Problem:
An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is 3, and the area of the trapezoid is 72. Let the parallel sides of the trapezoid have lengths r and s, with rξ =s. Find r2+s2.
Solution:
Suppose CD=r and AB=s in the diagram.
Note that the height of the trapezoid is the diameter of the circle - 6. By the area formula of a trapezoid:
2r+sββ6=72βr+s=24β
By Pitot's Theorem,
AB+CD=AD+BCβΉ24=AD+BCβΉAD=BC=12β
From Pythagorean theorem, BF=AE=AD2βAE2β=63β. We can now see that
s=AB=AE+EF+BF=r+123ββ
. Solving the system r+s=24 and s=r+123β gives r=12β63β and s=12+63β.
