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Question Number 57184 by Tawa1 last updated on 31/Mar/19

Commented by Tawa1 last updated on 31/Mar/19

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{portion}$$

Answered by mr W last updated on 31/Mar/19

$$R=radiusofbigcircle=5$$$$r=radiusofsmallcircle$$$$\sqrt{2}r+r+R=\sqrt{2}R$$$$\Rightarrow r=\frac{(\sqrt{2}-1)R}{\sqrt{2}+1}=(3-2\sqrt{2})R$$$$bluearea=R^2-\frac{\pi R^2}{4}-\pi r^2$$$$=R^2-\frac{\pi R^2}{4}-\pi (17-12\sqrt{2})R^2$$$$=\frac{4-(48\sqrt{2}-67)\pi }{4}{R}^2$$$$=0.307R^2$$$$=7.677{cm}^2$$

Commented by mr W last updated on 31/Mar/19

Commented by Tawa1 last updated on 31/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$$$\mathrm{But}\mathrm{sir},\mathrm{i}\mathrm{have}\mathrm{a}\mathrm{request}.\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{understand}\mathrm{your}\mathrm{steps}\mathrm{sir}.$$$$\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{explain}\mathrm{some}\mathrm{steps}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Mar/19

$$a=2R$$$$areaofshadedportion$$$$=\frac{a^2-\pi R^2}{4}-\pi r^2$$$$=\frac{a^2-\pi {\left(\frac{a}{2}\right)}^2}{4}-\pi r^2$$$$nowdiagonalofsquare=a\sqrt{2}$$$$a\sqrt{2}=2R+4r+2l$$$${(l+r)}^2=r^2+r^2$$$$l+r=r\sqrt{2}$$$$l=r(\sqrt{2}-1)$$$$a\sqrt{2}=2R+4r+2l$$$$a\sqrt{2}=a+4r+2r(\sqrt{2}-1)$$$$a\sqrt{2}-a=2r+2\sqrt{2}r$$$$a(\sqrt{2}-1)=r(2+2\sqrt{2})$$$$r=\frac{a(\sqrt{2}-1)}{2(\sqrt{2}+1)}$$$$soshadedarea$$$$=\frac{a^2-\pi {\left(\frac{a}{2}\right)}^2}{4}-\pi r^2$$$$=\frac{a^2-\pi {\left(\frac{a}{2}\right)}^2}{4}-\pi \times {\left\{\frac{a(\sqrt{2}-1)}{2(\sqrt{2}+1}\right\}}^2$$$$nowputa=10$$$$plscheckmistakeifany$$

Commented by Tawa1 last updated on 31/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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