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Question Number 94201 by $@ty@m123 last updated on 17/May/20

Commented by $@ty@m123 last updated on 17/May/20

Find the area of shaded region if side of square is a and radius of arc is a.

Answered by mr W last updated on 17/May/20

Commented by mr W last updated on 17/May/20

$$p=2a\sin \frac{\pi }{12}$$$$A_{shade}=\frac{\sqrt{3}{p}^2}{2}+(2-1)\times \frac{a^2}{2}(\frac{\pi }{6}-\sin \frac{\pi }{6})$$$$A_{shade}=\frac{4a^2\sqrt{3}}{2}{\sin }^2\frac{\pi }{12}+\frac{a^2}{2}(\frac{\pi }{6}-\sin \frac{\pi }{6})$$$$A_{shade}=a^2\sqrt{3}(1-\cos \frac{\pi }{6})+\frac{a^2}{2}(\frac{\pi }{6}-\sin \frac{\pi }{6})$$$$A_{shade}=\frac{a^2(2\sqrt{3}-3)}{2}+\frac{a^2(\pi -3)}{12}$$$$A_{shade}=\frac{a^2(\pi +12\sqrt{3}-21)}{12}\approx 0.24385a^2$$

Commented by $@ty@m123 last updated on 17/May/20

$$Thanksalot.$$

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