Thread: find ratio of areas

E divides DC such that DE/EC = x/y, find the ratio of the area of triangle a to that of parallelogram ABCD

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Originally Posted by oohhoohh

E divides DC such that DE/EC = x/y, find the ratio of the area of triangle a to that of parallelogram ABCD

123 :: 123.jpg picture by ooorandomooo - Photobucket

Let P be the point of intersection of AC and BD.
Let MPN be a perpendicular line to AB and DC.
Let PN = h1 and MP = h2.
Triangle ABP and DEP are similar. So
AB/EC = h2/h1. AB = x+y, and EC = y
So (x+y)/y = h2/h1........(1)
Area of ABCD is A = DC*(h1 + h2) = (x+y)(h1+h2).....(2)
Area PEC ia a = 1/2*EC*h1 = 1/2*y*h1.....(3)
Rewrite eq.(1) as
1 + (x+y)/y = 1 + h2/h1
(2y+x)/y = (h1+h2)/h1
h1 = (h1+h2)y/(x +2y)
Substitute this value of h1 in eq. 3.
a= 1/2*y*y(h1+h2)/(x+2y) .......(4)
Now find the ratio of a/A