In the figure ABCD is a 3 cm sided square, GC = FC = 1 cm, find the area of EFGC in square centimeter.

Guest May 22, 2020

#1**+1 **

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

CPhill May 22, 2020