straight line; and, for the like reason, the parallelopiped A M is equal to the same AP. Therefore (I. ax. 1.) the parallelopipeds AG, A M are equal to one another. Therefore, &c.

PROP. 24. (Euc. xi. 31.) Parallelopipeds upon equal bases, and between the same parallel planes, are equal to one another.

Let AG, KQ be two parallelopipeds upon equal bases ABCD, KLMN, and between the same parallel planes; the parallelopipeds AG, KQ shall be equal to one another.

The bases ABCD, KLMN are either similar, or equiangular, or not equiangular.

In the first case, since they are equal as well as similar, they may be made to coincide; and the parallelopipeds will then stand upon a common base, and between the same parallel planes: therefore they are equal to one another.

Secondly, let them be equiangular, and not similar, having the angle at B equal to the angle at L; and therefore, also, their other angles equal, each to each, (I. 15. and I. ax. 3.). Pro

KLMN to cut the sides. It may be observed, however, that a parallelopiped can always be completed when its base ABCD and one of its principal edges AE are given, viz. by drawing the side-planes EAB, EAD, and then through BC, CD the side-planes BG, CH parallel to EAD and FAB respectively, and, lastly, through the point E, the upper base EG parallel to ABCD. This operation is analogons to that by which a parallelogram is completed from two adjoining sides (see note, page 17); and will be some. times understood when it is directed to complete a parallelopiped in future propositions.

duce AB to R, so that BR may be equal to KL; complete the parallelogram BCSR; join SB, and produce it to meet D A produced in T; through T draw TU parallel to A B, to meet CB and SR produced in the points V and U; complete the parallelopiped DX upon the base DTUS, and with the edge D H; produce the planes ABFE, CBFG, in order to complete the parallelopiped BX, and draw the diagonal plane S T Y Z of the parallelopiped DX. Then the parallelopipeds BH and BX are equal to one another, because, together with the parallelopipeds BY and BZ about the diagonal plane, they complete the whole parallelopiped DX (22. Cor. 2.). Again, because the bases BU and BD are complements of the parallelogram DU, BU is equal to BD (I. 23.), that is, to LN; but BR is equal to KL, and the angle RBV to the angle ABC (I. 3.), that is, to KLM: therefore, also, B V is equal to L M, and the bases BU, LN are both equal and similar. Therefore, by the first case, the parallelopiped KQ is equal to the parallelopiped B X, that is, to B H or AG.

Lastly, let the bases A B C D, K L M N be equal, but not equiangular. Make the

angle ABS, in the plane ABCD, equal to the angle K L M, and complete the parallelogram ABST, and the parallelopiped A X, upon the base ABST and with the edge BF. Then it is evident that the triangular prisms D H EATY, CGFBSX have the edges DH, CG equal to one another (22.), and the dihedral angles at those edges equal (17. Scholium 2.), and the sides containing those angles equal, each to each, for the sides A H, B G are opposite faces of the parallelopiped AG (22.), and the sides TH, SG stand upon equal bases (I. ax. 1. and ax. 2. or 3.) TD, SC, and between the same parallels (I. 25.); therefore, the prisms are equal to one another (21.); and, these being taken from the whole solid AEYTCGFB, there remains the parallelopiped A G, equal to A X. But the parallelopiped K Q is equal to AX by the preceding case; because their bases are equal (I. 24. and

contain the same straight line M, five, three, and four times respectively: the parallelopiped shall contain the cube of M, 5 x 3x4, or 60 times.

In the straight line AB, take the five parts A b, bb', &c. each equal to M, in AC the three parts Ac, cc', &c. each equal to M, and in AD the four parts Ad, dd', &c., each equal to the same M. Through the points b, c, and d, let planes be drawn parallel to the planes cAd, b Ad, and Ac respectively, (11. Cor. 1.) and let them meet one another in the point k; then Ak is equal to the cube of M. Let the planes ck, dk be produced to meet the plane BN; and let them be cut by the planes b'k', &c. which are drawn through 5', &c. parallel to the plane b k, or the plane CAD. Then, because the lines A b, bb', &c. are equal to one another, the bases of the parallelopipeds Ak, bk', &c. are equal to one another (I. 25.), and the parallelopipeds have the same altitude; therefore (24.), they are equal to one another, and the whole row AK is equal to five times the cube of M.

