+269 In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$. Find the area of the trapezoid.
+129771 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
