principle of Continuity. By this principle two
or more theorems are made special cases of a single more
general theorem. An important aid in obtaining continuity
among geometric principles is the application of the con-
cept of negative quantity to geometric magnitudes.
Thus, a negative line is a line opposite in direction to
a given line taken as positive.
For example, if OA is +, OB is — ,
(485)
486
GEOMETKY. APPENDIX
Similarly, a negative angle is
an angle formed by rotating a line
in a plane in a direction opposite
from a direction of rotation taken
as positive. Thus, if the line OA
rotating from the position OA
forms the positive angle AOB, the
same line rotating in the opposite direction forms the
negative angle AOB'. Similarly, positive and negative
arcs are formed.
In like manner, if P and Pf are
on opposite sides of the line AB and
the area PAB is taken as positive,
the area P'AB will be negative.
As an illustration of the law of
continuity, we may take the theorem
that the sum of the triangles formed
by drawing lines from a point to the
vertices of a polygon equals the area
of the polygon.
Applying this to the quadrilateral ABCD, if the point
P falls within the quadrilateral, APA£+AP£C + APOD
+ APAD=ABCD (Ax. 6).
Also, if the point falls without the quadrilateral at P',
A P'AB + A P'BC + A P'CD + A P'AD = ABCD, since
AP'AD is a negative area, and hence is to be subtracted
from the sum of the other three triangles.
843. The Principle of Reciprocity, or Duality, is a
principle of relation between two theorems by which each
theorem is convertible into the other by causing the words
for the same two geometric objects in each theorem to ex-
change places.
MODERN GEOMETRIC CONCEPTS
487
Thus, of theorems VI and VII, Book I, either may be
converted into the other by replacing the word "sides" by
"angles," and "angles" by "sides." Hence these are
termed reciprocal theorems.
The following are other instances of reciprocal geometric
properties :
1. Two points determine a
straight line.
2. Three points not in the same
straight line determine a plane.
3. A straight line and a point
determine a plane.
1. Two lines determine a point.
2. Tliree planes not through the
same straight line determine a
point.
3. A straight line and a plane
determine a point.
The reciprocal of a theorem is not necessarily true.
Thus, two parallel straight lines determine a*plane, but
two parallel planes do not determine a line.
However, by the use of the principle of reciprocity,
geometrical properties, not otherwise obvious, are fre-
quently suggested.
844. Principle of Homology. Just as the law of reci-
procity indicates relations between one set of geometric
concepts (as lines) and another set of geometric concepts
(as points), so the law of homology indicates relations
between a set of geometric concepts and a set of concepts
outside of geometry: as a set of algebraic concepts, for
instance.
Thus, if a and 6 are numbers, by algebra (a + 5) (a — 5)
= a2— 62.
Also, if a and b are segments of a line, the rectangle
(a-j-ft)X(a — &) is equivalent to the difference between the
squares a2 and 62.
By means of this principle, truths which would be over-
looked or difficult to prove ill one department of thought
488 GEOMETRY. APPENDIX
are made obvious by observing the corresponding truth in
another department of thought.
Thus, if a and I are line segments, the theorem (a+ 6)2
+ (a — &)2=2(a2-f-62) is not immediately obvious in geo-
metry, but becomes so by observing the like relation
between the algebraic numbers a and &.
i
845. Non-Euclidean Geometry. Hyperspace. By vary-
ing the properties of space, as these are ordinarily stated,
different kinds of space may be conceived of, each having
its own geometric laws and properties. Thus, space, as
we ordinarily conceive it, has three dimensions, but it is
possible to conceive of space as having four or more
dimensions. To mention a single property of four dimen-
sional space, in such a space it would be possible, by
simple pressure, to turn a sphere, as an orange, inside
out without breaking its surface.
As an aid toward conceiving how this is possible, consider a plane
in which one circle lies inside another. No matter how these circles
are moved about in the plane, it is impossible to shift the inner circle
so as to place it outside the other without breaking the circumference
of the outer circle. But, if we are allowed to use the third dimension
of space, it is a simple matter to lift the inner circle up out of the
plane and set it down outside the larger circle.
Similarly if, in space of three dimensions, we have one spherical
shell inside a larger shell, it is impossible to place the smaller shell
outside the larger without breaking the larger. But if the use of a
fourth dimension be allowed,— that is, the use of another dimension
of freedom of motion,— it is possible to place the inner shell outside
the larger without breaking the latter.
846. Curved Spaces. By varying the geometric axioms
of space (see Art. '47), different kinds of space may be
conceived of. Thus, we may conceive of space such that
through a given point one line may be drawn parallel to a
given line (that is ordinary, or Euclidean space); or such
MODEKN GEOMETRIC CONCEPTS 489
that through a given point no line can be drawn parallel
to a given line (spherical space) ; or such that through a
given point more than one line can be drawn parallel to a
given line (pseudo- spherical space).
These different kinds of space differ in many of their
properties. For example, in the first of them the sum of
the angles of a triangle equals two right angles; in the
second, it is greater; in the third, it is less.
These different kinds of space, however, have many
properties in common. Thus, in all of them every point in
the perpendicular bisector of a line is equidistant from the
extremities of the line.
EXERCISES. GROUP 85
Ex. 1. Show by the use of zero and negative arcs that the princi-
ples of Arts. 257, 263, 258, 264, 265, are particular cases of the general
theorem that the angle included between two lines which cut or touch
a circle is measured by one-half the sum of the intercepted arcs.
Ex. 2. Show that the principles of Arts. 354 and 358 are particular
cases of the theorem that, if two lines are drawn from or through a
point to meet a circumference, the product of the segments of one line
equals the product of the segments of the other line.
Ex. 3. Show by the use of negative angles
that theorem XXXVIII, Book I, is true for a
quadrilateral of the form ABCD. [BCD is a
negative angle ; the angle at the vertex D is
the reflex angle
Ex. 4. What is the reciprocal of the state-
ment that two intersecting straight lines deter-
mine a plane ?
Ex. 5. What is the reciprocal of the statement that three planes
perpendicular to each other determine three straight lines perpen'
dicular to each other ?
H. HISTORY OF GEOMETRY
847. Origin of Geometry as a Science. The beginnings
of geometry as a science are found in Egypt, dating back
at least three thousand years before Christ. Herodotus
says that geometry, as known in Egypt, grew out of the
need of remeasuring pieces of land parts of which had been
washed away by the Nile floods, in order to make an equi-
table readjustment of the taxes on the same.
The substance of the Egyptian geometry is found in an
old papyrus roll, now in the British museum. This roll
is, in effect, a mathematical treatise written by a scribe
named Ahmes at least 1700 B.C., and is, the writer states,
a copy of a more ancient work, dating, say, 3000 B. C.
848. Epochs in the Development of Geometry. From
Egypt a knowledge of geometry was transferred to Greece,
whence it spread to other countries. Hence we have the
following principal epochs in the development of geometry;
1. Egyptian : 3000 B. C.— 1500 B. C.
2. Greek : 600 B. C.— 100 B. C.
3. Hindoo : 500 A. D.— 1100 A. D.
4. Arab : 800 A. D.— 1200 A. D.
5. European : 1200 A. D.
In the year 1120 A. D., Athelard, an English monk,
visited Cordova, in Spam, in the disguise of a Mohamme-
dan student, and procured a copy of Euclid m the Arabic
language. This book he brought back to central Europe,
where it was translated into Latin and became the basis of
all geometric study m Europe till the year 1633, when,
(490)
HISTORY OF GEOMETRY 491
owing to the capture of Constantinople by the Turks,
copies of the works of the Greek mathematicians in the
original Greek were scattered through Europe.
