GM

Because AF, AK are parallelograms, EF and IK are each equal to AB, and therefore equal to each other. Hence, if EF and IK be taken away from the same line EK, the remainders EI and D FK will be equal. Therefore the triangle AEI is equal to the triangle BFK. Also, the parallelogram EM is equal to the parallelogram FL, and AH to BG. Hence the solid angles at E and F are contained by three faces which are equal to each other and similarly situated; therefore the prism AEIM is equal to the prism BFK-L (Prop. III.).

Now, if from the whole solid AL, we take the prism AEI-M, there will remain the parallelopiped AL; and if from the same solid AL, we take the prism BFK-L, there will remain the parallelopiped AG. Hence the parallelopipeds AL, AG are equivalent to one another.

Case second. When their upper bases are not between the same parallel lines.

Let the parallelopipeds AG, AL have the same base AC and the same altitude; then will their opposite bases EG, IL be in the same plane. And, since the sides EF and IK are equal and parallel to AB, they are equal and parallel to each other. For the same reason FG is equal and parallel to KL. Produce the sides EH, FG, as also IK, LM, and let

them meet in the points N, O, P, Q; the figure NOPQ is a parallelogram equal to each of the bases EG, IL; and, consequently, equal to ABCD, and parallel to it.

Conceive now a third parallelopiped AP, having AC for its Lower base, and NP for its upper base. The solid AP will be equivalent to the solid AG, by the first Case, because they have the same lower base, and their upper bases are in the same plane and between the same parallels, EQ, FP. For the same reason, the solia AP is equivalent to the solid AL: hence the solid AG is equivalen. to the solid AL. Therefore, parallelopipeds, &c.

PROPOSITION VII. THEOREM.

Any parallelopiped is equivalent to a right parallelopipea having the same altitude and an equivalent base.

Let AL be any parallelopiped; it is equivalent to a right parallelopiped having the same altitude and an equivalent

base.

M

HM

ᏀᏞ

K

From the points A, B, C, D draw AE, BF, CG, DH, perpendicular to the plane of the lower base, meeting the plane of the upper base in the points E, F, G, H. Join EF, FG, GH, HE; there will thus be formed the parallelopiped AG, equivalent to AL (Prop. VI.); and its lateral faces AF, BG, CH, DE are rectangles. If the base ABCD is also a rectangle, AG will be a right parallelopiped, and it is equivalent to the parallelopiped AL. But if ABCD is not a rectangle, from A and B draw AI, BK perpendicular to CD; and from E and F draw EM, FL perpendicular to GH; and join IM, KL. The solid ABKI-M will be a right parallelopiped. For, by construction, the bases ABKI and EFLM are rectangles; so, also, are the lateral faces, because the edges AE, BF, KL, IM are perpendicular to the plane of D the base. Therefore the solid AL is a right parallelopiped. But the two parallelopipeds AG, AL may be regarded as having the same base AF, and the same altitude AI; they are therefore equivalent. But the parallelopiped AG is equivalent to the first supposed parallelopiped; hence this parallelopiped is equivalent to the righ parallelopiped AL, having the same altitude, and an equiva ent base. Therefore, any parallelopiped, &c.

A

B

PROPOSITION VIII. THEOREM.

Right parallelopipeds, having the same base, are to each other as their altitudes.

Let AG, AL be two right parallelopipeds having the same base ABCD; then will they be to each other as their altitudes AE, AI.

Case first. When the altitudes are in the ratio of two whole numbers.

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Suppose the altitudes AE, AI are in the vatio of two whole numbers; for example, as seven to four. Divide AE into seven equal parts; AI will contain four of those parts. Through the several points of division, let planes be drawn parallel to the base; these planes will divide the solid AG into seven small parallelopipeds, all equal to each other, having equal bases and equal altitudes. The bases are equal, because every section of a prism parallel to the base is equal to the base (Prop. II., Cor.); the altitudes are equal, for these altitudes are the equal divisions of the edge AE. But of these seven equal parallelopipeds, AL contains four; hence the solid AG is to the solid AL, as seven to four, or as the altitude AE is to the altitude AI.

Case second. When the altitudes are not in the ratio of two whole numbers.

