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Question Number 136377 by MJS_new last updated on 21/Mar/21

Commented by MJS_new last updated on 21/Mar/21

$$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}‘‘{\mathrm{square}}^{″}\mathrm{is}(\frac{\pi }{3}+1-\sqrt{3})a^2\mathrm{but}$$$$\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{this}without\mathrm{integration}?$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{afraid}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{a}\mathrm{blockhead}...$$

Answered by mr W last updated on 21/Mar/21

Commented by mr W last updated on 21/Mar/21

$$b=\frac{\sqrt{3}a}{2}-\frac{a}{2}=\frac{(\sqrt{3}-1)a}{2}$$$$areaB=\frac{b^2}{2}=\frac{(2-\sqrt{3})a^2}{4}$$$$areaA=\frac{a^2}{2}(\frac{\pi }{6}-\sin \frac{\pi }{6})=\frac{a^2}{12}(\pi -3)$$$$shadedarea=4\times (A+B)$$$$=(\frac{\pi -3}{3}+2-\sqrt{3})a^2$$$$=(\frac{\pi }{3}+1-\sqrt{3})a^2$$

Commented by MJS_new last updated on 21/Mar/21

$$\mathrm{typo}:\frac{b^2}{2}=\frac{(2-\sqrt{3})a^2}{4}$$$$\mathrm{thank}\mathrm{you}!$$

Commented by mr W last updated on 21/Mar/21

$$nowfixed.$$

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