HISTORY OF GEOMETRICAL METHODS
-•*. f-
849. Rhetorical Methods. By rhetorical methods in
the presentation of geometric truths, is meant the use of
definitions, axioms, theorems, geometric figures, the rep-
resentation of geometric magnitudes by the use of letters,
the arrangement of material in Books, etc. The Egyptians
had none of these, their geometric knowledge being re-
corded only in the shape of the solutions of certain numeri-
cal examples, from which the rules used must be inferred.
Thales (Greece 600 B.C.) first made an enunciation of
an abstract property of a geometric figure. He had a rude
Idea of the geometric theorem.
Pythagoras (Italy 525, B.C.) introduced formal defini-
tions into geometry, though some of those used by him
were not very accurate. F,or instance, his definition of a
point is "unity having position." Pythagoras also
arranged the leading propositions known to him in
something like logical order.
Hippocrates (Athens, 420 B. C.) was the first systemati-
cally to denote a point by a capital letter, and a segment
of a line by two capital letters, as the line AB, as is done
at present. He also wrote the first text -book on geometry.
Plato (Athens, 380 B. C.) made definitions, axioms
and postulates the beginning and basis of geometry.
To Euclid (Alexandria, 280 B. C.) is due the division
of geometry into Books, the formal enunciation of theo-
rems, the particular enunciation, the formal construction,
492 GEOMETKY. APPENDIX
proof, and conclusion, in presenting a proposition. He
also introduced the use of the corollary and scholium.
Using these methods of presenting geometric truths,
Euclid wrote a text -book of geometry in thirteen books,
which was the standard text -book on this subject for
nearly two thousand years.
The use of the symbols A, ZZ7 , || , etc., in geometric
proofs originated in the United States in recent years.
850. Logical Methods. The Egyptians used no formal
methods of proof. They probably obtained their few crude
geometric processes as the result of experiment.
The Hindoos also used no formal proof. One of their
writers on geometry merely states a theorem, draws a
figure, and says "Behold ! "
The use of logical methods of geometric proof is due to
the Greeks. The early Greek geometricians used experi-
mental methods at times, in order to obtain geometric
truths. For instance, they determined that the angles at
the base of an isosceles triangle are equal, by folding half
of the triangle over on the altitude as an axis and observ-
ing that the angles mentioned coincided as a fact, but
without showing that they must coincide.
Pythagoras (525 B. C.) was the first to establish geo-
metric truths by systematic deduction, but his methods
were sometimes faulty. For instance, he believed that the
converse of a proposition is necessarily true.
Hippocrates (420 B.C.) used correct and rigorous de-
duction in geometric proofs. He also introduced specific
varieties of such deduction, such as the method of reduc-
ing one proposition to another (Art. 296), and the reductio
ad absurdum.
HISTORY OF GEOMETRY 493
The methods of deduction used by the Greeks, however, were de-
fective in their lack of generality. For instance, it was often thought
necessary to have a separate proof of a theorem for each different
kind of figure to which the theorem applied.
Thus, the theorem that the sum of the an-
gles of a triangle equals two right angles
was proved,
(1) for the equilateral triangle by use
of the regular hexagon ;
(2) for the right triangle by the use of
a rectangle ;
(3) for a scalene triangle by dividing
the scalene triangle into two right triangles.
The Greeks appeared to fear that a
general proof might be vitiated if it were
applied to a figure in any way special or
peculiar. In other words, they had no conception of the principle of
continuity (Art. 842).
Plato (380 B. C.) introduced the method of proof by
analysis, that is, by taking a proposition as true and work-
ing from it back to known truths (see Art. 196).
To Eudoxus (360 B. C.) is virtually due proof by the
method of limits; though his method, known as the
method of exhaustions, is crude and cumbersome.
Apollonius (Alexandria, 225 B. C.) used projections,
transversals, etc., which, in modern times, have developed
into the subject of projective geometry.
851. Mechanical Methods. The Greeks, in demonstra-
ting a geometrical theorem, usually drew the figure em-
ployed in a bed of sand. This method had certain advan-
tages, but was not adapted to the use of a large audience.
At the time when geometry was being developed in Greece, the
interest in the subject was very general. There was scarcely a town
but had its lectures on the subject. The news of the discovery of a
494 GEOMETKY. APPENDIX
new theorem spread from town to town, and the theorem was redemon-
strated in the sand of each market place.
The Greek treatises, however, were written on vellum
or papyrus by the use of the reed, or calamus, and ink.
In Roman times, and in the middle ages, geometrical
figures were drawn in wax smeared on wooden boards,
called tablets. They were drawn by the use of the stylus,
a metal stick, pointed at one end for making marks, and
broad at the other for erasing marks. These wax tablets
were still in use in Shakespeare's time (see Hamlet Act I,
Sc. 5, 1. 107). The blackboard and crayon are modern
inventions, their use having developed within the last one
hundred years.
The Greeks invented many kinds of drawing instruments
f6r tracing various curves. It was due to the influence of
Plato (380 B. C.) that, in constructing geometric figures,
the use of only the ruler and compasses is permitted.
HISTORY OF GEOMETRIC TRUTHS. PLANE GEOMETRY
852. Rectilinear Figures. The Egyp-
tians measured the area of any four-
sided field by multiplying half the sum
of one pair of opposite sides by half the
sum of the other pair; which was equivalent to using the
a~rc^.b-\-d
formula, area = 0 X 0 •
_ —
This, of course, gives a correct result for the rectangle and square,
but gives too great a result for other quadrilaterals, as the trapezoid,
etc. Hence Joseph, of the Book of Genesis, in buying the fields of
the Egyptians for Pharoah in time of famine by the use of this
formula, in many cases paid for a larger field than he obtained.
The Egyptians had a special fondness for geometrical
constructions, probably growing out of their work as temple
HISTORY OF GEOMETRY 495
builders. A class of workers existed among them called
"rope -stretchers," whose business was the marking out of
the foundations of buildings. These men knew how to
bisect an angle and also to construct a right angle. The
latter was probably done by a method essentially the same
as forming a right triangle whose sides are three, four and
five units of length. Ahmes, in his treatise, has various
constructions of the isosceles trapezoid from different data.
Thales (600 B. C.) enunciated the following theorems:
If two straight lines intersect, the opposite or vertical
angles are equal;
The angles at the base of an isosceles triangle are equal;
Two triangles are equal if two sides and the included
angle of one are equal to two sides and the included angle
of the other;
The sum of the angles of a triangle equals two right
angles ;
Two mutually equiangular triangles are similar.
Thales used the last of these theorems to measure the
height of the great pyramid by measuring the length of
the shadow cast by the pyramid and also measuring the
length of the shadow of a post of known height at the same
time and making a proportion between these quantities.
Pythagoras (525 B. C.) and his followers discovered
correct formulas for the areas of the principal rectilinear
figures, and also discovered the theorems that the areas of
similar polygons are as the squares of their homologous
sides, and that the square on the hypotenuse of a right
triangle equals the sum of the squares on the other two
sides. The latter is called the Pythagorean theorem.
They also discovered how to construct a square equivalent
to a given parallelogram, and to divide a given line in
mean and extreme ratio,
496 GEOMETRY. APPENDIX
To Eudoxus (380 B. C.) we owe the general theory of
proportion in geometry, and the treatment of incommen-
surable quantities by the method of Exhaustions. By the
use of these he obtained such theorems as that the areas
of two circles are to each other as the squares of their
radii, or of their diameters.
In the writings of Hero (Alexandria, 125 B. C.) we first
find the formula for the area of a triangle in terms of its
sides, K=V s(s — a) (s — b) (s — c) - Hero also was the first
to place land-surveying on a scientific basis,,
It is a curious fact that Hero at the same time gives an incorrect
formula for the area of a triangle, viz., lT=ia(&-f-c), this formula
being apparently derived from Egyptian sources.
Xenodorus (150 B. C.) investigated isoperemetrical
figures.
