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Question Number 67623 by mr W last updated on 29/Aug/19

Commented by mr W last updated on 29/Aug/19

$$findtheshadedareaA=?$$

Commented by Prithwish sen last updated on 29/Aug/19

$$\because \mathrm{AL}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}$$$$\therefore \mathrm{AL}=\sqrt{7\times 8[1-{\left(\frac{9}{7+8}\right)}^2}=\frac{8\sqrt{14}}{5}\backsim 6$$$$\mathrm{CL}=\mathrm{s}-\mathrm{c}=\left(\frac{7+8+9}{2}\right)-8=4=\mathrm{CD}$$$$\mathrm{Let}\mathrm{the}\mathrm{radius}\mathrm{be}\mathrm{R}$$$$\therefore \mathrm{From}△\mathrm{AOD}$$$${(\mathrm{R}+6)}^2={\mathrm{R}}^2+{(4+7)}^2\Rightarrow \mathrm{R}=7.08\left(\mathrm{approx}\right)$$$$\mathrm{Area}\mathrm{of}\mathrm{LODC}=4\times 7.08\backsim 28.32$$$$\mathrm{Now}\mathrm{\angle }\mathrm{DCO}={\tan }^{-1}\frac{\mathrm{OD}}{\mathrm{DC}}={\tan }^{-1}\frac{7.08}{4}\backsim {60}^{°}$$$$\therefore \mathrm{\angle }\mathrm{COD}={30}^{°}$$$$\therefore \mathrm{\angle }\mathrm{LOD}={60}^{°}$$$$\therefore \mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{OLD}={(7.08)}^2\times \pi \times \frac{60}{360}\backsim 26.25$$$$\therefore \mathbf{Area}\mathbf{of}\mathbf{the}\mathbf{shaded}\mathbf{portion}=28.32-26.25=2.07$$$$\mathbf{S}\mathrm{ir}\mathrm{waiting}\mathrm{for}\mathrm{your}\mathrm{valuable}\mathrm{response}.$$

Commented by Prithwish sen last updated on 29/Aug/19

Commented by mr W last updated on 29/Aug/19

$$pleasecheckmyanswersir!$$

Commented by Prithwish sen last updated on 29/Aug/19

$$\mathrm{Your}\mathrm{solution}\mathrm{makes}\mathrm{the}\mathrm{problem}\mathrm{so}\mathrm{simple}.$$$$\mathrm{You}\mathrm{just}\mathrm{have}\mathrm{an}\mathrm{extraordinary}\mathrm{creative}$$$$\mathrm{mind}.\mathrm{Everyday}\mathrm{I}\mathrm{am}\mathrm{learning}\mathrm{something}$$$$\mathrm{new}\mathrm{from}\mathrm{you}.\mathrm{Thanks}\mathrm{a}\mathrm{lot}\mathrm{sir}.$$

Answered by mr W last updated on 29/Aug/19

Commented by mr W last updated on 29/Aug/19

$$seealsoQ67350,wehave$$$$r=\sqrt{\frac{s(s-b)(s-c)}{s-a}}=\sqrt{\frac{12\times 4\times 5}{3}}=4\sqrt{5}$$$$\alpha =\mathrm{\angle }C$$$$\cos \mathrm{\angle }C=\frac{7^2+9^2-8^2}{2\times 7\times 9}=\frac{11}{21}=\cos \alpha$$$$\Rightarrow \tan \frac{\alpha }{2}=\frac{\sqrt{5}}{4}$$$$\Rightarrow \frac{\alpha }{2}={\tan }^{-1}\frac{\sqrt{5}}{4}$$$$A_{shaded}=2\times \frac{r^2\tan \frac{\alpha }{2}}{2}-\frac{r^2\alpha }{2}=r^2(\tan \frac{\alpha }{2}-\frac{\alpha }{2})$$$$=80(\frac{\sqrt{5}}{4}-{\tan }^{-1}\frac{\sqrt{5}}{4})$$$$=3.942$$

Commented by Prithwish sen last updated on 29/Aug/19

$$\mathrm{Sir}\mathrm{is}\mathrm{there}\mathrm{any}\mathrm{possibility}\mathrm{that}\mathrm{OA}\mathrm{might}\mathrm{be}$$$$\mathrm{gone}\mathrm{through}\mathrm{P}?$$

Commented by TawaTawa last updated on 29/Aug/19

$$\mathrm{Great}.\mathrm{Thanks}\mathrm{sir}.$$

Commented by mr W last updated on 29/Aug/19

$$OApassesthroughPonlyifAC=AB.$$

Commented by Prithwish sen last updated on 29/Aug/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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