In the figure, $AB = 12$, $FE = 8$, $BC = CD = DE = 3$, $AB\perp BE$, $FE\perp BE$,
Then find the area of $\Delta GCD$.
+130551 Call the altitude of triangle GCD , GH
And triangle GHC is similar to triangle FEC
Call HC, x
So we have that
FE/CE = GH / x
8/6 = GH /x
4/3 = GH /x
GH = (4/3)x
And triangle GHD is similar to triangle ABD
So
AB/ BD = GH / HD
Let HD = 3 - x
So
12/6 = [ (4/3)x ] / ( 3 - x)
2 (3 - x) = (4/3)x
6 - 2x = (4/3)x
6 = ( 4/3 + 2) x
6 = ( 10/3) x
x = 18/10 = 9/5
So GH =(4/3) (9/5) = 36/15 = 12/5
So the area of GCD = (1/2) CD * GH = (1/2) (3) ( 12/5) = 18/5 units^2