The Romans, though they excelled in engineering, ap-
parently did not appreciate the value of the Greek geom-
etry. Even after they became acquainted with it, they
continued to use antiquated and inaccurate formulas for
areas, some of them of obscure origin. Thus, they used
the Egyptian formula for the area of a quadrilateral,
K=^- X— — . They determined the area of an equilat-
JJ —
eral triangle whose side is a, by different formulas, all
incorrect, as K=-r , £T=i(a2+a), and j£=Ja2.
853. The Circle. Thales enunciated the theorem that
every diameter bisects a circle, and proved the theorem
that an angle inscribed in a semicircle is a right angle.
To Hippocrates (420 B. C.) is due the discovery of
nearly all the other principal properties of the circle given
in this book.
HISTORY OF GEOMETRY 497
The Egyptians regarded the area of the circle as equiva-
lent to ff of the diameter squared, which would make
The Jews and Babylonians treated n as equal to 3.
Archimedes, by the use of inscribed and circumscribed
regular polygons, showed that the true value of n lies
between 3| and 3yf; that is, between 3.14285 and 3.1408.
The Hindoo writers assign various values to 7t, as 3, 3i,
1/10, and Aryabhatta (530 A. D.) gives the correct ap-
proximation, 3.1416. The Hindoos used the formula
1/2 _ 1/4— AB2 (^ee Art. 468) in computing the numeri-
cal value of 71.
Within recent times, the value of n has been computed
to 707 decimal places.
The use of the symbol n for the ratio of the circum-
ference of a circle to the diameter was established in
mathematics by Euler (Germany, 1750).
HISTORY OF GEOMETRIC TRUTHS. SOLID GEOMETRY
854. Polyhedrons. The Egyptians computed the vol-
umes of solid figures from the linear dimensions of such
figures. Thus, Ahmes computes the contents of an Egyp-
tian barn by methods which are equivalent to the use
3c
of the formula V=aXbX— . As the shape of these barns
is not known, it is not possible to say whether this formula
is correct or not.
Pythagoras discovered, or knew, all the regular poly-
hedrons except the dodecahedron. These polyhedrons were
supposed to have various magical or mystical properties.
Hence the study of them was made very prominent.
FF
498 GEOMETEY. APPENDIX
Hippasus (470 B.C.) discovered the dodecahedron, but
lie was drowned by the other Pythagoreans for boasting
of the discovery.
Eudoxus (380 B. C.) showed that the volume of a pyra°
mid is equivalent to one -third the product of its base by
its altitude.
E. F. August (Germany, 1849) introduced the prisma-
toid formula into geometry and showed its importance.
855. The Three Round Bodies. Eudoxus showed that
the volume of a cone is equivalent to one -third the area of
its base by its altitude.
Archimedes discovered the formulas for the surface and
volume of the sphere.
Menelaus (100 A. D.) treated of the properties of
spherical triangles.
Gerard (Holland, 1620) invented polar triangles and
found the formulas for the area of a spherical triangle and
of a spherical polygon.
856. Non-Euclidean Geometry. The idea that a space
might exist having different properties from those which
we regard as belonging to the space in which we live, has
occurred to different thinkers at different times, but
Lobatchewsky (Russia, 1793-1856) was the first to make
systematic use of this principle. He found that if, instead
of taking Geom. Ax. 2 as true, we suppose that through a
given point in a plane several straight lines may be drawn
parallel to a given line, the result is not a series of absur-
dities or a general reductio ad absurdum; but, on the con-
trary, a consistent series of theorems is obtained giving
the properties of a space.
III. REVIEW EXERCISES
EXERCISES. GROUP 86
REVIEW EXERCISES IN PLANE GEOMETRY
Ex. 1. If the bisectors of two adjacent angles are perpendiculai
to each other, the angles are supplementary.
Ex. 2. If a diagonal of a quadrilateral bisects two of its angles,
the diagonal bisects the quadrilateral.
Ex. 3. Through a given point draw a secant at a given distance
from the center of a given circle.
Ex. 4. The bisector of one angle of a triangle and of an exterior
angle at another vertex form an angle which is equal to one -half the
third angle of the triangle.
Ex. 5. The side of a square is 18 in. Find the circumference of
the inscribed and circumscribed circles.
Ex. 6. The quadrilateral ADBC is inscribed in a circle. The diag-
onals AB and DC intersect in the point F. Arc AD = 112°, arc A C =
108°, LAFC= 74°. Find all the other angles of the figure.
Ex. 7. Find the locus of the center of a circle which touches two
given equal circles.
Ex. 8. Find the area of a triangle whose sides are 1 m., 17 dm.,
210 cm.
Ex. 9. The line joining the midpoints of two radii is perpendicular
to the line bisecting their angle.
Ex. 10. If a quadrilateral be inscribed in a circle and its diag-
onals drawn, how many pairs of similar triangles are formed ?
Ex. 11. Prove that the sum of the exterior angles of a polygon
(Art. 172) equals four right angles, by the use of a figure formed by
drawing lines from a point within a polygon to the vertices of the
polygon.
(499)
500 GEOMETRY. APPENDIX
Ex. 12. In a circle whose radius is 12 cm., find the length of the
tangent drawn from a point at a distance 240 mm. from the center.
Ex. 13. If two sides of a regular pentagon be produced, find the
angle of their intersection.
Ex.|14. In the parallelogram ABCD, points are taken on the
diagonals such that AP=BQ=CR=DS. Show that PQRS is a
parallelogram.
Ex. 15. A chord 6 in. long is at the distance 4 in. from the center
of a circle. Find the distance from the center of a chord 8 in. long.
Ex. Ii6. If B is a point in the circumference of a circle whose*
center is \O, PA a tangent at any point P, meeting OB produced at A,
and PD perpendicular to OB, then PB bisects the angle APD.
Ex. 17. Construct a parallelogram, given a side, an angle, and a
diagonal.
Ex. 18. Find in inches the sides of an isosceles right triangle
whose area is 1 sq. yd.
Ex. 19. Given the line a, construct qvi/2— 1)^
o
Ex. 20. If two lines intersect so that the product of the segments
of one line equals the product of the segments of the other, a cir-
cumference may be passed through the extremities of the two lines.
Ex. 21. Find the locus of the vertices of all triangles on a given
base and having a given area.
Ex. 22. On the figure p. 206, prove thatl^2+Z^=ZB2+JW2.
Ex. 23. The area of a rectangle is 108 and the base is three times
th'i altitude. Find the dimensions.
Ex. 24. If, on the sides AC and BC of the triangle ABC, the
squares, AD and BF, are constructed, AF and DB are equal.
Ex. 25. If the angle included between a tangent and a secant is
half a right angle, and the tangent equals the radius, the secant
passes through the center of the circle.
REVIEW EXERCISES IN PLANE GEOMETRY 501
Ex. 26. The sum of the areas of two circles is 20 sq. yds., and the
difference of their areas is 15 sq. yds. Find their radii.
Ex. 27. Construct an isosceles trapezoid, given the bases and a
leg.
Ex. 28. Show that, if the alternate sides of a regular pentagon
be produced to meet, the points of intersection formed are the vertices
of another regular pentagon.
Ex. 29. If a post 2 ft. 6 in. high casts a shadow 1 ft. 9 in. long,
how tall is a tree which, at the same time, casts a shadow 66 ft. long ?
Ex. 30. If two intersecting chords make equal angles with the
diameter through their point of intersection, the chords are equal.
Ex. 31. From a given point draw a secant to a circle so that the
external segment is half the secant.
Ex. 32. Find the locus of the center of a circle which touches a
given circle at a given point.
Ex. 33. If one diagonal of a quadrilateral bisects the other
diagonal, the first diagonal divides the quadrilateral into two equi-
valent triangles.
