+837 The area of trapezoid ABCD is 80. One base is 12 units longer than the other, and the length of the diagonal is 14. Find the length of the median of the trapezoid.
+130081 Assuming an isosceles trapezoid
80 = (1/2) h (b + b + 12)
160 = h ( 12 + 2b)
80 = h ( 6 + b)
80 / ( 6 + b) = h
We have a right triangle
One leg = h
The hypotenuse = 14
One leg = b + 12 - 6 = b + 6
So
[ 80 / (b + 6)]^2 + (b + 6)^2 = 14^2 let (b + 6)^2 = x
6400/x + x = 14^2 multiply through by x
6400 + x^2 = 196x
x^2 -196x + 6400 = 0
x ≈ 154.6 or x = 41.936
(b + 6 )^2 = 154.6 → b ≈ 6.43
(b + 6)^2 = 41.936 → b ≈ .476
So
The median = ( b + b + 12) / 2 = b + 6 = either ≈ b + 6 ≈ 12.43 or ≈ 6.476
