these same sides ST, NO, are included between the parallels NS, OT, which are sides of the prism: hence NO is equal to ST. For like reasons, the sides OP, PQ, QR, &c., of the section NOPQR, are respectively equal to the sides TV, VX, XY, &c., of the section STVXY. And since the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, &c., of the first section

are respectively equal to the angles STV, TVX, &c., of the second. (13.6.) Hence the two sections NOPQR, STVXY are equal polygons. Therefore,

In every prism, sections formed by parallel planes, are equal polygons.

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

PROPOSITION V. THEOREM.

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

For, since the sides of a right prism are perpendicular to its base; (Def. 7. 7;) all F its lateral faces must be rectangles; hence its convex surface must be equal to the sum of all the rectangles AG, BH, &c., which compose it. Now AF, the altitude of the prism, is the common altitude of all these rectangles, and the perimeter of the prism is

A

B

K

E

H

equal to the sum of all their bases, AB+BC+CD+DE+EA.

And since the measure of a rectangle is the product of its base by its altitude, (4. 4. Sch.,) therefore the sum of all the rectangles which compose the convex surface of the prism is equal to (AB+BC+CD+DE+EA) × AF. Therefore,

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

If two parallelopipedons AG, AL, have a common base ABCD, and if their upper bases EFGH, IKLM lie in the same plane and between the same parallels EK, HL, those two parallelopipedons will be equivalent.

triangular prism AEIDHM is equal to the triangular prism BFKCGL.

Since AE is parallel to BF, and HE to GF, the angle AEI=BFK, HEI=GFK, and HEA=GFB. Of these six angles the first three form the solid angle E, the last three the solid angle F; therefore, the plane angles being respectively equal, and similarly arranged, the solid angles F and E must be equal. (21. 6. Sch.) Now, if the prism AEM is laid on the prism BFL, the base AEI being placed on the base BFK, will coincide with it because they are equal; and

since the solid angle E is equal to the solid angle F, the side EH will fall on its equal FG: and nothing more is required to prove the coincidence of the two prisms throughout their whole extent, for the base AEI and the edge EH determine the prism AEM, as the base BFK and the edge FG determine the prism BFL; (1.7;) hence these prisms are equal.

But if the prism AEM is taken from the solid AL, there will remain the parallelopipedon AIL; and if the prism BFL is taken from the same solid, there will remain the parallelopipedon AEG; hence those two parallelopipedons AIL, AEG are equivalent. Hence,

Any two parallelopipedons which have a common base, and their upper bases in the same plane and between the same parallels, are equivalent.

Two parallelopipedons AG, AL having the same base ABCD and the same altitude, are equivalent.

equal and parallel to LK. Let the sides EF, HG be produced, and likewise KL, IM, till by their intersections they

form the parallelogram NOPQ; this parallelogram will evidently be equal to either of the bases EFGH, IKLM. Now if a third parallelopipedon be conceived, having for its lower base the same ABCD, and NOPQ for its upper, this third parallelopipedon will be equal to the parallelopipedon AG, since with the same lower base, their upper bases lie in the same plane and between the same parallels GQ, FN. (6. 7.) For the same reason, this third parallelopipedon will also be equivalent to the parallelopipedon AL; hence the two parallelopipedons AG, AL, which have the same base and the same altitude, are equivalent. Therefore,

Two parallelopipedons having the same base and the same altitude, are equivalent.

PROPOSITION VIII. THEOREM.

Any parallelopipedon may be changed into an equivalent rectangular parallelopipedon having the same altitude and an equivalent base.

&c., rectangular. Hence if the base ABCD is a rectangle, AL will be a rectangular parallelopipedon equivalent to AG, the parallelopipedon proposed.

But if ABCD is not a rectangle, draw AO and BN perpendicular to CD, MQ and OQ and NP perpendicular to the base; you will then have the solid ABNOIKPQ, which will be a rectangu lar parallelopipedon: for by construction, the base ABNO and its opposite IKPQ are rectangles; so also are the lateral D faces, the edges AI, OQ, &c., being perpendicular to the plane of the base;

But

hence the solid AP is a rectangular parallelopipedon. the two parallelopipedons AP, AL may be conceived as having the same base ABKI and the same altitude AO: hence the parallelopipedon AG, which was at first changed into an equivalent parallelopipedon AL, is again changed into an equivalent rectangular parallelopipedon AP, having the same altitude AI, and a base ABNO equivalent to the base ABCD. Hence, Any parallelopipedon, &c.

Two rectangular parallelopipedons AG, AL, which have the same base ABCD, are to each other as their altitudes AE, AI.

First, suppose the altitudes AE, AI, to be to each other as two whole numbers, for example, as 15 is to 8. Divide AE into 15 equal parts; whereof AI will contain 8; and through x, y, z, &c. the points of division, draw planes parallel to the base.

E

Im

These planes will cut the solid AG into 15 partial parallelopipedons, all equal to A each other, because they have equal bases and equal altitudes,-equal bases, since

H

F

G

M

L

K

B