Ex. 34. In a given square inscribe # square having a given side.
Ex. 35. A field in the shape of an equilateral triangle contains
one acre. How many feet does one side contain ?
Ex. 36. If perpendiculars are drawn to a given line from the ver-
tices of a parallelogram, the sum of the perpendiculars from two
opposite vertices equals the sum of the other two perpendiculars.
Ex. 37. Any two altitudes of a triangle are reciprocally propor-
tional to the bases on which they stand.
Ex. 38. Construct a triangle equivalent to a given triangle and
having two given sides.
Ex. 39. The apothem of a regular hexagon is 20. Find the area
of the inscribed and circumscribed circles.
Ex. 40. M is the midpoint of the hypotenuse AB of e, right tri-
angle ABC. Prove 8 MC2=A£2+BC2-^AC2,
502 GEOMETRY. APPENDIX
Ex. 41. Transform a given triangle into an equivalent right tri-
angle containing a given acute angle.
Ex. 42. The area of a square inscribed in a semicircle is to the
area of the square inscribed in the circle as 2 : 5.
Ex. 43. If, on a diameter of the circle 0, OA = OB and AC is par-
allel to BD, the chord CD is perpendicular to AC.
Ex. 44. Find the radius of a circle whose area is equal to one-
third the area of the circle whose radius is 7 in.
Ex. 45. State and prove the converse of Prop. XXI, Book III.
Ex. 46. If, in a given trapezoid, one base is three times the other
base, the segments of each diagonal are as 1 : 3.
Ex. 47. If two sides of a triangle are 6 and 12 and the angle
included by them is 60°, fiad the length of the other side. Also find
this when the included angle is 45°; also, when 120°.
Ex. 48. How many sides has a polygon in which the sum of the
interior angles exceeds the sum of the exterior angles by 540°?
Ex. 49. If the four sides of a quadrilateral are the diameter of a
circle, the two tangents at its extremities, and a tangent at any other
point, the area of the quadrilateral equals one-half the product of the
diameter by the side opposite it in the quadrilateral.
Ex. 50. An equilateral triangle and a regular hexagon have the
same perimeter; find the ratio of their areas.
Ex. 51. To a circle whose radius is 30 cm. a tangent is drawn
from a point 21 dm. from the center. Find the length of the tangent.
Ex. 52. If two opposite sides of a quadrilateral are equal, and
the angles which they make with a third side are equal, the quad-
rilateral is a trapeaoid.
Ex. 53. If two circles are tangent externally and two parallel
diameters are drawn, one in each circle, a pair of opposite extremities
of the two diameters and the point of contact are collinear.
Ex. 54. If, in the triangle ABC, the line AD is perpendicular to
BD, the bisector of the angle B, a line through D parallel to BC
bisects AC.
REVIEW EXERCISES IN PLANE GEOMETRY 503
Ex. 55. Bisect a given triangle by a line parallel to a given line.
Ex. 56. If two parallelograms have an angle of one equal to the
supplement of an angle of the other, their areas are to each other as
the products of the sides including the angles.
Ex. 57. The sum of the medians of a triangle is less than the
perimeter, and greater than half the perimeter.
Ex. 58. If PARE is a secant to a circle through the center O, PT
a tangent, and TE perpendicular to PB, then PA : PR=PO : PB.
Ex. 59. Two concentric circles have radii of 17 and 15. Find the
length of the chord of the larger which is tangent to the smaller.
Ex. 60. On the figure, p. 244,
(a) Find two pairs of similar triangles;
(1} Find two dotted lines which are perpendicular to each other;
(c) Discover a theorem concerning points, not connected by lines
on the figure, which are collinear;
(d) Discover a theorem concerning squares on given lines.
Ex. 61. One of the legs, AC, of an isosceles triangle is produced
through the vertex, C, to the point F, and F is joined with D, the mid-
point of the base AB. DF intersects BC in E, Prove that CF is
greater than CE.
Ex. 62. The line of centers of two circles intersects their common
external tangent at P. PABCD is a secant intersecting one of the
two circles at A and B and the other at C and D. Prove
Ex. 63. Trisect a given parallelogram by lines drawn through a
given vertex.
Ex. 64. Find the area of a triangle the sides of which are the
chord of an arc of 120° in a circle whose radius is 1 ; the chord of an
arc of 90° in a circle whose radius is 2 ; and the chord of an arc of
60° in a circle whose radius is 3.
Ex. 65. Construct a triangle, given the median to one side and a
median and altitude on the other side.
Ex. 66. Two circles intersect at P and Q. The chord CQ is tan-
gent to the circle QPB at Q. APB is any chord through P. Prove
that AC is parallel to BQ.
504 GEOMETRY. APPENDIX
Ex. 67. In the triangle ABC, from D, the midpoint of BC, DE
and DF are drawn, bisecting the angles ADB and ADC, and meeting
AB at E and AC at ^. Prove EF \\ BC.
Ex. 68. Produce the side BC of the triangle ABC to a point P, so
that PBXPC=RA2.
Ex. 69. In a given circle inscribe a rectangle similar to a given
rectangle.
Ex. 70. In a given semicircle inscribe a rectangle similar to a
given rectangle.
Ex. 71. The area of an isosceles trapezoid is 140 sq. ft., one base
is 26 ft., and the legs make an angle of 45° with the other base. Find
the other base.
Ex. 72. Cut off one-third the area of a given triangle by a Hue
perpendicular to one side.
Ex. 73. Find the sides of a triangle whose area is 1 sq. ft., if
the sides are in the ratio 2:3:4.
Ex. 74. Divide a given line into two parts such that the sum of
the squares of the two parts shall be a minimum.
Ex. 75. If, from any point in the base of a triangle, lines are
drawn parallel to the sides, find the locus of the center of the paral-
lelogram so formed.
Ex. 76. Three sides of "a quadrilateral are 845, 613, 810, and the
fourth side is perpendicular to the sicies 845 and 810. Find the area.
Ex. 77. If BP bisects the angle ABC, and DP
bisects the angle CDA, prove that angle P=i sum
of angles A and C.
Ex. 78. Two circles intersect at P and Q.
Through a point A in one circumference lines APC
and AQD are drawn, meeting the other in C and D. Prove the tan-
gent at A parallel to CD.
Ex. 79. In a given triangle, draw a line parallel to the base and
terminated by the sides so that it shall be a mean proportional be-
tween thy segments of one side.
KEYIEW EXERCISES IN PLANE GEOMETRY 505
Ex. 80. Find the angle inscribed in a semicircle the sum of whose
Rides is a maximum.
Ex. 81. The bases of a trapezoid are 160 and 120, and the alti-
tude 140. Find the dimensions of two equivalent trapezoids into
which the given trapezoid is divided by a line parallel to the base.
Ex. 82. If the diameter of a given circle be divided into any two
segments, and a semicircumference be described on the two segments
on opposite sides of the diameter, the area of the circle will be di-
vided by the semicircumferences thus drawn into two parts having
the same ratio as the segments of the diameter.
Ex. 83. On a given straight line, AB, two segments of circles are
drawn, APB and AQB. The angles QAP and QBP are bisected by lines
meeting in R. Prove that the angle JR is a constant, wherever P and
Q may be on their arcs.
Ex. 84. On the side AB of the triangle ABC, as diameter, a cir-
cle is described. EF is a diameter parallel to EC. Show that EB
bisects the angle ABC.
Ex. 85. Construct a trapezoid, given the bases, one diagonal, and
an angle included by the diagonals.
Ex. 86. If, through any point in the common chord of two inter-
secting circles, two chords be drawn, one in each circle, through the
four extremities of the two chords a circumference may be passed.
Ex. 87. From a given point as center describe a circle cutting a
given straight line in two points, so that the product of the distances
of the points from a given point in the line may equal the square of a
given line segment.
