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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
avatar+109563 
-1

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 

 

 

cool cool cool

 Dec 21, 2019
 #2
avatar+531 
+2

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

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

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