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BOOK VII.

POLYEDRONS.

Definitions.

1. THE name solid polyedron, or simple polyedron, is given to every solid terminated by planes or plane faces; which planes, it is evident, will themselves be terminated by straight lines.

2. The common intersection of two adjacent faces of a polyedron is called the side, or edge of the polyedron.

3. The prism is a solid bounded by several parallelograms, which are terminated at both ends by equal and parallel polygons.

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To construct this solid, let ABCDE be any polygon; then if in a plane parallel to ABCDE, the lines FG, GH, HI, &c. be drawn equal and parallel to the sides AB, BC, CD, &c. thus forming the polygon FGHIK equal to ABCDE; if in the next place, the vertices of the angles in the one plane be joined with the homologous vertices in the other, by straight lines, AF, BG, CH, &c. the faces ABGF, BCHG, &c. will be parallelograms, and ABCDE-K, the solid so formed, will be a prism.

4. The equal and parallel polygons ABCDE, FGHIK, are called the bases of the prism; the parallelograms taken together constitute the lateral or convex surface of the prism; the equal straight lines AF, BG, CH, &c. are called the sides, or edges of the prism.

5. The altitude of a prism is the distance between its two bases, or the perpendicular drawn from a point in the upper base to the plane of the lower base.

6. A prism is right, when the sides AF, BG, CH, &c. are perpendicular to the planes of the bases; and then each of them is equal to the altitude of the prism. In every other case the prism is oblique, and the altitude less than the side.

7. A prism is triangular, quadrangular, pentagonal, hex. agonal, &c. when the base is a triangle, a quadrilateral, a pentagon, a hexagon, &c.

8. A prism whose base is a parallelogram, and which has all its faces parallelograms, is named a parallelopipedon.

The parallelopipedon is rectangular when all its faces are rectangles.

9. Among rectangular parallelopipedons, we distinguish the cube, or regular hexaedron, bounded by six equal squares.

10. A pyramid is a solid formed by several triangular planes proceeding from the same point S, and terminating in the different sides of the same polygon ABCDE.

The polygon ABCDE is called the base of the pyramid, the point S the vertex; and the triangles ASB, BSC, CSD, &c. form its convex or lateral surface.

11. If from the pyramid S-ABCDE, the pyramid S-abcde be cut off by a plane parallel to the base, the remaining solid ABCDE-d, is called a truncated pyramid, or the frustum of a pyramid.

E

F

A

S

B

12. The altitude of a pyramid is the perpendicular let fall from the vertex upon the plane of the base, produced if necessary.

13. A pyramid is triangular, quadrangular, &c. according as its base is a triangle, a quadrilateral, &c.

14. A pyramid is regular, when its base is a regular polygon, and when, at the same time, the perpendicular let fall from the vertex on the plane of the base passes through the centre of the base. That perpendicular is then called the axis of the pyramid.

15. Any line, as SF, drawn from the vertex S of a regular pyramid, perpendicular to either side of the polygon which forms its base, is called the slant height of the pyramid.

16. The diagonal of a polyedron is a straight line joining the vertices of two solid angles which are not adjacent to each other.

17. Two polyedrons are similar when they are contained by the same number of similar planes, similarly situated, and having like inclinations with each other.

PROPOSITION 1. THEOREM.

The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude.

Let ABCDE-K be a right prism: then will its convex surface be equal to (AB+BC+CD+DE+EA) × AF.

For, the convex surface is equal to the sum of all the rectangles AG, BH, CI, DK, EF, which compose it. Now, the altitudes AF, BG, CH, &e. of the rectangles, are equal to the altitude of the prism. Hence, the sum of these rectangles, or the convex surface of the prism, is equal to

(AB+BC+CD+DE+ÉA)×

F

K

H

E

A

B

AF; that is, to the perimeter of the base of the prism multiplied by its altitude.

Cor. If two right prisms have the same altitude, their convex surfaces will be to each other as the perimeters of their bases.

PROPOSITION II. THEOREM.

