PROPOSITION V.

Parallelopipeds on the same base and between the same parallel planes, have equal volumes.

(1.) Let the parallelopipeds AG, AG' be upon the same base AC, and between the same parallel planes; and first let EF and HG be in the same straight lines with E'F' and H'G'.

The student must now prove that the triangular pyramid having HDH' and E'AE for its opposite triangular faces is equal to the triangular prism having G'GC and F'FB for its opposite faces.

The remaining step is analogous to Euc. 1. 35, and the conclusion is attained.

S

H'

(2.) Let the parallelopipeds be denoted, in the second place, by the same letters as before, but not have EF and GH coincident with E'F' and G'H', nor yet the other pairs of sides coincident.

Produce the sides H'E', F'G', HG, FE to meet. They will form a parallelogram PQRS, having its sides parallel and equal to those of the base ABCD; as the student must prove. Whence drawing AP, BQ, CR, and DS, these will be the edges of another parallelopiped on the base ABCD, and between the same parallel planes as the other two.

Now by the former case, the parallelopiped AG is equal to AR, by reason of PQ, RS being in the same lines with EF, GH.

And, similarly, the parallelopiped AG' is equal to AR by reason of PS, QR being in the same lines with H'E', F'G'.

The parallelopipeds AG, AG' therefore being each equal to AR, they are equal to one another.

'PROPOSITION VI.

Triangular prisms which have one face common, and the edges opposite situate in a plane parallel to that face, have equal volumes."

Let A'PSRB'Q and APSRBQ be two prisms which have the face PQRS common, and the edges AB, A'B' opposite to PR in a plane parallel to PR.

Complete the parallelopipeds SC, SC' on the face PR.

Then it is to be shown that the faces AC, A'C' are in that plane, and hence that the parallelopipeds are equal.

The application of Prop. III. will then establish the equality of the prisms.

PROPOSITION VII.

Prisms upon equal polygonal bases and between the same parallel planes are equal to one another in volume.

(1.) Every prism may be transformed into a parallelopiped.

For, let ABCDE, A'B'C'D'E' be corresponding portions of the opposite ends of a prism on the poly

gonal base. Join BD, B'D', and through C, C' draw CG, C'G' parallel to them; produce AB, A'B' to meet CG, C'G' in G and G'; and join GG'.

Then show that B'G' is equal to BG, and thence infer that GG' is parallel to BB', and therefore also to CC'.

The next step is to show that the plane GCC'G' is parallel to the common face BDD'B'; and hence (Prop. VI.) that the prism on the triangular base BGD is equal to that on BCD.

Wherefore, exchanging the former for the latter, the original prism on the polygonal base ABCDE is transformed into one on the base AGDE, having one side less than the original base; and consequently the new prism whilst it is equal in volume to the original one, has one face less.

We may proceed thus till the base is reduced to a quadrilateral, or to a triangle; and whichever we choose we can make a parallelogram equal to it; and consequently the process enables us to form a parallelopiped equal in volume to a given polygonal prism. We may, moreover, make the base of this parallelopiped similar to any specified parallelogram.

(2.) If a second polygonal prism be given, so that their bases have equal areas and the prisms are between the same parallel planes, then the same process may be pursued with the second one, and the bases of both are ultimately reduced to parallelograms equal in all respects.

(3.) We have hitherto supposed the prisms to retain the original parallelism of their edges. We might now suppose one base placed upon the other, and show that the opposite faces are in one plane; and thence infer this equality, and consequently the equality of the original prisms, by means of Prop. v. The following, however, is equally simple and more usual, being Euclid's own way, and analogous to his treatment of 1. 36.;

Let ABCDEFGH and A'B'C'D'E'F'G'H' be the two transformed parallelopipeds between the same parallel planes; and ABCD, A'B'C'D' the bases equal in all respects, and having the equal sides parallel each to each. Join AE', BF', CG', DH': then the figure ABCDE'F'G'H' is to be shown to be a parallelopiped. And by the application of Prop. VI., it is to be proved equal to each of the parallelopipeds AG, A'G'; and thence the inference drawn.

SCHOLIUM. As the parallelism of the edges may be under any angle whatever with the base of a prism, the edges are for the purposes of calculation generally taken perpendicular to the base. This perpendicular is generally called the altitude of the prism. When in a problem the conditions are not so given in a direct form, the preliminary step most frequently is, to reduce it to this state from the given relations; and, moreover, to reduce the base of the prism to an equivalent rectangle in other words, reducing the prism to its equivalent rectangular parallelopiped.

