+228 In trapezoid ABCD, \overline{AB} \parallel \overline{CD}. Find the area of the trapezoid.
+129837 Draw AE and BF perpemdicular to BC
Let DE = x
Let CF = [ 104 - ( 78 + x ) ] = 26 - x
We have rwo right triangles ADE and BCF
Let the height of the trapezoid = h
So
h = sqrt [ 10^2 - x^2]
h = sqrt [ 24^2 - (26 -x)^2 ]
So
sqrt [ 10^2 -x^2] = sqrt [ 24^2 - (26 -x)^2 ] square both sides
10^2-x^2 = 24^2 - (26 -x)^2
100 - x^2 = 576 - x^2 + 52x - 676 simplify
52x - 200 = 0
13x - 50 = 0
x = 50/13
h = sqrt [ 10^2 - (50/13)^2] = 120/13
Area of trapezoid = (1/2) (120/13) (78 + 104) = 840
