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Question Number 103318 by I want to learn more last updated on 14/Jul/20

Commented by som(math1967) last updated on 14/Jul/20

$$\frac{\mathrm{AT}}{\mathrm{AB}}=\cos \mathrm{\angle }\mathrm{BAT}=\cos 30°$$$$\mathrm{AT}=15\times \frac{\sqrt{3}}{2}\mathrm{cm}$$$$\therefore \mathrm{BP}=\mathrm{QD}=(15-\frac{15\sqrt{3}}{2})\mathrm{cm}$$$$\mathrm{perimeter}\mathrm{of}\mathrm{shaded}\mathrm{part}$$$$=\{\frac{1}{6}\times 2\pi \times \frac{15\sqrt{3}}{2}+2(15-\frac{15\sqrt{3}}{2})+2\times 15\}\mathrm{cm}$$

Commented by I want to learn more last updated on 14/Jul/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by Tawa11 last updated on 15/Sep/21

$$\mathrm{nice}$$

Answered by 1549442205 last updated on 14/Jul/20

$$\mathrm{The}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{equal}\mathrm{to}\mathrm{the}$$$$\mathrm{altitude}\mathrm{of}\mathrm{equilateral}\mathrm{ABD},\mathrm{so}$$$$\mathrm{R}=\frac{15\sqrt{3}}{2}.\mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{arc}\mathrm{PQ}\mathrm{is}$$$$\mathrm{l}=\frac{\mathrm{R}\pi }{3}=\frac{15\pi \sqrt{3}}{6}=\frac{5\pi \sqrt{3}}{2}.\mathrm{QD}=\mathrm{PB}=15-\mathrm{R}=15-\frac{15\sqrt{3}}{2}$$$$=\frac{15(2-\sqrt{3}}{2}\Rightarrow \mathrm{PB}+\mathrm{QD}=15(2-\sqrt{3})$$$$\mathrm{The}\mathrm{perimeter}\mathrm{of}\mathrm{shaded}\mathrm{part}\mathrm{is}$$$$\mathrm{BC}+\mathrm{CD}+\mathrm{QD}+\mathrm{arc}\left(\mathrm{PQ}\right)+\mathrm{PB}=$$$$15\times 2+15(2-\sqrt{3})+\frac{5\pi \sqrt{3}}{2}$$$$=\frac{5(24-6\sqrt{3}+\mathit{\pi }\sqrt{3})}{2}$$

Commented by I want to learn more last updated on 14/Jul/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by I want to learn more last updated on 14/Jul/20

$$\mathrm{Sirs},\mathrm{what}\mathrm{of}\mathrm{the}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}.$$$$\mathrm{Am}\mathrm{i}\mathrm{going}\mathrm{to}\mathrm{say}:$$$$\mathrm{Area}\mathrm{of}\mathrm{shaded}={\mathrm{l}}^2\sin \left(\theta \right)-\mathrm{Area}\mathrm{of}\mathrm{sector}?$$

Commented by som(math1967) last updated on 14/Jul/20

$$\mathrm{Area}\mathrm{of}\mathrm{rhombus}-\mathrm{Area}\mathrm{of}$$$$\mathrm{sector}$$

Commented by I want to learn more last updated on 14/Jul/20

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

Commented by 1549442205 last updated on 14/Jul/20

$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{rhombs}\mathrm{is}\mathrm{S}=\frac{\mathrm{AC}\times \mathrm{BD}}{2}=\frac{15\sqrt{3}\times 15\sqrt{3}}{4}=\frac{675}{4}$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{secror}\mathrm{is}{\mathrm{S}}_0=\frac{\pi {\mathrm{R}}^2}{6}=\frac{\pi }{6}\times {\left(\frac{15\sqrt{3}}{2}\right)}^2=\frac{675\pi }{24}$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{part}\mathrm{is}:$$$${\mathbf{S}}_{\mathbf{sh}}=\mathbf{S}-{\mathbf{S}}_0=\frac{675}{4}-\frac{675\mathit{\pi }}{24}=\frac{675(6-\mathit{\pi })}{24}$$$$$$

Commented by I want to learn more last updated on 14/Jul/20

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

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