MATHEMATICS-MENSURATION. 



Co*, a If A B C D abed be a parallelepiped* (i. e., a 

 prism on a parallelogram for a base), it r\g. ss. 



can be divided into two equal prisms 

 by a pUttie A Cca passing through the 

 diagonals of its bases ; for it u ob- 

 vious that the edges of these prisms 

 are eq.ua!, each to each, and also that 

 the plane angles containing the solid 

 angles at D and B are equal (See 

 Fig. 32). 



Cot. 4. Again (in Fig. 31), if we 

 suppose the base DEF of the one 

 prism to be in all respects equal to 

 that of the other def, and if the angles 

 at D are equal to the angles at d, each to each, then it 

 ia plain that if D B is > db the prism BDFE is > 

 prism Mft, and if D B < db. the prism BDFE < 

 1'risin baft. Also if BD is double of bd, the prism 

 IJ D F E is double of the prism bdfe ; and generally if 

 B D is any multiple of bd, then BDFE is the same 

 multiple of ''//. 



COK. 6. If we suppose (Fig. 31) BD produced to K 

 so that D 1C is any multiple of D B, the prism K D F E 

 is the same multiple of BDFE; and if we suppose db 

 produced to k so that kd is any multiple of bd, then 

 lidft is the same multiple of bdfe. But if KD > kd, 

 KDFE > kdfe; if equal, equal; if less, less .". 

 (Euclid, V., p. 130). 



BDFE : bdfe : : BD : bd. 



COR. 6. The results proved in Cors. 4 and 5 to be true 

 of prisms, are manifestly true of paralw". 



(2). Parallelepipeds on the same Base and of the same 



Altitude are equal to one another. 

 If the paral"" are on the same base and of the same 

 altitude, it is manifest that the ends opposite the 

 common base are in the same piano. 



(a) Suppose two of the edges of the ends opposite to 

 the base of each figure to be in the same straight line. 

 Fig. S3. 



.f ff 



Let D B E G, D Be<7, be the parallelepipeds on tlie 

 same base B D, and of the same altitudes, and suppose 

 that E H, eh, are in the same straight line, and also F G, 

 /</, in the same straight line. Then it is obvious that lie 

 = lih, H C = E D, and angle TiH C = angle eE D. 



.'.the base e K D = base ftH C, also HG = FE, and 

 the plane angles that form the solid angles at H and K 

 are equal each to each, .'. the prism 7iC HG = the prism 

 f i> I'. F. Hence if we suppose the former prism taken 

 fn.in the whole figure fcCDEF, and the latter to be 

 t ;ilii'ii from the name figure, the remainders will be equal 



(6) Suppose that no two of the edges of the side 

 opposite to the base are in the same straight line. 



Let D BEG, DBej (Fig. 34), be the paral"" 1 on the 

 same base BD, and being of the same altitude, their 

 ends E G, eg, are in the same plane. Produce EH, F G, 

 ft, gh, to meet, they will evidently form a parallelogram ; 

 let this parallelogram be KLMN, then LK, MN, and 

 FE, GH -AD, BC, each to each, and K N, LM = 

 eh, fg D C, A B. each to each, and angle M L K = 

 angle G F E = angle BAD. 



.'. LKMN is equal to ABCD. Join AL, BM, 

 OK, D K. Then D B M K is a parallelepiped ; and by 



We iball UM throughout the abbreviation paral 1 " 1 * 1 for parallelepiped. 



[CONTESTS OF Sol HIS. 



the first part of this proposition, B D M K is equal to 

 each of EDGE, and B Dye. .'. BD G E-BD*/*. Q. E.D. 



Fig. 31. 



(3). Solid ParaUekpipeds on eqtud Bases and of the same 

 Altitudes are equal to each other. 



(a) Suppose the edges to be perpendicular to the bases. 



Fig. 35. 

 J, 



To avoid a complicated figure wo will only letter the 

 bases, and will call the edgea of figure that are perpen- 

 dicular to the bases by the letters at the angles of the 

 base. Thus the edge B means the edge perpendicular to 

 the base at the point B i. e., B6. Let A C, D F be the 

 bases where we suppose the solids to be so placed as to 

 have a common edge D, and tho sides A D, D E in the 

 same straight line ; produce C D, and F E to meet in H ; 

 through G draw L G K parallel to C D, and produce B C 

 to meet K G in L 



Now since 1) H = G K, and H E is evidently = K F 

 and the angle D H E = G K F /. the base D H E = base 

 G K F ; and since edges H and K are perpendicular to 

 base, the solid angle at H is contained by piano angles 

 respectively equal to those containing the angle K ; and 

 hence the prisms whose bases are DUE and G K F are 

 equal. Add the solid on base D E K G to both, .'. 

 paral"* 1 on D K. = paral' 1 "* 1 D F. 

 Now paral"p"BD : paral"> d DL : : AD : : DG : : BD : 



DL 

 and paral"* D L : paral'i* DK::CD:DH::DL: 



DK 



.'. (Ex equali). 



paral"" 1 BD : paral"^ D K : : B D : D K 

 But B ]> = D K, since (Eucl. I., 35) D K = D F 

 .-. paral"" 1 B D = paral"* D K = paral"i ld D F. 



(6) If the edges are not perpendicular to the bases, 

 call the paral"" 1 P and Q. If on P's base a paral"" 1 ;) be 

 described, whose edges are perpendicular to bcose, and 

 whose altitude is same as P, then ;> = P by last Proposi- 

 tion ; and if q be in like manner described on same buso 

 as Q, but having its edges perpendicular to the base, 

 then (last Proposition) q Q, and by the former part 

 of this Proposition, q = ;>,.'. Q = P. Q. E. D. 



COB. 1. Hence if there be two paral 1 ' 1 " 1 of the same 

 altitude, but the base of the one be double of the base 

 of the other, the former is double of the latter ; and 

 generally if the base of the one be any multiple of tho 

 base of the other, the former paral"* 1 is the same multiple 



