Question

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

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)


`72= (1)/(2) (2x)(y)+(1)/(2)(y)(x)+(1)/(2)(x)(2y)+A(AEF)`


`72= xy+(1)/(2)xy+xy+A(AEF)`


`72= 2.5xy+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)


Answer: 25.75 (units squared)