In every prism, the sections formed by parallel planes, are equal

polygons.

Let the prism AH be intersected by the parallel planes NP, SV; then are the polygons NOPQR, STVXY equal.

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For, the sides ST, NO, are parallel, being the intersections of two parallel planes with a third plane ABGF; these same sides, ST, NO, are included between the parallels NS, OT, which are sides of N the prism: hence NO is equal to ST. For like reasons, the sides OP, PQ, QR, &c. of the section NOPQR, are equal to the sides TV, VX, XY, &c. of the section STVXY, each to each. And since

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the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, &c. of the first section, are equal to the angles STV, TVX, &c. of the second, each to each (Book VI. Prop. XIII.). Hence the two sections NOPQR, STVXY, are equal polygons.

Cor. Every section in a prism, if drawn parallel to the base, is also equal to the base.

PROPOSITION III. THEOREM.

If a pyramid be cut by a plane parallel to its base, 1st. The edges and the altitude will be divided proportionally. 2d. The section will be a polygon similar to the base.

Let the pyramid S-ABCDE, of which SO is the altitude, be cut by the plane abcde; then will Sa: SA:: So: So, and the same for the other edges: and the polygon abcde, will be similar to the base ABCDE.

B

D

First. Since the planes ABC, abc, are parallel, their intersections AB, ab, by a third plane SAB will also be parallel (Book VI. Prop. X.); hence the triangles lar, and we have SA: Sa :: SB: Sb; we have SB Sb :: SC: Sc; and so on. SA, SB, SC, &c. are cut proportionally in a, b, c, &c. The altitude SO is likewise cut in the same proportion, at the point o; for BO and bo are parallel, therefore we have

SO: So :: SB : Sb.

SAB, Sab are simlfor a similar reason, Hence the edges

Secondly. Since ab is parallel to AB, bc to BC, cd to CD, &c. the angle abc is equal to ABC, the angle bed to BCD, and so on (Book VI. Prop. XIII.). Also, by reason of the similar triangles SAB, Sab, we have AB: ab :: SB: Sb; and by reason of the similar triangles SBC, Sbc, we have SB: Sb: BC: bc; hence AB: ab: BC: bc; we might likewise have BC: bc CD: cd, and so on. Hence the polygons ABCDE. abcde have their angles respectively equal and their homologous sides proportional; hence they are similar.

K

Cor. 1. Let S-ABCDE, S-XYZ be two pyramids, having a common vertex and the same altitude, or having their bases situated in the same plane; if these pyramids are cut by a plane parallel to the plane of their bases, giving the sections abcde, xyz, then will A the sections abcde, xyz, be to each other as the bases ABCDE,

XYZ.

For, the polygons ABCDE, abcde, being similar, their surfaces are as the squares of the homologous sides AB, ab; but AB ab: SA Sa; hence ABCDE: abcde :: SA2: Sa2. For the same reason, XYZ : xyz :: SX2 : Sx2. But since abc and xyz are in one plane, we have likewise SA : Sa

SX

Sx (Book VI. Prop. XV.); hence ABCDE: abcde :: XYZ xyz; hence the sections abcde, xyz, are to each other as the bases ABCDE, XYZ.

Cor. 2. If the bases ABCDE, XYZ, are equivalent, any sections abcde, xyz, made at equal distances from the bases, will be equivalent likewise.

PROPOSITION IV. THEOREM.

The convex surface of a regular pyramid is equal to the perimeter of its base multiplied by half the slant height.

For, since the pyramid is regular, the point O, in which the axis meets the base, is the centre of the polygon ABCDE (Def. 14.); hence the lines OA, OB, OC, &c. drawn to the vertices of the base, are equal.

In the right angled triangles SAO, SBO,
the bases and perpendiculars are equal:
since the hypothenuses are equal: and
it may be proved in the same way that
all the sides of the right pyramid are
equal. The triangles, therefore, which E
form the convex surface of the prism are
all equal to each other. But the area of
either of these triangles, as ESA, is equal

F

S

B

A

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