posted by Anonymous .
a trapezoid abcd has parallel sides ab and dc of lengths 8 and 22. If both diagonals AC and BD are of length 17, what is the area of the trapezoid
the excess length of dc over ab is 22-8=14
SO, there are two triangles of base 7 and hypotenuse 17 at the ends of the trapezoid.
The height h of the trapezoid is thus
h^2 = 17^2 - 7^2 = 240
h = √240
The area of the trapezoid
A = (8+22)/2 * √240
= 15*4√15 = 60√15
fd - correction -
Oops. I was considering the 17 as the length of the edges, not the diagonals.
The height h is given by
15^2 + h^2 = 17^2
A = (8+22)/2 * 8 = 120