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Two 2*8 rectangles overlap, as shown below. Find the area of the overlapping region (which is shaded)

 

 

 Jun 24, 2021
 #1
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Let  the 2  right triangles  contained  within ABCD  have legs  of   2 and  x

Their  combined  areas =   (2) (1/2) ( 2x)  =  2x

The shaded  area is a parallelogram with  a height of 2  and bases of  sqrt (2^2 + x^2)  = sqrt (4 + x^2)

 

So....these combined areas = 2 * 8    =  16

 

So

 

2x   +  sqrt  (4 + x^2)  * 2    =  16

 

x   +  sqrt ( 4  + x^2)    = 8

 

sqrt ( 4 + x^2)  = 8 - x               square both sides

 

4 + x^2   =  x^2  - 16x  +  64

 

16x   - 64  + 4  =   0

 

16x  - 60  =   0

 

4x  -  15 = 0

 

4x  = 15

 

x = 15/4

 

So....the  area of  the  parallelogram  (shaded area )   =   

 

sqrt  [  4 +  (15/4)^2 ] * 2  =

 

sqrt  [ 64  + 225 ]  / 4   *   2   =

 

17  /  2   =

 

8.5  units ^2   

 

cool cool cool

 Jun 24, 2021

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