Ex. 88. AB is any chord in a given circle, P any point on the
circumference, PM is perpendicular to AB and is produced to meet
the circle at Q ; AN is drawn perpendicular to the tangent at P.
Prove the triangles NAM and PAQ similar.
Ex. 89. If two circles ABCD and EBCF intersect in B and C and
have common exterior tangents AE and DF cut by EC produced at G
and H, then GH2=£C2-\-'A£!2,
506 GEOMETRY. APPENDIX
EXERCISES. CROUP 87
REVIEW EXERCISES IN SOLID GEOMETRY
Ex. 1. A segment of a straight line oblique to a plane is greater
than its projection on the plane.
Ex. 2. Two tetrahedrons are similar if a dihedral angle of one
equals a dihedral angle of the other, and the faces forming these
dihedral angles are similar each to each.
Ex. 3. A plane and a straight line, both of which are parallel to
the same line, are parallel to each other.
Ex. 4. If the diagonal of one face of a cube is 10 inches, find the
volume of the cube.
Ex. 5. Construct a spherical triangle on a given sphere, given the
poles of the sides of the triangle.
Ex. 6. Given AB _L MN,
AE and BF _L ME;
prove EF J. PM.
Ex. 7. The diagonals of a rectangular paral-
lelepiped are equal.
Ex. 8. What portion of the surface of a sphere
is a triangle each of whose angles is 140°?
Ex. 9. Through a given point pass a plane parallel to two given
straight lines.
Ex. 10. Show that the lateral area of a cylinder of revolution is
equivalent to a circle whose radius is a mean proportional between
the altitude of the cylinder and the diameter of its base.
Ex. 11. The volumes of polyhedrons circumscribed about equal
spheres are to each other as the surfaces of the polyhedrons.
Ex. 12. Find /Sand T of a regular square pyramid an edge of
whose base is 14 dm., and whose lateral edge is 250 cm.
Ex. 13. If two lines are parallel and a plane be passed through each
line, the intersection of these plane? is parallel to the given lines.
REVIEW EXEECISES IN SOLID GEOMETRY 507
If
Ex. 14. Given PH J_ plane AD,
/.PEH= /.PFH;
prove Z PEF= Z PFE.
Ex. 15. If a plane be passed
through the midpoints of three <?Z
edges of a parallelepiped which
meet at a vertex, the pyramid thus formed is what part of the
parallelepiped ?
Ex. 16. Find a point in a plane such that the sum of its distances
from two given points on the same side of the plane is a minimum.
Ex. 17. Given the points A, B, C, D in a plane and P a point
outside the plane, AB perpendicular to the plane PBD, and AC per-
pendicular to the plane PCD; prove that PD is perpendicular to
the plane ABCD.
Ex. 18. In a sphere whoso radius is 5, find the area of a zone the
radii of whose upper and lower bases are 3 and 4.
Ex. 19. Two cylinders of revolution have equal lateral areas.
Show that their volumes are as R : R' .
Ex. 20. The midpoints of two opposite sides of a quadrilateral in
space, and the midpoints of its diagonals, are the vertices of a
parallelogram.
Ex. 21. How many feet of two-inch plank are necessary to con-
struct a box twice as wide as deep and twice as long as wide (on the
inside), and to contain 216 cu. ft.?
Ex. 22o If two spheres with radii R and r are concentric, find the
area of the section of the larger sphere made by a plane tangent to
the smaller sphere.
Ex. 23. In the frustum of a regular square pyramid, the edges
of the bases are denoted by &i and &•_> and the altitude by H; prove
that i=li(61— 62)2-f4JH"2.
Ex. 24. If the opposite sides of a spherical quadrilateral are equal
the opposite angles are equal.
508 GEOMETRY. APPENDIX
Ex. 25. Obtain the simplest formula for the lateral surface of a
truncated triangular right prism, each edge of whose base is a, and
whose lateral edges are p, q, and r.
Ex. 26. The area of a zone of one base is a mean proportional
between the remaining surface of the sphere and its entire surface.
Find the altitude of the zone.
Ex. 27. The lateral edges of two similar frusta are as 1 : a. How
do their areas compare ? Their volumes ?
Ex. 28. Construct a spherical surface with a given radius, r, which
shall be tangent to a given plane, and to a given sphere, and also pass
through a given point.
Ex. 29. The volume of a right circular cylinder equals the area
of the generating rectangle multiplied by the circumference generated
by the point of intersection of its diagonals.
Ex. 30. On a sphere whose radius is 8i inches, find the area of a
zone generated by a pair of compasses whose points are 5 inches
apart.
Ex. 31. The perpendicular to a given plane from the point where
the altitudes of a regular tetrahedron intersect equals one -fourth the
sum of the perpendiculars from the vertices of the tetrahedron to the
same plane.
Ex. 32. Two trihedral angles are equal or symmetrical if their
corresponding dihedral angles are equal.
Ex. 33. On a sphere whose radius is a, a zone has equal bases
and the sum of the bases equals the area of the zone. Find the alti-
tude of the zone.
Ex. 34. A plane which bisects two opposite edges of a tetrahedron
bisects the volume of the tetrahedron.
Ex. 35. Find the locus of all points in space which have their
distances from two given parallel lines in a given ratio.
Ex. 36. If a, b, c are the sides of a spherical triangle, a', b', cf
the sides of its polar triangle, and a>&>c, then a'<b'<d '.
Ex. 37. A cone of revolution has a lateral area of 4 sq. yd. and
an altitude of 2 ft. How much of the altitude must be cut off by a
plane parallel to the base, in order to leave a frustum whose lateral
area is 2 sq. ft. ?
REVIEW EXERCISES IN SOLID GEOMETRY 509
Ex. 38. The total area of an equilateral cone is to the area of the
inscribed sphere as 9 : 4.
Ex. 39. Construct a sphere of given radius, r, whose surface shall
be tangent to three given spheres.
Ex. 40. The volume of the frustum of an equilateral cone is 300
cu. in. and its altitude is 20 in. Show how to find the radii of the bases.
Ex. 41. On each base of a cylinder of revolution a cone is placed,
with its vertex at the center of the opposite base. Find the radius of
the circle of intersection of the two conical surfaces.
Ex. 42. The volume of a frustum of a cone of revolution equals
the sum of a cylinder and a cone of the same altitude as the frustum,
and with radii which are respectively the half sum and the half differ-
ence of the radii of the frustum.
Ex. 43. A square whose side is a revolves about a line through
one of its vertices and parallel to a diagonal, as axis; find the surface
and volume generated.
Ex. 44. If a cone of revolution roll on another fixed cone of revo-
lution so that their vertices coincide, find the kind of surface gen-
erated by the axis of the rolling cone.
Ex. 45. An equilateral triangle whose side is a revolves about an
altitude as an axis; find the surface and volume generated by the
inscribed circle, and also by the circumscribed circle.
Ex. 46. Find the locus of the center of a sphere which is tan-
gent to three given planes.
Ex. 47. If an equilateral triangle whose side is a be rotated about
a line through one vertex and parallel to the opposite side, as an axis,
find the surface and volume generated.
Ex. 48. What other formulas of solid geometry may be regarded
as special cases of the formula for the volume of a prismatoid ?
Ex. 49. Through a given point pass a plane which shall bisect the
volume of a given tetrahedron.
Ex. 50. In an equilateral cone and a cone whose opposite ele-
ments are perpendicular at the vertex, show that the ratio of tli§
Vertical solid angles is as 2— j/3 : 2 — j/2.
APPLICATIONS OF SOLID GEOMETRY TO MECHANICS
AND ENGINEERING
EXERCISES. GROUP 93
(BOOKS VI AND VII)
1. A carpenter tests the flatness of a surface by applying a e traight
edge to the surface in various directions. How does a plasterer test
the flatness of a wall surface? What geometric principle is used by
these mechanics?
