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Question

$$ABC$$ is an isosceles right angled triangle with $$\angle ABC =90^o$$. A semi-circle is drawn with $$AC$$ as the diameter. If $$AB = BC = 7\ cm$$, find the area (in $$cm^2$$) of shaded region. $$\left(\text{Take}\, \pi = \dfrac{22}{7}\right)$$
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Solution

$$\triangle ABC$$ is a right angled triangle. So, by Pythagoras theorem
$$AC^2 = AB^2 + BC^2 = 7^2+7^2 = 98$$
$$\therefore AC = \sqrt{98} = 7\sqrt2\ cm$$
Arc $$A-B-C$$ is a semicircle. $$\therefore AC = 7\sqrt2\ cm$$ is the diameter of the semicircle.
Area of the shaded portion $$=$$ Area of semi-circle $$(A{-}B{-}C)\, - $$ Area of $$\triangle ABC$$
$$=\cfrac12\times \cfrac{\pi d^2}4 - \cfrac12\times AB\times BC$$
$$= \cfrac{22\times98}{2\times7\times4} - \cfrac{7\times7}2$$
$$= 14 \ cm^2$$

Mathematics

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