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Question Number 96870 by john santu last updated on 05/Jun/20

Answered by Sourav mridha last updated on 05/Jun/20

$$\mathit{A}=\frac{1}{4}\mathit{\pi }{\left(2\right)}^2-{\int }_0^1\sqrt{4-{\mathit{x}}^2}\mathit{d}\mathit{x}$$$$=\mathit{\pi }-{[\frac{\mathit{x}\sqrt{4-{\mathit{x}}^2}}{2}+{2\sin }^{-1}\frac{\mathit{x}}{2}]}_0^1$$$$=\mathit{\pi }-\frac{\sqrt{3}}{2}-\frac{\mathit{\pi }}{3}=\frac{2\mathit{\pi }}{3}-\frac{\sqrt{3}}{2}{\mathit{c}\mathit{m}}^2$$$$\mathit{A}\mathit{n}\mathit{s}-\left(\mathit{D}\right)$$

Commented by john santu last updated on 05/Jun/20

$$\mathrm{yes}.\mathrm{thank}\mathrm{you}\mathrm{both}$$

Commented by Awouboye last updated on 05/Jun/20

$$Verygood$$

Answered by mr W last updated on 05/Jun/20

$$r=2$$$$a=\sqrt{3}$$$$\sin \theta =\frac{a}{r}=\frac{\sqrt{3}}{2}$$$$\Rightarrow \theta =\frac{\pi }{3}=60°$$$$A_{shaded}=\frac{\pi \times 2^2}{6}-\frac{1\times \sqrt{3}}{2}=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$$$$\Rightarrow \left(D\right)$$

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