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Question 72
What is the area of the figure below, if ABDC is a rectangle and BDE is an isosceles right triangle ?
Solution
Area of figure = area (ABCD) + area (BDE)
= $$(ab)+(\frac{1}{2}\times b\times b)$$
= $$b(a+\frac{b}{2})$$
=> Ans - (C)
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