Again, through the points c', &c., let the planes d'1, &c. be drawn parallel to the plane c k, or (11, Cor. 2.) the plane BAD,

to cut the plane dk produced. Then, because the lines A c, cc', &c. are equal to one another, the bases of the parallelopipeds A K, cl, &c. are equal to one another (I. 25.); and the parallelopipeds have the same altitude; therefore (24.) they are equal to one another, and the whole tier A L is equal to three times AK; that is, to 3 x 5, or fifteen times the cube of M.

Lastly, through the points d', &c., let planes be drawn parallel to the plane Then, as before, because Ad, dd', &c. dk, or (11. Cor. 2.) the plane BAC. are equal, the parallelopipeds A L, dn, &c. are likewise equal to one another; and, therefore, the whole parallelopiped is to 4 x 3 x 5, or sixty times the cube AN is equal to four times AL, that of M.

And, it is evident that a similar demonstration may be applied, whatever other numbers be taken in place of the numbers 5, 3, and 4. Therefore, &c.

Scholium.

Hence the solid content of a rectangular parallelopiped is said to be equal to the product of its three dimensions, that is to ABX ACX AD, if AB, A C, AD are the three edges; this expression being interpreted in the same sense with the product of the two dimensions or sides, which is said to constitute the area of a rectangle, viz., that the number of cubical units in the parallelopiped is equal to the product of the numbers which denote how often the corresponding linear unit is contained in the three edges. Thus, if the linear unit be a foot, and the edges 3, 4, and 5 feet respectively, the solid content will be 3 x 4 x 5, or 60 cubic feet.

It is likewise said, in a similar sense, that the solid content of a rectangular parallelopiped is equal to the product of its base and altitude: thus, in the example just stated, the number of square feet in the base is 4 × 5, or 20; and this, being multiplied by 3, the number of linear feet in the altitude, gives 60 for the number of cubic feet in the parallelopiped, as before.

The cube is also the unit in the mensuration of all other solids; their contents being the same with the contents of rectangular parallelopipeds equal to them. Thus, since every parallelopiped (24. Cor.) is equal to a rectangular parallelopiped of equal base and altitude, the solid content of every pa

rallelopiped is equal to the product of its base and altitude. The prism and the pyramid will presently furnish new examples.

PROP. 26. (Euc. xi. 32.) Parallelopipeds, which have equal altitudes, are to one another as their bases; and parallelopipeds, which have equal bases, are to one another as their altitudes; also, any two parallelopipeds

are to one another in the ratio, which is compounded of the ratios of their bases and altitudes.

First, let A B, A C be two rectangular parallelopipeds, having equal altitudes, and let A E, A F be their bases; the parallelopipeds being so placed that a solid angle of the one may coincide, as at A, with a solid angle of the other. Then, the parallelopipeds will have a common part AD, which is likewise a rectangular parallelopiped, having the same altitude AL with them, and the two adjoining edges AG, AH common to it with AB, AC respectively. Let the base AK of this parallelopiped be divided into any number L of equal rectangles by lines parallel to AH; and, there- A fore, the parallelo

G

C

D

F

B

E

piped into the same number of equal parallelopipeds, by planes passing through these lines parallel to the planes LAH (24.) Then, if one of the first parts be contained in the base A E any number of times, exactly, or with a remainder; one of the others will be contained in the parallelopiped A B the same number of times, exactly, or with a corresponding remainder (24.). Therefore (II. def. 7.) the parallelopiped A B is to AD as the base A E to the base A K. And in the same manner it may be shewn that the parallelopiped AD is to AC as the base AK to the base AF. Therefore, ex æquali (II. 24.), the parallelopiped A B is to the parallelopiped A C as the base A E to the base A F.

