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

 

 Mar 21, 2024
 #1
avatar+128794 
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Draw AE , BF  perpendicular to CD

 

Let DE =  x

Let CF =  104 - 78 - x  =  26 - x

 

We can solve this  to  help find  the height

 

AD^2  - DE^2   = BC^2  -  CF^2

 

10^2  - x^2  = 24^2 - ( 26-x)^2

 

100 - x^2  = 576  - x^2 + 52x  - 676

 

100 = 52x - 100

 

200   = 52x

 

x  =  200 /  52  =   50  / 13

 

The height is   sqrt   [ 24^2  - (50/13)^2 ]    ≈  23.7

 

Area =   (1/2) height (sum of bases)  =  (1/2) (23.7) (78 + 104)  =   2156.7

 

cool cool cool

 Mar 21, 2024
edited by CPhill  Mar 21, 2024

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