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In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$. Find the area of the trapezoid.

 

 Apr 30, 2024
 #1
avatar+128772 
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From A,B  draw  AE  and BF perpendicular to BC

Let FC = x

By the PythagoreanTheorem

sqrt [24^2 - x^2 ]  =  height of  trapezoid

 

And DE = 104 - (78 + x)  =  26  - x

And, again,

sqrt [ 10^2  - ( 26 - x)^2 ] = height of trapezoid

 

So

 

sqrt [ 24^2 - x^2 ] = sqrt [ 10^2 - (26-x)^2 ]    square both sides

 

24^2 - x^2  = 10^2 - (26 - x)^2

 

576 - x^2  = 100  - x^2 + 52x - 676

 

52x  =  1152

 

x =  1152 / 52   = 288/13

 

height of trapezoid = sqrt [ 24^2  - (288/13)^2 ]    =  120 /  13

 

Area =  (1/2) (120 / 13) ( 104 + 78)  =  (1/2) (120/13) ( 182)   =  (91) (120 / 13)  =   840

 

 

 

cool cool cool

 Apr 30, 2024

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