Iift 2021 - 5th Dec Question 11
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IIFT 2021 (5th Dec)
Area of shaded region = area of semicircle on BC + area of semicircle of AC - area of unshaded semicircle regions
Area of unshaded semicircle regions = area of semicircle on AB - area of triangle ABC
Given,
AB = 50, BC = 14
In triangle ABC,
Solving we get, AC2+BC2=AB2
AC = 48
Area of unshaded semicircle regions = π (25)2 -1/2 X 48 X 14 = 625π - 336
Area of shaded region = 576π + 49π - 625π + 336 = 336
The answer is option A.