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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?

Respuesta :

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)   

[tex]72= \frac{1}{2} (2x)(y)+\frac{1}{2}(y)(x)+\frac{1}{2}(x)(2y)+A(AEF)[/tex]

[tex]72= xy+\frac{1}{2}xy+xy+A(AEF)[/tex]

[tex]72= 2.5xy+A(AEF)[/tex]

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)
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