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Question Number 118448 by peter frank last updated on 17/Oct/20

Answered by Olaf last updated on 17/Oct/20

$$\mathrm{A}=\pi \times 7^2-2{\int }_{-7}^{+7}(7-\sqrt{49-x^2})dx$$$$\mathrm{A}=49\pi -4{\int }_0^{+7}(7-\sqrt{49-x^2})dx$$$$\mathrm{A}=49\pi -196+28{\int }_0^{+7}\left(\sqrt{1-{\left(\frac{x}{7}\right)}^2}\right)dx$$$$x=7\sin \theta$$$$\mathrm{A}=49\pi -196+28{\int }_0^{\frac{\pi }{2}}\left(\cos \theta \right)\times 7\cos \theta d\theta$$$$\mathrm{A}=49\pi -196+196{\int }_0^{\frac{\pi }{2}}\frac{1+\cos 2\theta }{2}d\theta$$$$\mathrm{A}=49\pi -196+98{[\theta +\frac{\sin 2\theta }{2}]}_0^{\frac{\pi }{2}}$$$$\mathrm{A}=49\pi -196+98\left(\frac{\pi }{2}\right)$$$$\mathrm{A}=98\pi -196$$

Commented by peter frank last updated on 17/Oct/20

$$\mathrm{ans}33.6{\mathrm{cm}}^2$$

Answered by mr W last updated on 18/Oct/20

$$shadedarea=circle-4segments$$$$\pi R^2-4\times \frac{R^2}{2}(\frac{2\pi }{3}-\frac{\sqrt{3}}{2})$$$$=(\sqrt{3}-\frac{\pi }{3})R^2$$$$withR=7cm$$

Commented by peter frank last updated on 17/Oct/20

$$\mathrm{ans}33.{6\mathrm{cm}}^2$$

Commented by mr W last updated on 18/Oct/20

$$putR=7,yougetanswer33.558{cm}^2$$

Commented by mr W last updated on 18/Oct/20

$$itismeantthattheradiusofcircle$$$$isequaltheradiusofsegment.$$$$thediagramwasjustnotwellmade.$$

Commented by peter frank last updated on 18/Oct/20

$$\mathrm{more}\mathrm{explanation}.\mathrm{i}\mathrm{did}$$$$\mathrm{not}\mathrm{understood}$$$$(\frac{2\pi }{3}-\frac{\sqrt{3}}{2})?$$

Commented by mr W last updated on 18/Oct/20

$$areaofsegment$$$$A=\frac{R^2}{2}(\theta -\sin \theta )$$$$herewehave\theta =120°=\frac{2\pi }{3}$$

Commented by mr W last updated on 18/Oct/20

Commented by mr W last updated on 18/Oct/20

Commented by 1549442205PVT last updated on 18/Oct/20

$$\mathrm{But}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{R}.\mathrm{The}\mathrm{given}$$$$\mathrm{segment}\mathrm{on}\mathrm{the}\mathrm{figure}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{the}\mathrm{radius}$$$$\mathrm{of}\mathrm{the}\mathrm{circle}$$

Commented by mr W last updated on 18/Oct/20

Commented by peter frank last updated on 18/Oct/20

Commented by peter frank last updated on 18/Oct/20

$$\mathrm{i}\mathrm{think}\mathrm{this}\mathrm{diagram}$$$$\mathrm{clear}$$

Commented by peter frank last updated on 18/Oct/20

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

Commented by 1549442205PVT last updated on 18/Oct/20

$$\mathrm{If}\mathrm{R}=7\mathrm{then}{\mathrm{S}}_{\mathrm{shaded}}=\pi {\mathrm{R}}^2-4(\frac{\pi {\mathrm{R}}^2}{3}-\frac{{\mathrm{R}}^2\sqrt{3}}{4})$$$$={\mathrm{R}}^2(\sqrt{3}-\frac{\pi }{3})\approx 33.55780$$

Commented by peter frank last updated on 18/Oct/20

$$\mathrm{thank}\mathrm{you}$$

Answered by 1549442205PVT last updated on 18/Oct/20

Commented by 1549442205PVT last updated on 18/Oct/20

$$\mathrm{This}\mathrm{problem}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{enough}$$$$\mathrm{data}\mathrm{to}\mathrm{solve}.\mathrm{Indeed},\mathrm{we}\mathrm{prove}$$$$\mathrm{that}\mathrm{can}\mathrm{construct}\mathrm{many}\mathrm{the}\mathrm{figures}$$$$\mathrm{satisfying}\mathrm{the}\mathrm{condition}\mathrm{of}\mathrm{problem}$$$$\mathrm{Put}\alpha =\widehat{\mathrm{ASQ}},\mathrm{R}-\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circles}$$$$\mathrm{with}\mathrm{the}\mathrm{centers}\mathrm{at}\left(\mathrm{A}\right)\mathrm{and}\left(\mathrm{C}\right).\mathrm{Then}$$$$\mathrm{AQ}=2\mathrm{Rsin}\alpha ,\mathrm{AO}=\mathrm{R}.\mathrm{Hence},\mathrm{from}\mathrm{the}$$$$\mathrm{hypothesis}\mathrm{we}\mathrm{have}\mathrm{PQ}\mathrm{\perp }\mathrm{SN}\mathrm{at}\mathrm{A}\mathrm{and}$$$$\mathrm{NP}=\mathrm{OQ}=7.\mathrm{From}{\mathrm{Pithagorean}}^{\prime }\mathrm{s}$$$$\mathrm{theorem}\mathrm{we}\mathrm{have}:{\mathrm{AQ}}^2+{\mathrm{AO}}^2={\mathrm{OQ}}^2=49$$$$\iff {4\mathrm{R}}^2{\sin }^2\alpha +{\mathrm{R}}^2=49\Rightarrow \mathrm{R}=\frac{7}{\sqrt{{4\sin }^2\alpha +1}}$$$$\mathrm{or}\sin \alpha =\sqrt{\frac{49-{\mathrm{R}}^2}{{4\mathrm{R}}^2}}=\frac{\sqrt{49-{\mathrm{R}}^2}}{2\mathrm{R}}$$$$\mathrm{Therefore},\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{valuesR}\mathrm{satisfying}$$$$4.95\approx \frac{7}{\sqrt{2}}⩽\mathrm{R}<7\mathrm{we}\mathrm{determine}\mathrm{one}\mathrm{value}$$$$\mathrm{of}\mathrm{the}\mathrm{angle}\alpha \mathrm{as}\mathrm{on}\mathrm{the}\mathrm{figure}\mathrm{and}\mathrm{the}$$$$\mathrm{segments}\mathrm{NP}=\mathrm{OQ}=7\mathrm{satisfying}\mathrm{the}$$$$\mathrm{cionditions}\mathrm{of}\mathrm{given}\mathrm{problem}!$$

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