1. In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE = 7 $$cm^2$$ ; EC = 3(BE). The area of ABCD (in $$cm^2$$) is





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  • By: anil on 05 May 2019 02.30 am
    Let AB = BE = x Area of triangle ABE = $$x^2/2$$ = 14; we get x = $$sqrt{14}$$ So we have side BC = 4*$$sqrt{14}$$ Now area is AB*BC = 14 *4 = 56 $$cm^2$$
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