Question Number 96870 by john santu last updated on 05/Jun/20
Answered by Sourav mridha last updated on 05/Jun/20
![A=(1/4)π(2)^2 ββ«_0 ^1 (β(4βx^2 )) dx =πβ[((x(β(4βx^2 )))/2)+2sin^(β1) (x/2)]_0 ^1 =πβ((β3)/2)β(π/3)=((2π)/3)β((β3)/2) cm^2 Ansβ(D)](https://www.tinkutara.com/question/Q96895.png)
$$\boldsymbol{{A}}=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{\pi}\left(\mathrm{2}\right)^{\mathrm{2}} β\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{4}β\boldsymbol{{x}}^{\mathrm{2}} }\:\boldsymbol{{dx}} \\ $$$$\:\:\:\:\:=\boldsymbol{\pi}β\left[\frac{\boldsymbol{{x}}\sqrt{\mathrm{4}β\boldsymbol{{x}}^{\mathrm{2}} }}{\mathrm{2}}+\mathrm{2sin}^{β\mathrm{1}} \frac{\boldsymbol{{x}}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:=\boldsymbol{\pi}β\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}β\frac{\boldsymbol{\pi}}{\mathrm{3}}=\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{3}}β\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{{cm}}^{\mathrm{2}} \\ $$$$\boldsymbol{{Ans}}β\left(\boldsymbol{{D}}\right) \\ $$
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
$$\:\:\:{Very}\:{good} \\ $$
Answered by mr W last updated on 05/Jun/20
$${r}=\mathrm{2} \\ $$$${a}=\sqrt{\mathrm{3}} \\ $$$$\mathrm{sin}\:\theta=\frac{{a}}{{r}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{3}}=\mathrm{60}Β° \\ $$$${A}_{{shaded}} =\frac{\piΓ\mathrm{2}^{\mathrm{2}} }{\mathrm{6}}β\frac{\mathrm{1}Γ\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{2}\pi}{\mathrm{3}}β\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\left({D}\right) \\ $$