Nor are we obliged to confine this process to the mere practice of calculation; for from what has already appeared, we are equally entitled to do so in our theoretical investigations. The generality of our conclusions is not affected by this restriction, since the preceding theorem reduces (when it may be deemed advantageous) every prism whatever to a rectangular parallelopiped. It will be occasionally adopted here, not to evade any difficulties in the reasoning, but from

its simplifying the verbal enunciations of some of the theorems; and from its being at the same time in strict accordance with the ordinary language and notions of practical mensuration.

PROPOSITION VIII.

Prisms of equal altitude are to one another as their bases ; and prisms having equal bases are to one another as their altitudes.

(1.) Let the parallelopipeds KC, AG be taken, so that KB, AF their bases be equiangular, and the side AB of the bases and the face AC be common to both; and let them have the same altitude, namely, AD or BC the distance between the two planes KF, NG. Then will the parallelopiped KC be to AG as the base KB is to the base AF.

Take any number of segments in the line AE produced, viz., EU, UW, WX, etc., each equal to AE; from the points U, W, X, etc., draw parallels to EF, meeting BF (in points not lettered), and complete the parallelopipeds ET, UV, WY, etc.

Proceed similarly by taking KQ, QS, etc., each equal to AK, and completing the parallelopipeds QM, SP, etc.

To complete the demonstration, show:

(a.) That each of the parallelopipeds ET, UV, WY, etc., is equal to AG; and each of the parallelopipeds QM, SP, etc., is equal to KC. (b.) That whatever multiple AX is of AE, the same multiple is the base BX of the base BE, and the parallelopiped CX of the parallelopiped CE; and similarly for the parallelopipeds AM and AR.

(c.) That if AX be greater than AS, the base BX is greater than the base BS, and the parallelopiped AY than AR; but if equal, equal, and if less, less.

(d.) Apply Euc. v. Def. 5, which will establish

pard. AG: pard. AM::AE: AK

:: base AF: base AL.

(2.) Let the prisms be reduced to rectangular parallelopipeds; as in the figure, where AG and AG' are the rectangular parallelopipeds equivalent to the two prisms. In this it is to be shown:

(a.) That the construction is possible. (b.) That the parallelopipeds AG, AG' being viewed as standing on the bases BG and BG', they are to one another as BF to BF'.

(c.) To modify the expression of this conclusion to agree with the enunciation.

H

F

PROPOSITION IX.

All prisms are to one another in a ratio compounded of the ratio of their bases and the ratio of their altitudes.

This is easily deduced from the preceding, and the composition of proportions. It is left for the exercise of the student, as are likewise the two following.

PROPOSITION X.

Parallelopipeds having one trihedral angle of the one either equal or symmetrical to a trihedral angle of the other, are to one another in a ratio compounded of the ratios of the homologous edges terminating in those angles.

PROPOSITION XI.

Similar parallelopipeds are to one another as the cubes of their like linear dimensions;

Or again,

As the cubes of any homologous lines of the two figures.

PROPOSITION XII.

Triangular pyramids on equal bases and between the same parallel planes have equal volumes.

The demonstration of this proposition has never been effected in a manner analogous to the preceding ones: namely, by supraposition, the addition and subtraction of figures previously proved to be equal, or the comparison of each with some subsidiary figure. Several forms of proof have been published; but they all involve the same principle. We have, therefore, only to select that which appears to be most simple in detail; though even this is sufficiently complicated. We shall break it up as far as possible into distinct steps; dwelling, however, only on those which seem to require more than casual remark to an attentive f and intelligent student.

(a) Innumerable prisms may be constituted on the base of a given pyramid, having their opposite ends in a plane through the vertex of the pyramid, and parallel to the base.

Let S be the vertex, and ABC the base of the pyramid; take any point R within the base ABC, and join RS: then the edges of the prism AA', BB', CC' being drawn parallel to RS, the prism ABC A'B'C' itself will wholly enclose the pyramid ABCS.*

The same thing will occur if R, instead of being taken within the base, be taken in one of the sides, or at an angular point, the junction of two sides. In complex figures this is sometimes convenient; and in the former case to take the face upon which RS falls for the plane of one face of the prism, and in the other edge with which RS coincides for one of the edges of the prism.