ANSWERS. 



179 



x-- r * + 2 r ^ + 3 ^ 

 ~ 



4 r^-r, 



(38) Let Pi be the volume of the whole pyramid, F 2 that of the top cut 

 off, V that of the frustum ; x it x< 2 , x the distances of their centroids from 

 the lower base ; h^ h^ h z , their heights. Then the equation of moments is 

 ( F, y z )x = V&i y^c 2 . By geometry, we have V^j V t = r^/r} ; hence, 

 (r 1 3 rf)x = r 1 3 xi r 2 3 x 2 j also, x^= \h^ x z = h + %h 2 , h l -h^ = h, 

 hi/h z = r 1 /r 2 . By means of these relations, we find 



(~* r*\^- k r *~ 4*1*2* + 3^ - ^ ^ + 2 f^g + 3 ^ 



(n ' T "^^ -- ' r X = 4 - 



, . - ,(2a-hY 



(39) " = ii ' 



(40) Let A, B, C, D be the vertices of the tetrahedron, G its 

 centroid, GI that of the face ABC \ let a, b, c, d t x, Xi be the dis- 

 tances of these points from the plane ; and let the projections of these 

 points on the plane be denoted by A', B', C, >', G', GJ. Then, 

 since GG l /J)G l = 1/4, and GG l /DG l = (x ~x^/(d x^, we have 

 (x x\)l(d Xi)= 1/4; hence x = ^($x 1 + d). Let E be the mid- 

 dle point of AB, e its distance from the plane ; then, applying a similar 

 method to the triangle ABC,wt find Xi = ^(2 e -f c) = ^(a -f b + c) . 

 Hence, finally, x = \(a + b + c + d). 



(41) x = \h. (43) 3' = 



(42) y = ^yi- (44) ^ = f, ^=1^, 2 = f^. 



(45) (a) x=j=a. (d) x = ^ 7r ' +I28 a. 



9O 7T 



16 _ _ 2(11: TT 8) 



a, y=a4tf. (<?) ^= ^^ -- ^^z. 



15(3^-4) 



(46) Take as element a hemispherical shell of radius r and thick- 

 ness dr; x = ** 3 a. 



(47) x = \(H+h). 



(48) Compare problems (40) and (5), and apply the propositions 

 of Pappus, Arts. 30 and 42 ; V J-TT(/ -f- q -f r)A, where A is the area 

 of the triangle ; 6"= -rr\a(q-\- r} -f (r + /)-f- <:(/-!- ^)]. ../ 



