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Question 14
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
Solution
Let AB = BE = x
Area of triangle ABE = $$\dfrac{x^2}{2}$$ = 7; 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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