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?

Let the sides of the rectangle be |AB|=|DC|=2x units, and |BC|=|AD|=2y units, 

then

2x*2y = 72
4xy=74
xy=74/4=18. 5 (units squared)

The Area of rectangle: 

ABCD= A(ABE)+A(ECF)+A(AFD)+A(AEF)  (check the picture)

72 = A2+A3+A1+A(AEF)   







substituting xy=18.5:

72= 2.5*18.5+A(AEF)

72=46.25+A(AEF)

then, A(AEF)=72-46.25=25.75 (units squared)

25.75 (units squared)