Next, let AB, CD be two rectangular parallelopipeds, having equal bases; and let AE, CF be their altitudes. Then, if the altitude C F be divided into any number of equal parts; it may easily be shown that the parallelopiped CD will be divided into the same number of equal parallelopipeds, by planes

bases A E, CF, and the altitudes AG, CH. Take A L equal to CH, and let AK be a third parallelopiped having the base AE and the altitude AL. Then, because the parallelopiped A B has to CD the ratio which is compounded of the ratios of A B to A K, and A K to CD (II. def. 12.); which ratios, by what has been just demonstrated, are the same with those of AG to AL or CH, and AE to CF; the parallelopiped A B has to the parallelopiped CD, the ratio which is compounded of the ratios of the altitude AG to the altitude GH, and of the base A E to the base C F.

And what has here been demonstrated of rectangular parallelopipeds is true also with regard to any two parallelopipeds whatever, because these (24. Cor.) are equal to rectangular parallelopipeds having equal bases and altitudes with them.

Therefore, &c.

PROP. 27.

Any two rectangular parallelopipeds are to one another in the ratio, which is

compounded of the ratios of their edges.

Let A E, A e be two rectangular parallelopipeds, the first having the edges AB, AC, AD, the other the edges A6, Ac, Ad; the parallelopiped A E shall be to the parallelopiped AB in the ratio which is compounded of the

A

D

d

For, the parallelopipeds being placed so that a solid angle of the one may coincide with a solid angle of the other as at A, they will have a common part A F, which is likewise a rectangular parallelopiped; and the face of this parallelopiped, which is opposite to Db, being produced to meet the plane de, will cut off from the parallelopiped Ae, the part A G, which is also a rectangular parallelopiped. Now, the parallelopipeds AE, AF have a common base DC; therefore (26.) they are to one another as their altitudes AB, Ab; and, for the like reason, the parallelopipeds AF, AG are to one another as their altitudes A D, Ad, and the parallelopipeds AG, A e are to one another as their altitudes AC, Ac. But the parallelopiped A E has to the parallelopiped Ae the ratio which is compounded of the ratios of AE to AF, AF to AG, and AG to Ae. Therefore, the parallelopiped A E has to the parallelopiped Ae, the ratio which is compounded of the ratios of AB to Ab, AD to Ad, and A C to Ac; or, which is the same thing, which is compounded of the ratios of AB to Ab, AC to Ac, and AD to Ad. (II. 27.)*

This demonstration is analogous to that of II. 36. The proposition may, however, be otherwise, and more concisely established by the aid of Prop. 26. ; for, since the parallelopipeds are to one another in the ratio which is compounded of the ratios of their bases and altitudes, and that the bases, which are rectangles under two of the conterminous edges of each, are to one another in the ratio which is compounded of the ratios of those edges (II. 36.), and the altitudes equal to the third edges, it follows that

bC (26.), that is, as AB to Ab (II. 35. Cor.); and in like manner AF is to AG as AD to Ad, and AG to Ae as AC to Ac; therefore, as in the proposition, the parallelopiped AE is to the parallelopiped Ae, in the ratio which is compounded of the ratios of A B to Ab, AD to Ad, and AC to Ac; that is, in a ratio which is compounded of the ratio of the edges about the common angle A.

Cor. 2. Cubes are to one another in the triplicate ratio of their edges (II. 27. Cor. 2.); therefore, the triplicate ratio of two straight lines A B, ab is the same with the ratio of their cubes.

Cor. 3. If one straight line be to another as a third to a fourth, the cube of the first shall be to the cube of the second as the cube of the third to the cube of the fourth (II. 27.).

having its base EG equal to the base ABC of the prism, and its altitude FG the same with the altitude of the prism: the prism Abc shall be equal to the parallelopiped E F.

Let the bases A B C, a b c of the prism be completed into the parallelograms A D, ad, and let D d be joined. Then it is evident that the solid Ad is a parallelopiped (def. 11.). But the prism Abc is equal to half the parallelopiped A d (21.); and the parallelopiped EF is also equal to half the parallelopiped Ad, because the base E G is equal to AB C, that is, to half of the base A D, and the altitude FG is the same with the altitude Aa (26.). Therefore (I. ax. 5.) the prism Abc is equal to the parallelopiped E F Therefore, &c.

PROP. 29.

Every prism is equal to a rectangular

which is compounded of the ratios of their edges. And the first corollary (analogous to II. 36. Cor. 1.) admits of a demonstration little different.

and the same altitude.

Let AdC be a prism, having the