PROPOSITION II. THEOREM.

In any prism, the sections made by parallel planes are polygons equal in all their parts.

Let the prism AH be intersected by the parallel planes NP, SV: then are the sections NOPQR,

equal polygons.

For, the sides NO, ST, are parallel, being the intersections of parallel planes with a third plane ABGF; these sides, NO, ST, are included between the parallels NS, OT: hence, NO is equal to ST (B. I., P. XXVIII., C. 2). For like reasons, the sides OP, PQ, of NOPQR, are TV, VX, &c., of each; and since the

QR, &c.,

equal to the sides
STVXY,

each to

equal sides are par

allel, each to each, it follows that the

N

F

STVXY,

K

III

Y

angles NOP, OPQ, &c., of the first section, are equal to the angles STV, TVX, &c., of the second section, each to each (B. VI., P. XIII.): hence, the two sections NOPQR, STVXY, are equal in all their parts; which was to be proved.

Cor. The bases of a prism, and every section of a prism, parallel to the bases, are equal in all their parts.

If a pyramid be cut by a plane parallel to the base • 1°. The edges and the altitude will be divided proportionally: 2°. The section will be a polygon similar to the base.

Let the pyramid S-ABCDE, whose altitude is SO, be cut by the plane abcde, parallel to the base ABCDE.

1o. The edges and altitude will be divided proportionally. For, conceive a plane to be passed through the vertex S,

parallel to the plane of the base; then

will the edges and the altitude be cut by three parallel planes, and consequently they will be divided proportionally (B. VI., P. XV., C. 2); which was to be proved.

2o. The section abcde, will be similar to the base ABCDE. For, ab is parallel to AB, and be to BC (B. VI., P. X.) hence, the angle abc is equal to the angle ABC. In like manner, it may

be shown that each angle of the polygon abcde is equal to the corresponding angle of the base: hence, the two polygons are mutually equiangular.

Again, because ab is parallel to AB, we have,

ab AB :: sb : SB;

and, because bc is parallel to BC, we have,

bc : BC :: sb : SB;

hence (B. II., P. IV.), we have,

ab: AB :. bc : BC.

In like manner, it may be shown that all the sides of abcde are proportional to the corresponding sides of the polygon ABCDE: hence, the section abcde is similar to the base ABCDE (B. IV., D. 1); which was to be proved.

S-XY2,

€

Cor. 1. If two pyramids S-ABCDE, and having a common vertex S, and their bases in the same plane, be cut by a plane abc, parallel to the plane of their bases, the sections will be to each other as the bases.

For, the polygons abcd and ABCD, being similar, are

to each other as the squares of their homologous sides ab and AB (B. IV., P. XXVII); but,

abcde : ABCDE :: xyz: XYZ.

Cor. 2. If the bases are equal, any sections at equal dis tances from the bases will be equal.

Cor. 3. The area of any section parallel to the base, is proportional to the square of its distance from the vertex.

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

Let S be the vertex, ABCDE the hase, and SF, perpendicular to EA, the slant height of a right pyramid: then will the convex surface be equal to,

(AB+ BC + CD + DE + EA) × †SF.

Draw SO perpendicular to the plane of the base.

E

S

From the definition of a right pyramid, the point is the centre of the base (D. 11): hence, the lateral edges, SA, SB, &c., are all equal (B. VI., P. V.); but the sides of the base are all equal, being sides of a regular polygon : hence, the lateral faces are all equal, and consequently their altitudes are all equal, each being equal to the slant height of the pyramid.

Now, the area of any lateral face, as SEA, is equal to its base EA, multiplied by half its altitude SF: hence, the sum of the areas of the lateral faces, or the convex surface of the pyramid, is equal to,

(AB + BC + CD + DE + EA) × †SF ;

which was to be proved.

Scholium. The convex surface of a frustum of a right pyramid is equal to half the sum of the perimeters of its upper and lower bases, multiplied by the slant height.

Let ABCDE-e be a frustum of a right pyramid, whose vertex is S: then will the section abcde be similar to the base ABCDE, and their homologous sides will be parallel, (P. III.). Any lateral face of the frustum, as AEea, is a trapezoid, whose altitude is equal to Ff, the slant height of the frustum;

hence, its area is equal to

(B. IV., P. VII.).

(EA + ea) × Fƒ

But the area of the con

vex surface of the frustum is equal to the sum of the areas of its lateral faces; it is, therefore, equal to the half sum of the perimeters of its upper and lower bases, multiplied by the slant height.

PROPOSITION V. THEOREM.

If the three faces which include a triedral angle of a prism are equal in all their parts to the three faces which include a triedral angle of a second prism, each to each, and are like placed, the two prisms are equal in all their parts.

Let B and b be the vertices of two triedral angles, included by faces respectively equal to each other, and similarly placed then will the prism ABCDE-K be equal to the prism abcde-k, in all of its parts.

sides fg and gh, of one upper base, will coincide with the homologous sides of the other upper base; and because the upper bases are equal in all their parts, they must coincide throughout; consequently, each of the lateral faces of one prism will coincide with the corresponding lateral face of the other prism the prisms, therefore, coincide throughout, and ere therefore equal in all their parts; which was to be proved.

Cor. If two right prisms have their bases equal in all their parts, and have also equal altitudes, the prisms themselves will be equal in all their parts. For, the faces which include any triedral angle of the one, will be equal in all their parts to the faces which include the corresponding triedral angle of the other, each to each, and they will be similarly placed.