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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- Apr 13th 2010, 08:24 PMoohhoohhfind 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

123 :: 123.jpg picture by ooorandomooo - Photobucket - Apr 14th 2010, 09:43 PMsa-ri-ga-ma
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