Given the rectangle ABCD has a total area of 72. E is in the midpoint of BC and F is the midpoint of DC. What is the area of the inscribed triangle AEF?

Answer 1

If BC=2y and CD=2x, note that triangles
ADF, ECF, and ABE are all right triangles with one leg half of a side of the rectangle. We know that ABCD has area (2x)(2y) = 72, so xy=18.

ADF has area (2y)(x)/2 = xy = 18
ECF has area (y)(x)/2 = xy/2 = 9
ABE has area (y)(2x)/2 = xy = 18

So, AEF has area 72-18-18-9 = 27