Pressure of Fluids. 299 



the two planes ABCD, FECD, since CD is their common inter- Geom 

 section ; lastly, from the point K, where HK meets AB, let fall 355. 

 the perpendicular KI upon the plane 7CJ>, this line must be 

 perpendicular to HI. 



These steps being taken, I decompose the force p into two 

 others, both being in the plane KIH produced, and of which one 

 GL is horizontal or perpendicular to the plane FECD, and the 

 other GM vertical. Calling these two forces q and r, and form- 

 ing the parallelogram GMNL upon* the line GJV, taken arbitra- 

 rily as the diagonal, we shall have 



p : q : r : : GJV : GL : GM, 

 :: GJV : GL : LN. 



But as the triangle GLJVhas its sides perpendicular respectively 

 to those of the triangle KIH, these two triangles are similar, and 

 we have Geom, 



209. 



GJV : GL : ZJV : : HK : HI : IK '; 



and consequently 



p : q : r :: HK : HI : IK. 



nn j /7/1 



Multiplying the three last terms by -X X GG', which 

 does not change the ratio, we obtain, 



p : q : r 



: HK X + CD X GG' : HI 



ft D I /^i T\ 



We observe now, 1. That HK X - -~ --- is the surface of 

 the trapezoid ABCD; 2. That since CE, DF are parallel, as 



a D [ (77) 



also CD, F, CD is equal to FE ; whence IK X -* - is 



Geom, 

 /] 7? i Ff 178 



equivalent to IK X -~4 , and consequently is the sur- 

 face of the trapezoid AFEB ; 3. As we suppose the altitude of 



