+1557 In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$. Find the area of the trapezoid.
+129881 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)
