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Question Number 176946 by Ar Brandon last updated on 28/Sep/22

Answered by mr W last updated on 29/Sep/22

Commented by mr W last updated on 29/Sep/22

$$R=radius,K=diameterofcircle$$$$\sin \alpha =\frac{a}{2R}=\frac{a}{K}$$$$\sin \beta =\frac{b}{2R}=\frac{b}{K}$$$$\alpha +\beta =\theta =60°=\frac{\pi }{3}$$$$\cos (\alpha +\beta )=\cos \theta =\frac{1}{2}$$$$\sqrt{(1-\frac{a^2}{K^2})(1-\frac{b^2}{K^2})}-\frac{a}{K}\times \frac{b}{K}=\cos \theta$$$$(1-\frac{a^2}{K^2})(1-\frac{b^2}{K^2})={(\frac{ab}{K^2}+\cos \theta )}^2$$$$1-\frac{a^2}{K^2}-\frac{b^2}{K^2}=2\cos \theta \frac{ab}{K^2}+{\cos }^2\theta$$$$K^2=4R^2=\frac{a^2+b^2+2ab\cos \theta }{{\sin }^2\theta }$$$$\Rightarrow R=\frac{\sqrt{a^2+b^2+2ab\cos \theta }}{2\sin \theta }$$$$=\frac{\sqrt{2^2+5^2+2\times 2\times 5\times \frac{1}{2}}}{2\times \frac{\sqrt{3}}{2}}=\sqrt{13}$$$$A_{segment,a}=\alpha R^2-\frac{a}{2}\sqrt{R^2-\frac{a^2}{4}}$$$$A_{segment,b}=\beta R^2-\frac{b}{2}\sqrt{R^2-\frac{b^2}{4}}$$$$A_{shaded}=R^2\theta -\frac{a}{2}\sqrt{R^2-\frac{a^2}{4}}-\frac{b}{2}\sqrt{R^2-\frac{b^2}{4}}$$$$=13\times \frac{\pi }{3}-\frac{5}{2}\sqrt{13-\frac{5^2}{4}}-\frac{2}{2}\sqrt{13-\frac{2^2}{4}}$$$$=\frac{13\pi }{3}-\frac{23\sqrt{3}}{4}$$$$\approx 3.654$$

Commented by Ar Brandon last updated on 29/Sep/22

Thank you, Sir!

Commented by Tawa11 last updated on 02/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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