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Question Number 61169 by Tawa1 last updated on 29/May/19

Commented by Tawa1 last updated on 29/May/19

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{Black}$$

Answered by mr W last updated on 30/May/19

Commented by mr W last updated on 30/May/19

$$squaresidelength=a=14cm$$$$R=a$$$$r=\frac{a}{2}$$$$OC=\frac{a}{\sqrt{2}}$$$$\cos \alpha =\frac{a^2+{\left(\frac{a}{\sqrt{2}}\right)}^2-{\left(\frac{a}{2}\right)}^2}{2a\left(\frac{a}{\sqrt{2}}\right)}=\frac{\frac{5}{4}}{\frac{2}{\sqrt{2}}}=\frac{5\sqrt{2}}{8}$$$$\sin \alpha =\frac{\sqrt{64-25\times 2}}{8}=\frac{\sqrt{14}}{8}$$$$\Rightarrow \alpha ={\sin }^{-1}\frac{\sqrt{14}}{8}\approx 27.89°$$$$\frac{\sin \mathrm{\angle }COE}{a}=\frac{\sin \alpha }{\frac{a}{2}}$$$$\sin \mathrm{\angle }COE=\sin \beta =2\sin \alpha =\frac{\sqrt{14}}{4}$$$$\Rightarrow \beta ={\sin }^{-1}\frac{\sqrt{14}}{4}\approx 69.3°$$$$\frac{1}{4}{A}_{black}=\frac{\beta r^2}{2}-(\frac{\alpha R^2}{2}-\frac{R\times \frac{a}{\sqrt{2}}\times \sin \alpha }{2})$$$$=\frac{\beta a^2}{8}-\frac{\alpha a^2}{2}+\frac{\sqrt{7}{a}^2}{16}$$$$=\frac{a^2}{16}(2{\sin }^{-1}\frac{\sqrt{14}}{4}-8{\sin }^{-1}\frac{\sqrt{14}}{8}+\sqrt{7})$$$$\Rightarrow A_{black}=\frac{a^2}{4}(2{\sin }^{-1}\frac{\sqrt{14}}{4}-8{\sin }^{-1}\frac{\sqrt{14}}{8}+\sqrt{7})$$$$\approx 0.293a^2=57.38{cm}^2$$

Commented by Tawa1 last updated on 30/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}$$

Commented by Tawa1 last updated on 30/May/19

$$\mathrm{Sir},\mathrm{any}\mathrm{basis}\mathrm{to}\mathrm{learn}\mathrm{all}\mathrm{this}\mathrm{shapes}?.\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{start}\mathrm{learning}$$$$\mathrm{it}\mathrm{from}\mathrm{the}\mathrm{begining}$$

Commented by mr W last updated on 30/May/19

$$sorryi{can}^{\prime }tansweryourquestion,$$$$becausei{don}^{\prime }tknowwhathavealready$$$$learntandwhatnotyet.accordingto$$$$thequestionsyouputhere,ithink,$$$$youhavealltheknowledgewhichis$$$$necessarytosolvesuchquestions.$$

Commented by Tawa1 last updated on 30/May/19

$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by mr W last updated on 30/May/19

Commented by Tawa1 last updated on 30/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}$$

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