CHAP. III.] CENTRE OF GRAVITY. 31 



the height. of the equivalent rectangle. Then 



bfi k as 

 ax= or-=-^. 

 2, a $ b 



Now a, b, and h being given, it is required to find x graphi- 

 cally. 



Let A B C be the triangle, and D the middle of the base. 

 [Fig. 17, PI. 5.] Lay off A E = h and A F = a. Draw F D, 

 and parallel to F D draw E x. Then A a? is the required 

 height. 



-,-, A a; A E x h 



For : = - or ^ =.. 



AD A F b a 



As to the centre of gravity of the triangle, it is at the inter- 

 section of the lines from each apex to the centre of the opposite 

 side ; since these are medial lines 



33. Reduction off Trapezoid to equivalent Rectangle. 



In the trapezoid A B C D, Fig. 18, PI. 5, draw through the 

 middle points of A D and B C perpendiculars to D C, and pro- 

 duce to intersections E and F with A B produced. 



Then lay off F g a = the given reduction base, and draw 

 g E intersecting D C in a?. Then H a? is the required height. 



T, EF H x EF x 



For - = ==-= or 



F g HE a HE' 



hence <zoj = EFxHE area. 



To find the centre of gravity, draw a line through the mid- 

 dle points of the parallel sides A B and D C. Prolong A B and 

 CD and make Ca = AB and A b = CD and join'a and b. 

 Then the intersection of a b with the axis of symmetry gives 

 the centre of gravity. 



The construction for the reduction of a parallelogram is pre- 

 cisely similar. [Fig. 18 (b).] 



The points F and E here coincide with A and B, and we 

 have 



A a; AB , . _ 



- 7 = , or a x = h x A B = area. 



h B tf 



The same construction also holds good, of course, for a rect- 

 angle or square. The centre of gravity in each case is at the 

 intersection of two diameters, since these are axes of symmetry. 



34. Reduction of Quadrilaterals Generally. In general 



