Centre of Gravity in particular Bodies. 53 



94. Hence, In order to find the centre of gravity G of a trape- 

 zoid, we draw KL through the middle points of the two parallel Fig- 34. 

 sides, and from these same points K, Z/, we draw the lines KA, 

 LD, to the vertices of the opposite angles A, D ; then having 

 taken 



we join EF, which will cut KL in G, the point sought. 



For, by reasoning as we have done in the case of the triangle, 

 we shall see that the centre of gravity G must be in KL. More- 

 over, since , F, are the centres of gravity respectively, of the 

 triangles CAD, ADB, which compose the trapezoid ABDC, the 93. 

 common centre of gravity of the two triangles or of the trapezoid 

 must be in EF ; it follows, therefore, that it is at the intersec- 77. 

 tion G. 



To find the distance LG, which we shall have occasion to use 

 hereafter, we draw the lines EH, FI, parallel respectively to AE ; 

 and since 



KE = \KA, and LF = -| LD, 

 we shall have 



EH=^AL, and FI = -* KD, 

 or 



EH = J AB, and FI = J CD. 



For the same reason, 



therefore 



HI = | KL. 



Now the similar triangles GHE, GF7, give 



EH : GH :: FI : G/; 

 whence 



EH+FI : GH + GI or HI :: FI : G/$ 

 that is, 



^5 + | CD : | ,STL : : {. CD : G/; 

 therefore 



jffLx -J CD __ $ ff L x CD 

 - i ^jy -f A CD " AB+ CD ' 



and, because LG = L/ + G/, if we substitute for i/ and GI tke 

 values above found, we shall have 



