COXTKSTS OF SOLIDS.] 



MATHEMATICS. MENSURATION. 



651 



of the latter ; and hence, by reasoning similar to that 

 employed in Cor. 5, Prop. I. of the present article, it 

 appears that paral' 1 * 1 ' of equal altitudes are as their bases. 

 And again, if they have equal bases, they are to one an- 

 other as their altitudes. 



COR. 2. Let there be two paral^' P, P , the base and 

 height of one of which are A and A, and one of the 

 other A' and A', where AA A'A' are in numbers. And 

 suppose S to be a third paral'"" 1 on the base A' and the 

 height h. 



Then P : S : : A : A' 



S : P' : : A : A' 

 .'. P : F : : AA : A'A' 



COR. 3. If we suppose F to be a cube which has one 

 of its edges equal to unity, then A' = 1. and A' = 1. 



.'.P :F : : AA : 1. 



hence if we consider P' to be the unit of solid measure 

 (a cubic inch, or foot, for instance) then 



P= AA. 



Hence the volume of a paral'-f is found by multiplying 

 the area of one face by the distance between that face 

 and the opposite one. 



COR. 4. We have seen that a prism on a triangular 

 base is half of a paral l>1>d of the same altitude, and on a 

 base which is double of the triangle. Hence if A be 

 the area of the triangle, and A the height of the prism, 

 the volume of this paral""" 1 = 2A X A, and .*. the vol- 

 ume of the prism = A x A ; or the volume of a prism 

 on a triangular base is found by multiplying the area of 

 the base by the altitude. 



COR. 5. It is plain that a prism on a polygonal base 

 can be divided by planes passing through one edge of 

 the prism into a number of prisms on triangular bases, 

 each having the same altitude as the original prism. 



Let A, A 3 A, ... be the areas of these triangles, and 

 A the common altitude, the volumes of all these prisms 

 = (A t + A a + A s + ... ). A 



But the polygon is equal to all the triangles ; hence 

 if A be the area of the polygon, 



.'. The volume of the prism on a polygonal base = AA, 

 or is equal to the base multiplied by the altitude. 



COK. 6. It is plain that Cor. 5 is true of a regular 

 polygon of any number of sides, and therefore is true 

 in limit ; now when the number of sides of the polygon 

 is increased, its limit is the circumscribing circle, and 

 the limit of the corresponding prism is the circumscribing 

 cylinder. Hence the volume of a cylinder = AA, or 

 base X height. 



N.B. When we speak of the volume of a solid, a 

 prism for instance, being equal to the base multiplied by 

 the height, it is of course understood that all the mea- 

 surements are referred to the same unit Thus, if we 

 were asked what is the volume of a prism whose base is 

 2 square feet, and height 18 inches 1 the answer is, not 



18 

 2 X 18, but 2 X ^ = 3, and the 3 is iu CCTBIO MET. 



The cubic foot and the square foot being the unit of con- 

 tent and of area corresponding to the linear unit one foot. 



(4). If P, A B C be any pyra- ^ x 



mid on a triangular bate, and if 

 through the middle point (D) of 

 one of the tidet (AP) we draw 

 planet (D G H and D E F) pa- 

 rallel to one of the facet (P B C) 

 and to the bate (A B C) of the 

 pyramid, then if we ntppote a 

 plane to be drawti through D E B 

 and DH, cutting off a pritm 

 EFD, HKC, thefigureKDG 

 H C B it double of the pritm 

 EDF, HKC. 



Since the plane E D F is parallel to A B C, E D ia 



parallel to A B. Now A P is bisected in D. .'. P B is 

 bisected in E. Similarly A C is bisected in H. Also 

 since plane D E F is parallel to A B C. .'. E D is paral- 

 lel to K H, and .'. K H is parallel to A B. Draw C N 

 perpendicular to A B, meeting K H in M ; then since 

 C H : H A : : C M : M N (Euclid VI., 2), and C H = 



i H A .'. C M = M N. Hence, if from N and C per- 

 per.diculars are drawn on the plane, D E K H, they are 



i equal Hence paral^ 1 on the base D E K H, with 

 these perpendiculars respectively for altitudes, are equal. 

 Now the prism B E K H D G is half the former, and 

 the prism K C H D F E is half the latter. Hence the 

 prisms are equal, and they are together double of one 

 of them. But the two prisms make up the figure 

 E D G H C B, which is, therefore, double of the prism 

 E D F H K C. 



(5). Pyramids on triangular Bases and of equal Altitudes 

 are to one another as their Bases. 



Let P, ABC, p, abc be the two pyramids, and let 

 them be divided by planes as in the last proposition. 

 Fig. 37. 



Now efd : bca in the duplicate ratio of fd : ca ; but 

 fd : ca : : 1 : 2 .'. efd : bca : : 1 : 4. Similarly EFD : 

 BCA : : 1:4 .'. efd : bca : : E F D : B C A or efd : 

 E F D : : fccrt : B C A. But the prism efdkch : prism 

 EFDKCH ::efd :EFD : : abc : ABC. .'. (by last 

 proposition) 

 The double prism edghcb : E D G H C B : : a&c : A B C. 



Now if we suppose pd and P D, and also da and D A 

 to be equal, figures will be formed p,dfe, and P, D F E of 

 the same kind as in p,a6c and P, A B C, and also in d,agh 

 and D, A G H, and these figures will have to each other 

 the ratios efd : E F D and agh : A G H respectively ; 

 which ratios are each equal to the ratio abc : A B C. 

 \nd the name will be true, however often we continue 

 to bisect the bisections of the sides, and hence 



All the double prisms in 



p, abc : all those in P, A B C : : abc : A B C. 



And the number of successive bisections being as great 

 as we please, this is true in the limit. Now the pyramids 

 are the limits of the sum of these double prisms. Hence 



p,abc : P, A B C : : abc : A B C. Q. E. D. 



Cor. : Hence pyramids on equal bases and of the same 

 altitude are equal. 



(6). Every Pritm on a triangular Base can be divided into 



three equal Pyramids, 

 Let ABE, DFC be the prism; 

 draw a plane through D E and 0, cut- 

 ting off the pyramid E, DFC, and 

 through E D and B draw a plane di- 

 viding the remainder into two pyra- 

 mids E, ABD and E, BCD. Now 

 since these two latter have their ver- 

 tices coincident at E, and since their 

 bases ABD, B D E are manifestly 

 equal, being halves of the parallelogram 

 A C, the two pyramids E, ABD and 

 E, B D E are equal Now the pyramid 

 E, A B D is clearly the same as the 

 pyramid D, A B E. But the base ABE 

 = base DBF, and the perpendicular from E on D F E, 

 is equal to that from D on ABE since these planes aro 

 parallel ; hence the pyramid D, A B E = pyramid E, 

 D F C, and therefore the three pyramids are equal. 



Fig. 38. 



