Let $$ABCD$$ be a rectangle in which $$AB + BC + CD = 20$$ and $$AE = 9$$ where $$E$$ is the mid-point of the side $$BC$$. Find the area of the rectangle.

Put $$AB = CD = x$$ and $$BC = y$$, so $$2x + y = 20$$. From the right triangle $$ABE$$ we get $$x^2 + \left(\frac{y}{2}\right)^2 = 81$$, and substituting $$y = 20 - 2x$$ gives $$x^2 + (10-x)^2 = 81$$, that is $$2x^2 - 20x + 19 = 0$$. The area is $$xy = 20x - 2x^2$$, which equals 19 by the same equation.