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Question Number 178001 by Tawa11 last updated on 11/Oct/22

Commented by Tawa11 last updated on 11/Oct/22

$$\left(\mathrm{b}\right)\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{part}.$$

Answered by aleks041103 last updated on 11/Oct/22

$$BDistangenttothecircleatT$$$$\Rightarrow AT\mathrm{\perp }BDandAT=r=AQ=AP$$$$S_{ABCD}=2S_{ABD}=2(\frac{1}{2}BD.AT)=BD.AT$$$$ButAB=ADand\measuredangle ABD=60°\Rightarrow BD=AB$$$$S_{ABCD}=AB.AD.sin\left(A\right)={AB}^2\frac{\sqrt{3}}{2}=AB.AT$$$$\Rightarrow AT=\frac{\sqrt{3}}{2}AB$$$$\left(a\right)P_{sh}=4AB-2r+\frac{\pi }{3}r=(4-(2-\frac{\pi }{3})\frac{\sqrt{3}}{2})AB$$$$\Rightarrow P_{sh}=(4+\frac{\pi }{2\sqrt{3}}-\sqrt{3})15$$$$\left(b\right)S_{sh}=S_{rhomb}-\frac{\pi }{6}{r}^2=$$$$=\frac{\sqrt{3}}{2}{AB}^2-\frac{\pi }{6}\frac{3}{4}{AB}^2=$$$$=\frac{{AB}^2}{2}(\sqrt{3}-\frac{\pi }{4})$$$$\Rightarrow S_{sh}=(\sqrt{3}-\frac{\pi }{4})\frac{225}{2}$$

Commented by Tawa11 last updated on 12/Oct/22

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

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