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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) = \(6/\sqrt{2} *(6/\sqrt{2} + 6*\sqrt{2}) = 54\)

Re: If in the figure above AE=AD, FB=6 and the area of the parallelogram D [#permalink]

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02 Aug 2017, 14:01

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