reto wrote:

Attachment:

T8925.png

If in the figure above AE=AD, FB=6 and the area of the parallelogram DEBF is 36, what is the area of the rectangle ABCD?

A. 45

B. 54

C. 72

D. 90

E. 108

DEBF is a parallelogram. So if BF = 6, ED = 6 too.

Now, AE = AD so AED forms a right isosceles triangle in which ratio of the sides is \(1:1:\sqrt{2}\).

Hence, AE = AD = \(6/\sqrt{2}\)

Note that the altitude of parallelogram BEDF is AD and base is BE.

Area of parallelogram = 36 = AD * BE = \(6/\sqrt{2} * BE\)

So \(BE = 6*\sqrt{2}\)

Area of rectangle = AD * (AE + BE) = \(6/\sqrt{2} *(6/\sqrt{2} + 6*\sqrt{2}) = 54\)

Answer (B)

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