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Question Number 141608 by cherokeesay last updated on 21/May/21

Commented by cherokeesay last updated on 21/May/21

$$theareahatched?$$

Answered by MJS_new last updated on 21/May/21

$$\mathrm{the}\mathrm{circle}\mathrm{is}\mathrm{the}\mathrm{incircle}\mathrm{of}\mathrm{an}\mathrm{equilateral}$$$$\mathrm{triangle}(\alpha =2\times 30°=60°)$$$$r=\frac{a\sqrt{3}}{6}=2\Rightarrow a=4\sqrt{3}\Rightarrow h=\frac{a\sqrt{3}}{2}=6$$$$\mathrm{the}\mathrm{cut}-\mathrm{off}\mathrm{triangle}\left(\mathrm{missing}\mathrm{below}\mathrm{the}\mathrm{circle}\right)$$$$\mathrm{has}\frac{1}{9}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{triangle}$$$$\Rightarrow$$$$\mathrm{searched}\mathrm{area}=$$$$=(\frac{8}{9}\left(\mathrm{area}\mathrm{of}\mathrm{triangle}\right)-\left(\mathrm{area}\mathrm{of}\mathrm{circle}\right))/2=$$$$=\frac{16\sqrt{3}}{3}-2\pi$$

Commented by cherokeesay last updated on 21/May/21

$$verynice!$$$$thankyou!$$

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