+0  
 
0
101
2
avatar

The diagram shows three squares.  If EF + BF = 15, then find the area of the red region.

 

 Jan 11, 2020
 #1
avatar+109739 
+2

I know there is an easier way, but using coordinate geometry....

EF + BF  =15

BF  = 2EF

EF + 2EF = 15

3EF  =15

EF  =5

BF = 10

 

Let C  = (0,0)    Let D = (0,5)   F =  (5,0)   A = (-5, 5)  G = (-5, -5)

 

The line through AF  has the slope  (5-0) / ( -5.-5) = 5/-10  = -1/2

And the equation of this line  is   y  =  (-1/2)( x - 5)

 

And the line through GD  has the slope   (-5 - 5) /( -5 - 0)  = -10/ -5  = 2

So the equation of this line is y = 2x + 5  

 

And we can find the x coordinate of the intersection of these lines as

 

(-1/2)(x - 5)  = 2x + 5

(-1/2)x + 5/2 = 2x + 5

5/2 - 5 =  (2 + 1/2) x

-5/2  = (2.5)x

-2.5 = 2.5x           divide both sides by 2.5

-1  = x

 

So....the altitude of  the red triangle    l  -5 - -1 l  =  l -4 l    = 4

And the base  = AG = BF   =10

 

So....the red area =  (1/2) (10) (4)  = 20 units^2

 

cool cool cool

 Jan 11, 2020
 #2
avatar+531 
+1

Let us denote a crosspoint of AF and DG with a letter "M" and an angle AGD fith ''q''.

 

EF + BF = 15       BF = AG                   AG = 10  

q = ?         sin(q) = AM/AG                   q = 26.565°  

AM = ?             AM = sin(q) * AG          AM = 4.472  

GM = ?      GM = sqrt(AG)² - (AM)²       GM = 8.944  

Area (AGM) = (AM * GM)*1/2               Area = 20 u²    indecision

 Jan 11, 2020
edited by Dragan  Jan 11, 2020

34 Online Users

avatar
avatar