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Re: In square ABCD above, E is the midpoint of side BC. If DE = 8√5, what [#permalink]

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10 Oct 2017, 06:58

Let each side of the square be x. CD=x, CE=x/2 (Since E is midpoint of the side) In right angled triangle CDE, by Pythagoras theorem \(x^2+(x/2)^2\)=(8\(\sqrt{5})\)^2=64*5 \(\frac{5x^2}{4}\) = 64*5 x^2 = 64 * 4 x=8*2 = 16 Answer A.
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