A parallelogram ABCD has perimeter equal to 124 . Let E be the foot of the perpendicular from A to BC, and let F be the foot of the perpendicular from A to CD. If AE= 7 and AF=20, what is the area of the parallelogram?
I recommend drawing a diagram.
I'm not completely confident about my answer, but here's what I got.
We know that DF+FC+CE+EB = 124/2 = 62, and that 20(DF + FC) = 7(CE + EB), which are just 2 different ways to calculate the area of the parallelogram.
We need to calculate either 20(DF + FC) or 7(CE + EB).
From our first equation, we get DF + FC = 62 - (CE + EB), so 20(62-(CE+EB)) = 7(CE+EB). Solving, CE+EB = 1240/27 and so the area is 7 * 1240/27 = 8680/27.