In the figure ABCD is a 3 cm sided square, GC = FC = 1 cm, find the area of EFGC in square centimeter.
+129852 Let D = (0,0) Let F = (3,1)
So the line containing segment DF has the equation y = (1/3)x
Let G = (2,0) Let B = ( 3,3)
The slope between these points = (3- 0)/ (3 - 2) = 3/1 = 3
So...the line containing segment GB has the equation y = 3(x-2) = 3x - 6
The x coordinate of the intersection of these lines (point E ) is
(1/3)x = 3x - 6
x = 9x - 18
18 = 8x
x = (18/8) = 9/4 = DG
And the y coordiate of E is (1/3)(9/4) =3/4 = altitude of triangle DEG
So...the area of triangle DFC = (1/2)(DC)(FC) = (1/2)(3)(1) = (3/2) cm^2 (1)
And the area of triangle DEG = (1/2) ( DG)(3/4) = (1/2)(9/4)(3/4) = ( 27/32 ) cm^2 (2)
So the area of EFGC = (1) - (2) = (3/2) - (27/32) = (48/32) - (27/32) = (21/32) cm^2
