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EXERCISES.

1. Given a pyramid whose base is a rectangle 80 feet by 60 feet, and whose lateral edges are each 130 feet; find its volume, and its entire surface.

2. Given the frustum of a pyramid whose bases are heptagons; each side of the lower base being 10 feet, and of the upper base 6 feet, and the slant height 42 feet; find the convex surface in square yards.

3. Given a stick of timber 30 feet long, the greater end being 18 inches square, and the smaller end 15 inches square ; find its volume in cubic feet. .

4. Given a stone obelisk in the form of a regular quadrangular pyramid, having a side of its base equal to 25 decimetres, and its slant height 12 metres. The stone weighs 2.5 as much as water. What is its weight in kilogrammes ?

5. Given the frustum of a pyramid whose bases are squares ; each side of the lower base being 35 decimetres, each side of the upper base 25 decimetres, and the altitude 15 metres ; find its volume in steres.

6. Given a right hexagonal pyramid whose base is inscribed in a circle 30 feet in diameter, and whose altitude is 20 feet; find its convex surface, and its volume.

7. Given a right pentagonal pyramid whose base is inscribed in a circle 20 feet in diameter, and whose slant height is 30 feet; find its convex surface, and its volume.

8. Find the difference between the volume of the frustum of a pyrainid, and the volume of a prism of the same altitude whose base is a section of the frustum parallel to its bases and equidistant from them.

9. Given a stick of timber 32 feet long, 18 inches wide, 15 inches thick at one end, and 12 inches at the other; find the number of cubic feet, and the number of feet board measure it contains. Find equivalents for the results in the metric system. On SIMILAR POLYHEDRONS. 584. DEF. Similar polyhedrons are polyhedrons which have the same form. They have, therefore, the same number of faces, respectively similar and similarly placed, and their corresponding polyhedral angles equal.

585. DEF. Homologous faces, lines, and angles of similar polyhedrons are faces, lines, and angles similarly placed.

[blocks in formation]

I. The homologous edges of similar polyhedrons are proportional.

Since the faces S A B, SAC, SBC and A B C are similar respectively to S' A' B', S' A' C', S' B'C' and A' B'C', we have SA S B A B

-, etc. . SA ESB = A'B'

§ 278

II. Any two homologous faces of similar polyhedrons are proportional to the squares of any two homologous edges.

SAB. SA SAC SC2 SBC
Thus, S' A' B' SA2 S A' C S C12 S B' C' 9 342

III. The entire surfaces of two similar polyhedrons are proportional to the squares of any two homologous edges.

SAB SAC
Thus, since

S' A' B=S A' CI, etc.,

SA B + S A C, etc. SAB
S A' B' + S' A' C', etc. - SA' B =

4,= SA

42

$ 266 §

Place two horoloyous faces, A BEDE and A B C D E

518

GEOMETRY. — BOOK VII.

PROPOSITION XXII. THEOREM. 586. Two similar polyhedrons may be decomposed into the same number of tetrahedrons similar, each to each, and similarly placed.

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Let ABCDE-O P Q R S and A' B'C' D' E-OʻP Q R S be

two similar polyhedrons of which P and P are homologous vertices.

We are to prove that A B C D E-OPQRS and A'B'C'D' E'O' P'Q'R' S' can be decomposed into the same number of tetrahedrons similar and similarly placed.

· Place these polyhedrons in the same plane, having any two homologous edges, as A B and A' B' || and lying in the same direction.

On any two corresponding faces not adjacent to P and P', as A B C D E and A' B'C'D' E', from two homologous vertices, as E and E', draw diagonals dividing these faces into A, similar and similarly placed.

From the homologous vertices P, P' of the polyhedrons draw straight lines to the vertices of these A.

Repeat this construction for each of the faces not adjacent to P, P.

Then the polyhedrons will be divided into the same number of tetrahedrons ;

that is, into as many tetrahedrons as there are A in these faces.

Now, any two corresponding tetrahedrons, as P-A B E and P'-A' B' E', are similar ;

for the faces E A B and P A B are similar respectively to the faces E' A' B' and P'A' B',

§ 294 (being similarly situated o of similar polygons). In the A PBE and P' B' E'

PB is 1l to P' B', and B E to B' E', (since they make equal ts respectively with the II lines A B and A' B');

.LPBE=< P'B' E', $ 462 (two É not in the same plane having their sides II and lying in the same

direction are equal);

and

PB (AB) BE

§ 278
P' B = A' B') = B' E'
..face P B E is similar to face P' B' E'. § 284
Also, in the A P A E and P'A' E'

PE (PB) PA AB) A E 278.
PE =(PB) = PA = GA' B) = A' E' $ ?

(being homologous sides of similar A).

i face P A E is similar to face P'A' E'. § 282 Moreover, since any two corresponding trihedral of these tetrahedrons are formed by three plane s which are equal, each to each, and similarly situated, they are equal. . § 492

.:: P-A B E and P-A' B' E' are similar. $ 584 In like manner we may show that any other two tetrahedrons similarly situated are similar.

That is, the two similar polyhedrons have the same number of tetrahedrons similar each to each, and similarly situated.

Q. E. D.

587. COROLLARY. Any two homologous lines in two similar polyhedrons have the same ratio as any two homologous edges.

PROPOSITION XXIII. THEOREM. 588. Similar tetrahedrons are to each other as the cubes of their homologous edges.

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Let S-B C D and S-B'C'D' be two similar tetrahedrons

having for bases the similar faces BCD and B'C'D', and for altitudes SO and S O'.

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Apply the tetrahedron S'-B'C' D' to the tetrahedron S-BC D, so that the polyhedral S shall coincide with S.

Then the base B'C' D' will be ll to the face B C D,

since thetr ptanes make equal t-wible the face SB-0), and the I so, I to B C D, will also be I to B'C' D'. SO will be the altitude of the tetrahedron S-B'C' D'. NS-B C D BCDXSO BCD.S0

S-B'C' DIB'C' D' X SO=B'C' Di so'9070 (any two tetrahedrons are to each other as the products of their bases and

altitudes).
Since the bases are similar,

BC D В С2
.

§ 343

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