2. Explain why an object with three legs, as a stool or tripod,
always rests firmly on the floor while an object with four legs, as a table,
does not always rest so. Why do we ever use four-legged pieces of
30' furniture?
3. How can a carpenter
get a corner post of a house
in a vertical position by use of
a carpenter's square? What
geometrical principle does he
use?
4. The diagram is the plan
of a hip roof. The slope of
each face of the roof is 30°.
Find the length of a hip rafter
as AB.
20
1 2
[SUG.
C
Draw a triangle DBC representing a section of the roof at
20
DBC on the plan. Hence it may be shown that BC = — V3. In
o
like manner by taking a section through
BF, it is found that AC = 10. Hence in
The triangle ABC, AB may be found.]
6. Find the area of the entire roof
represented in Ex. 4.
6. Make drawings showing at- what
angle the two ends of a rafter like BC in Ex. 4 must be cut.
510
10'
PRACTICAL APPLICATIONS 511
7. Make drawings showing at what angle a jack rafter like 12 in
Ex. 4 must be cut.
[Sue. To determine how the end 1 of the jack rafter must be cut
use the principle that two intersecting straight lines determine a plane
(Art. 501). The cutting plane at 1 'must make an angle at the side
of the jack rafter equal to angle CBH, and on the top of the jack rafter
equal to angle ABC.]
8. What is a gambrel roof? Make up a set of examples concerning
a gambrel roof similar to Exs. 4-7.
9. By use of Art. 645, show that a page of this book held at twice
the distance of another page from the same lamp receives one fourth
the light the first page receives.
10. The supporting power of a wooden beam varies directly as the
area of the cross section times the height of the beam and inversely as
the length of the beam. Compare the supporting power of a beam
12 ft. long, 3 in. wide, and 6 in. high with that of a beam 18 ft. long, 4 in.
wide, and 10 in. high. Also compare the volumes of the two beams.
EXERCISES. GROUP 94
(BOOKS VIII AND IX)
1. A hollow cylinder whose inside diameter is 6 in. is partly filled
with water. An irregularly shaped piece of ore when placed in the
water causes the top surface of the water to rise 3.4 in. in the cylinder.
Find the volume of the ore.
2. What is a tubular boiler? What is the advantage in using a
tubular boiler as compared with a plain cylindrical boiler? If a tubular
boiler is 18 ft. long and contains 32 tubes each 3 in. in diameter, how
much more heating surface has it than a plain cylindrical boiler of the
same length and 36 in. in diameter? (Indicate both the long method and
the short method of making this computation and use the short method.)
3. If a bridge is to have its linear dimensions 1000 times as
great as those of a given model, the bridge will be how many times
as heavy as the model?
Why, then, may a bridge be planned so that in the model it will sup-
port relatively heavy weights, yet when constructed according to the
model, falls to pieces of its own weight?
Show that this principle applies to other constructions, such as
buildings, machines, etc., as well as to bridges.
512 GEOMETRY
4. Work again Exs. 23-25, 27, p. 473.
5. Make and work for yourself an example similar to Ex. 24, p. 473.
6. Sound spreads from a center in the form of the surface of an
expanding sphere. At the distance of 10 yd. from the source, how will
the surface of this sphere compare with its surface as it was at 1 yd.?
How, then, does the intensity of sound at 10 yd. from the source com-
pare with its intensity at a distance of 1 yd.?
Does this law apply to all forces which radiate or act from a center
as to light, heat, magnetism, and gravitation? Why is it called the
law of inverse squares?
7. If a body be placed within a spherical shell, the attractive forces
exerted upon the body by different parts of the shell will balance or
cancel each other. Hence a body inside the earth, as at the foot of a
mine, is attracted effectively only by the sphere of matter whose radius
is the distance from the center of the earth to the given body. Hence,
prove that the weight of a body below the surface of the earth varies
as the distance of the body from the center of the earth.
[Sua. If W denote the weight of the body at the surface, and w its
weight when below, R the radius of the earth, and r the distance of the
body from the center when below
the surface, show that
8. The light of the sun falling
on a smaller sphere, as on the earth or the moon, causes that body to
cast a conical shadow. Denoting the radius of the sun by R, the
radius of the smaller sphere by r, the distance between the two spheres
dr
by d, and the length of the shadow by Z, show that I = — •
ri — r.
Find I when d = 92,800,000 mi., R = .433,000 mi., r = 4000 mi.
9. If in a lunar eclipse the moon's center should pass through the
axis of the conical shadow, and the moon is traveling at the rate of
2100 mi. an hour, how long would the total eclipse of the moon last?
flow long if the moon's center passed through the earth's conical
shadow at a distance of 1000 mi. from the axis of the cone?
PRACTICAL APPLICATIONS 513
10. If the moon's diameter is 2160 mi., find the length of the moon's
shadow as caused by sunlight.
11. If the distance of the moon from the earth's center varies from
221,600 mi. to 252,970, show how this explains why some eclipses of the
sun are total and others annular.
Why, also, at a given point on the earth's surface is an eclipse of the
sun a so much rarer sight than an eclipse of the moon? Why, also, is its
duration so much briefer?
12. Prove that the latitude of a place on the
earth's surface equals the elevation of the pole
(that is, on the diagram, prove /_ QEA = /_PAO).
^ Z 13. Given the sun's declina-
tion (i.e. distance north or south
of the celestial equator), show
how to determine the latitude
H •* of a place by measuring the zenith distance of the
sun. Also by measuring the altitude of the sun above the horizon.
14. How was Peary aided by the principles of Exs. 26 and 27 in
determining whether he had arrived at the North Pole?
16. If on April 6 (the day of the year on which Peary was at the
North Pole) the sun was 6° 7' north of the celestial equator, how high
above the horizon should the sun have been as observed by Peary?
At what hour of the day was this?
16. What is the sextant? Explain and prove the principle of the
sextant.
17. Explain and prove the principle of the angle-meter.
FORMULAS OF PLANE GEOMETRY
SYMBOLS
a, 6, c= sides of triangle ABC. K= radius of a circle.
s=$ (a-f 6 + c). D= diameter of a circle.
7ic=altitude on side c. C= circumference of a circle.
me= median on side c. r= radius of an inscribed
/c=bisector of angle opposite circle.
side c. T=^7-approx. (or 3.1416—).
I and m = line segments. .fir=area.
P=perimeter. &=base of a triangle.
>S»=side of a regular polygon li — altitude of a triangle.
of n sides. 61 and &2=bases of a trapezoid.
LENGTHS OF LINES
1. In a right triangle, C being the right angle,
c2=a2 + 62. Art. 346.
2. In a right triangle, I and m being the projections of a and b
on c, and h, the altitude on c,
a2 = ZXc,WmXc. Art. 342.
3. In an oblique triangle, tn being the projection of & on c,
if a is opposite an acute Z , a'2 = 62 -j- c2 — 2 cX m> Art. 349.
if a is opposite an obtuse L , a2 = &2 + c2 + 2 c X w. Art. 350.
4. fcc=tl/s(s — a) (s — 6) (s — c). Art. 393.
5. wc=4-|/2 (a2 + 62) — c2. Art. 353.
6. fc=— — f/a&s (s — c). Art. 363.
7. If I and m are the segments of c made by the bisector of the
angle opposite, a:b = l:m. Arts. 332, 336.
(514}
FORMULAS OF PLANE GEOMETKY 515
8. If I and m are the segments of a line, a, divided in extreme and
mean ratio, and I > m, a :l=l :m. Art. 370.
(P:Pf =«:«'. Art. 341.