Let AG, AL be two parallelopipeds whose altitudes have any ratio whatever; we shall still have the proportion

Solid AG: solid AL:: AE : ÀI.

For if this proportion is not true, the first three terms remaining the same, the fourth term must be greater or less than AI. Suppose it to be greater, and that we have

Solid AG: solid AL:: AE: AO.

Divide AE into equal parts each less than OI; there will be at least one point of division between O and I. Designate that point by N. Suppose a parallelopiped to be constructed, having ABCD for its base, and AN for its altitude; and represent this parallelopiped by P. Then, because the altitudes AE, AN are in the ratio of two whole numbers, we shall have, by the preceding Case,

Solid AG: P::AE: AN.

But, by hypothesis, we have

Solid AG: solid AL:: AE: AO.

Hence (Prop. IV., Cor., B. II.),

Solid AL: P::AO: AN.

But AO is greater than AN; hence the solid AL must be greater than P (Def. 2, B. II.); on the contrary, it is less, which is absurd. Therefore the solid AG can not be to the solid AL, as the line AE to a line greater than AI.

In the same manner, it may be proved that the fourth term of the proportion can not be less than AI; hence it must be AI, and we have the proportion.

Solid AG: solid AL:: AE: AI.

Therefore, right parallelopipeds, &c.

PROPOSITION IX. THEOREM.

Right parallelopipeds, having the same altitude, are to each other as their bases.

Let AG, AN be two right parallelopipeds having the sam altitude AE; then will they be to each other as their bases; that is,

G

Solid AG: solid AN:: base ABCD: base AIKL. Place the two solids so that their M surfaces may have the common angle BAE; produce the plane LKNO till it meets the plane DCGH in the line PQ; a third parallelopiped AQ will thus be formed, which may be compared with each of the paral-I lelopipeds AG, AN. The two solids AG, AQ, having the same base AEHD, are to each other as their altitudes AB, AL (Prop. VIII.); and

the two solids AQ, AN, having the same base ALOE, are to each other as their altitudes AD, AI. Hence we have the two proportions

Solid AG: solid AQ:: AB: AL;

Solid AQ: solid AN:: AD: AI.

Hence (Prop. XI., Cor., B. II.),

Solid AG: solid AN:: ABXAD: ALXAI.

But ABXAD is the measure of the base ABCD (Prop. IV., Sch., B. IV.); and ALXAI is the measure of the base AIKL; hence

Solid AG: solid AN:: base ABCD: base AIKL. Therefore, right parallelopipeds, &c.

Any two right parallelopipeds are to each other as the prod ucts of their bases by their altitudes.

Let AG, AQ be two right parallelopipeds, of which the bases are the rectangles ABCD, AIKL, and the altitudes, the perpenaiculars AE, AP; then will the solid AG be to the solid AQ, as the product of ABCD by AE, is to the product of AIKL by AP.

Place the two solids so that their surfaces may have the common angle BAE; produce the planes necessary to form the third parallelo

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piped AN, having the same base with AQ, and the same altitude with AG. Then, by the last Proposition, we shall have Solid AG: solid AN :: ABČD : AIKL.

But the two parallelopipeds AN, AQ, having the same base AIKL, are to each other as their altitudes AE, AP (Prop. VIII.); hence we have

Solid AN: solid AQ :: AE: AP. Comparing these two proportions (Prop. XI., Cor., B. II.), we have

Solid AG solid AQ:: ABCD XAE: AIKL×AP.

If instead of the base ABCD, we put its equal ABXAD, and instead of AIKL, we put its equal AIX AL, we shall have Solid AG: solid AQ :: AB× AD × AE: AIXALXAP. Therefore, any two right parallelopipeds, &c.

Scholium. Hence a right parallelopiped is measured by the product of its base and altitude, or the product of its three dimensions.

It should be remembered, that by the product of two or more lines, we understand the product of the numbers which represent those lines; and these numbers depend upon the linear unit employed, which may be assumed at pleasure. If we take a foot as the unit of measure, then the number of feet in the length of the base, multiplied by the number of feet in its breadth, will give the number of square feet in the base. If we multiply this product by the number of feet in the altitude, it will give the number of cubic feet in the parallelopiped. If we take an inch as the unit of measure, we shall obtain in the same manner the number of cubic inches in the parallelopiped.