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# help needed

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

Dec 20, 2019

#1
+109563
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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

Dec 21, 2019
#2
+531
+2

My answer is somewhat different.  The area  GEFC = 0.75 cm2

Dragan  Dec 23, 2019
edited by Dragan  Mar 23, 2020