9. In similar polygons, ( ^ ; ^ ; &/ ^ ^
10. In circles, C : C' = R : R' ; also C : C'=>D : D1 . Art. 442.
11. C=2 irR, or C=irD. Art. 444.
12. An arc = eB*r*™*leX*B. Art. 445.
J-oU
13. In inscribed regular polygons,
S2n=V R ( 2 R — 1/4 R>— Sn2) . Art. 467 .
AREAS OF PLANE FIGURES
1. In a triangle, K=l Ih. Art. 389.
2. In a triangle, K=\/s (s—a) (s — b) (s — c). Art. 393.
3. In an equilateral triangle, K=C?^. Ex. 4, p. 257.
4. In a parallelogram, K=by^li. Art. 385.
5. . In a trapezoid, K=i h (61 + &2). Art. 394.
6. In a regular polygon, K=$ rXP. Art. 446.
7. In a circle, K^I? or JT=i vD\ Art. 442.
8. In a sector of a circle, Z"=i i?X arc, Art. 453.
9. In a segment of a circle, jK^= sector ± A formed by the chord
and radii of the segment.
10. In a circular ring, E=Tr (R2 — Rn). Art. 449.
11. In any two similar plane figures,
K:K'=a* :a/2; Art. 399,
also a : a' = /K : /!7. Art. 314.
12. In two circles, K : K'=R2 : .R/2 = Z>2 : Z>/2=C2 : C/2; Art. 452.
also R : R'=D : D> = C : C'=/K : /tf'. Art. 314.
FOEMULAS OF SOLID GEOMETRY
SYMBOLS
B, &=areas of the lower and up- P=perimeter of right section.
per bases of a frustum. P, p = perimeters of lower and up-
l£=lateral edge (or element); per bases of a frustum.
or r, rf = radii of bases.
= spherical excess. £= area of lateral surface; or
H= altitude. =area of surface of sphere,
I, b, 7& = length, breadth, height. etc.
i=slant height. T=area of total surface.
<Jf=area of midsection. V— volume.
FORMULAS FOR AREAS
1. In a prism, S=EXP> Art. 608.
2. In a regular pyramid, S=i LX?. Art. 641.
3. In a frustum of a regular pyramid, S=i(P-\-p) L. Art. 643.
4. In a cylinder of revolution, S=2 ^BH. Art. 697.
T=27rB(R + H). Art. 697.
5. In a cone of revolution, S=TRL. Art. 721.
T=TTK (L + E). Art. 721.
6. In a frustum of a cone of revolution, S=^L (B-\- r). Art. 727.
7. In a sphere, S=4 irg2, Or S=irIP. Art. 810.
8. In a zone, £=2 ^EH. Art. 813.
7J-J?2 A
9. In a lune, 8=-~~ . Art. 817.
10. In a spherical triangle, S^-. Art. 822.
loO
11. In a spherical polygon, s=-~ Art. 824.
180
(516)
VJi-' ' • * 2
FOKMULAS OF SOLID GEOMETRY 517
FORMULAS FOR VOLUMES
1. In a prism, V=B^H. Art. 628.
2. In a parallelepiped, F=lX^Xh. Art. 626.
3. In a pyramid, F=i BX H. Art. 651.
4. In a frustum of a pyramid, F=i H (B + 1 -\-\/ Bb). Art. 656.
5. In a prismatoid, V=\ H (B + 6 +4 M). Art. 663.
6. In a cylinder, V=BXH- Art. 698.
7. In a cylinder of revolution, V=irR'iH. Art. 699.
8. In a cone, F=i BXH. Art. 722.
9. In a circular cone, F=t TJ22^. Art. 723.
10. In a frustum of a cone, V=% H (B-\-l -\-\/ Bb). Art. 729.
11. In a frustum of a cone of revolution,
V=\TTR( U2 + r2 + Rr) . Art . 730 .
12. In a sphere, F=£ ^.B3, or F=|7rDs. Art. 832.
13. In a spherical sector, F=t ^-B2^. Art. 836.
14. In a spherical segment of two bases,
F=i (Trr2 + 7T/2) Jff + i 7T#3. Art. 837.
15. In a spherical segment of one base, F=TJH2 (R — ^H).
Art. 838.
CONSTANTS
1 aere = 43,560 sq. ft. ^2 = 1.2599 +
1 bushel = 2150.42 cu. in. ^3 = 1.4422 +
1 gallon = 231 cu. in, ^=-3183 +
l/2 = 1.4142+ v/7r=1.7725-
!/3 = 1.7321 - ^=0.5642 +
GG
513 GEOMETRY. APPENDIX
SUMMARY OF METRIC SYSTEM
TABLE FOR LEHGTH
10 millimeters (mm.) =1 centimeter (cm.)
10 cm. =1 decimeter (dm.)
10 dm. =1 meter (m.).
10 m. =1 Dekameter (Dm.)
10 Dm. =1 Hektometer (Hm.)
10 Hm. =1 Kilometer (Km.)
10 Km. = 1 Myriameter (Mm.)
Similar tables are used for the unit of weight, the gram ; for the
unit of capacity, the liter ; for the unit of land measure, the are; and
for the unit of wood measure, the stere.
TABLE FOR SQUARE MEASURE
100 sq. mm. = l sq. cm.
100 sq. cm. =1 sq. dm., etc.
TABLE FOR CUBIC MEASURE
1000 cu. mm. = l cu. cm.
1000 cu. cm. =1 cu. dm., etc.
A liter =1 cu. dm.
A gram — weight of 1 cu. cm. of water at 39.2° Fahrenheit.
An are =100 sq. m.
A stere =1 cu. m.
EQUIVALENTS
1 meter =39.37 inches.
1 liter =1.057 liquid quarts,
or .9581 dry quarts.
1 kilogram = 2. 2046 Ibs. av.
1 hektare =2.471 acres.
1 sq. m. — 1550— sq. in.
