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

 

 Dec 16, 2023
 #1
avatar+126978 
+1

Let the height of the trapezoid = h

We can form  two right triangles on each side of the trapezoid 

One will have a height of h and a hypotenuse of 10   and the other will have a height of h and a  hypotenuse of 24

 

The length of the remaining legs  are  sqrt [ 10^2 - h^2 ] and [ 24^2 - h^2 ] 

 

So......looking at the bottom base  we have that

 

sqrt [ 10^2 - h^2 ] + 78 + sqrt [ 24^2 - h^2] = 104       simplify

 

sqrt [ 100 - h^2 ] + sqrt [ 576 - h^2 ]  =  26         square both sides

 

[ 100 - h^2 ] + 2 sqrt [ (100- h^2) (576 -h^2) ]  + [576 -h^2] =  676

 

676 - 2h^2  + 2sqrt [ (100 - h^2) ( 576 -h^2 ) ]    = 676    simplify

 

h^2  =  sqrt [ (100-h^2) ( 576 -h^2) ]      square both sides again

 

h^4  =  (100 - h^2) (576 - h^2)

 

h^4  = h^4 - 676h^2 + 57600

 

676h^2  = 57600

 

h^2  =  57600 / 676  

 

h =  sqrt [ 57600 / 676 ]  =  120 /13

 

 

Now just use the formula for the area =   (1/2) (height) (sum of the bases)

 

 

cool cool cool

 Dec 16, 2023

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