In the figure ABCD is a 3 cm sided square, GC = FC = 1 cm, find the area of EFGC in square centimeter.
+129849 Let D = (0,0)
Let C = (3,0)
Let G = (2,0)
Let F= ( 3,1)
Let B = (3,3)
The slope of DF = [ 1 - 0] / [ 3 - 0 ] = 1/3
So the equation of the line containing DF = (1/3)x
The slope of BG = [ 3-0] / [3 - 2] = 3/1 = 3
So the equation of the line containing BG = 3(x - 3) + 3 = 3x - 6
We can find the x coordinate of the intersection of these lines can be found as
(1/3)x = 3x - 6 rearrange as
6 = [ 3 - 1/3]x
6 = (8/3] x
x = (3 * 6 / 8] = 18/8 = 9/4 = x coordinate of E
The y coordinate of E = (1/3)(9/4) = 9/12 = 3/4 = height of triangle DGE
Area of triangle DGE = (1/2)DG * height = (1/2) (2) * (3/4) = 3/4
Area of triangle DFC = (1/2) DC * CF = (1/2)(3) (1) = 3/2
Area of EFGC = area of triangle DFC - area of triangle DGE = (3/2) - (3/4) = (3/4)cm^2 = .75 cm^2