INDEX OF DEFINITIONS AND FORMULAS
PAGE
Altitude of cone . . . .412
of cylinder 403
of frustum of cone . .414
of frustum of pyramid . 378
of prism 3d
of pyramid .... 377
of spherical segment . . 402
of zone 452
Angle, dihedral . . . .337
formed by two curves . 433
of lune ...... 452
polyhedral 349
spherical 433
tetrahedral 349
trihedral 349
Angles of spherical polygon 438
Appolonius 493
Arabs 490
Archimedes . . . 497,498
Aryabhatta 497
August, E. F 498
Axis of circle of sphere . 427
of circular cone . . .413
of regular pyramid . . 378
of sphere 427
Babylonians 497
Base of cone . 412
of pyramid 377
of spherical pyramid . .461
of spherical sector . . .461
PAGE
Bases of cylinder . . .403
of frustum of cone . .414
of frustum of pyramid . 378
of prism 361
of spherical segment . . 462
of zone 452
Bodies, The Three Round . 492
Center of sphere .... 425
Circle 496
great 427
small .427
Cone 412
altitude of 412
axis of 413
base of 412
circular 413
circular, formula for vol-
ume of 418
circular, formulas for lat-
eral, a total area of .417
elements of 412
lateral surface of . . .412
oblique circular . . .413
of revolution . . . .413
right circular . . . .413
vertex of ..... 412
Cones, similar . . . .413
Conical surface . . . .412
directrix of . . . .412
element of .... 412
519
520
INDEX OF DEFINITIONS AND FORMULAS
PAGE
Conical surface, generatrix
of 412
nappes of 412
vertex of 412
Constants 513
Continuity, principle of . 485
Cube 363
Curved spaces .... 488
Cylinder 403
altitude of .... 403
bases of 403
circular 404
circular, properties of . 405
elements of 403
lateral surface of ... 403
oblique 404
of revolution .... 404
of revolution, formulas for
lateral and total areas
of 409
right 404
right circular .... 404
right section of ... 405
section of 405
Cylinders of revolution,
similar 404
Cylindrical surface . . . 403
directrix of 403
element of 403
generatrix of .... 403
Degree, spherical .... 452
Determined plane , 319,320-
Diagonal of polyhedron . 360
Diameter of sphere . . . 425
Dihedral angle . . . .337
edge of 337
faces of 337
plane angle of . . . .338
right 337
PAGE
Dihedral angles, adjacent . 337
equal 337
vertical 337
Directrix of conical surface 412
of cylindrical surface . 403
Distance from point to plane 328
on surface of sphere . . 428
polar, of great circle . . 429
polar, of small circle . . 429
Dodecahedron . . . .360
Duality, principle of . . 486
Edge of dihedral angle . . 337
Edges of polyhedral angle . 349
of polyhedron .... 360
Egyptians . 490, 491, 492,494, 497
Elements of conical surface 412
of cylindrical surface . 403
Equivalent solids . . . 363
Euclid 491
Eudoxus . . . 493,496,498
Eulor 497
European 490
Faces of dihedral angle . . 337
of polyhedral angle . . 349
of polyhedron .... 360
Figures, rectilinear . . . 494
Formulas of plane geometry
for lengths of lines 514, 515
for plane geometry, for
areas of plane figures . 515
of solid geometry, for
areas 516
of solid geometry, for vol-
umes 517
Foot of line 320
Frustum of cone . . .414
altitude of 414
bases of . . 414
INDEX OF DEFINITIONS AND FORMULAS
521
PAGE
Frustum of cone, formula for
lateral area of ... 420
formula for volume of . 422
lateral surface of . . .414
slant height of . . .414
Frustum of pyramid . .378
altitude of ..... 378
bases of ...... 378
slant height of . . .378
Generatrix of conical surface 412
of cylindrical surface . 403
Geometry, epochs in develop-
ment of . . . . .490
history of ... 490-498
modern ...... 485
Non-Euclidean . . 488,496
origin of ..... 490
projecti ve ..... 485
solid ...... 319
Gerard
Greeks .
498
490,492,493,494
Hero ....... 496
Hexahedron ..... 360
Hindoos . . . 490,492,497
Hippasius ...... 498
Hippocrates . . 491,492,496
History of geometry . 490-498
Homology, principle of . 487
Hyperspace ..... 488
Icosahedron ..... 360
Inclination of line to plane 347
Jews
497
Lateral area of frustum of
pyramid, formula for . 379
Lateral area of frustum
prism
of pyramid ....
Lateral edges of -prism .
of pyramid ....
Lateral faces of prism .
of pyramid ....
Lateral surface of cone
of cylinder ....
of frustum of cone
Line parallel to plane
perpendicular to plane
Lobatchewsky
Logical methods .
Lune
angle of
formula for area ofj
spherical degrees
formula for area of,
square units of area
PAGE
of
. 361
. 377
. 361
. 377
. 361
. 377
. 412
. 403
. 414
. 321
. 320
. 498
. 492
. 452
. 452
in
. 457
in
. 457
Mechanical methods . . . 493
Menelaus 498
Methods, logical . . . .492
mechanical 493
rhetorical . . . . .491
Metric system, summary of 514
Modern geometry . . .485
Nappes of cone . . . .412
Negative quantities . 485, 486
Non-Euclidean geometry
488, 498
Octahedron
Origin of geometry
. . . 360
. . . 490
Parallelepiped
right
. . . 362
362
rectangular
. . . 363
522
INDEX OF DEFINITIONS AND FORMULAS
PAGE
Parallel planes . . . .321
Perpendicular planes . . 338
Plane 319
determination of . 319,320
Planes, parallel . . . .321
Plato 491,493
Polar distance of circle . 429
triangle 441
Pole 427
Polygon, spherical . . . 438
Polyhedral angle . . .349
convex 349
edges of 349
face angles of .... 349
faces of 349
vertex of 349
Polyhedral angles, equal 350
symmetrical .... 350
vertical . . . . . .350
Polyhedron 360
convex 360
diagonal of 360
edges of 360
faces of 360
regular 391
section «)f 360
vertices of 360
Polyhedrons 497
classification of ... 360
similar 396
Postulate of solid geometry 320
Prism 361
altitude of 361
bases of 361
circumscribed about cyl-
inder 405
inscribed in cylinder . . 405
lateral area of . . . .361
lateral edges of . . .361
.lateral faces of ... 361
PAGE
Prism, oblique 362
quadrangular .... 362
regular 362
right 362
right section of . . .361
triangular 362
truncated 362
Prismatoid 389
bases of 389
formula for volume of . 390
Prismoid 389
Projection of line on plane 338
of point on plane . . . 338
Pyramid 377
altitude of . . . . . . 377
axis of 378
base of 377
circumscribed about cone 414
frustum of .... 378
inscribed in cone . . .414
lateral area of . . . . 377
lateral edges of . . .377
lateral faces of ... 377
quadrangular .... 377
regular 377
regular, slant height of . 378
spherical 461
triangular . . . . . 377
truncated 378
vertex of 377
Pythagoras . 491, 492, 495, 497
Radius of sphere . . . 425
Reciprocity, principle of . 486
Rectilinear figures . . . 494
Rhetorical methods . . .491
Right section of cylinder . 405
of prism 361
Romans .... 494,496
Round Bodies, The Three 498
INDEX OF DEFINITIONS AND FORMULAS
523
PAGE
Section of Polyhedron . . 360
of sphere 461
Segment of sphere . . . 462
Similar cones of revolution 413
cylinders of revolution . 404
Sectors of circles . . .275
Segments of circles . . . 275
Slant height of cone . .413
of frustum of cone . .414
of frustum of pyramid . 378
of regular pyramid . .377
Solid geometry . . ' . .319
Solids, equivalent . . . 363
Spaces, curved .... 488
Sphere 425
axis of 427
center of 425
circumscribed about poly-
hedron 432
diameter of .... 425
formula for area of sur-
face of 455
formula for volume of, . 463
great circle of .... 427
inscribed in polyhedron . 432
poles of 427
radius of 426
small circle of ... 427
Spherical angle .... 433
degrees 452
excess .... 444, 460
Spherical polygon . . . 438
angles of 438
sides of 438
vertices of 438
Spherical pyramid . . . 461
base of 461
vertex of 461
Spherical sector . . . .461
base of 461
PAGE
Spherical sector, formula for
volume of .... 464
Spherical segment . . . 462
altitude of 462
bases of 462
formulas for volumes of . 465
of one base .... 462
Spherical triangle . . . 438
bi-rectangular .... 443
formula for area of, in
square units of area . 460
tri-rectangular .... 443
Spherical triangles, supple-
mental 442
symmetrical .... 444
Spherical wedge . . . .461
Spherid 452
Surface, conical . . . .413
cylindrical 403
Tangent, line to sphere
plane to cone .
plane to cylinder .
plane to sphere
spheres ....
Tetrahedral angle
Tetrahedron .
Thales . . .
Triangle, polar
spherical
Trihedral angle
bi-rectangular
isosceles
rectangular
tri-rectangular
491,
. 425
. 413
. 404
. 426
. 426
. 349
. 360
495, 496
. 441
, . 438
. 349
. 350
. 350
. 350
. 350
Ungula 461
Unit of volume .... 363
Units of spherical surface . 452
524
INDEX OF DEFINITIONS AND FORMULAS.
Vertex of cone
of polyhedral angle .
of pyramid
PAGE
. 412
. 349
. 377
Wedge, spherical
Xenodorus
PAGE
. . . 461
. . . 496
of spherical pyramid .
Vertices of polyhedron
. 461
. 360
. . . 452
of spherical polygon
. 438
altitude of .
452
bases of
452
Volume of solid .
. 363
of one base .
. . . 452
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