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Question Number 79121 by M±th+et£s last updated on 22/Jan/20

Answered by mr W last updated on 23/Jan/20

Commented by mr W last updated on 23/Jan/20

$$R=radiusofsemicircle$$$$2a=sidelengthofequilateraltriangle$$$$\lambda =\frac{a}{R}$$$$\frac{\sin \frac{\pi }{3}}{R}=\frac{\sin (\pi -\beta -\frac{\pi }{3})}{a}=\frac{\cos (\beta -\frac{\pi }{6})}{a}$$$$\cos (\beta -\frac{\pi }{6})=\frac{\sqrt{3}a}{2R}=\frac{\sqrt{3}\lambda }{2}$$$$\Rightarrow \beta ={\cos }^{-1}\frac{\sqrt{3}\lambda }{2}+\frac{\pi }{6}$$$$\sin \beta =\frac{1}{2}\times \frac{\sqrt{3}\lambda }{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{4-3{\lambda }^2}}{2}=\frac{\sqrt{3}(\lambda +\sqrt{4-3{\lambda }^2})}{4}$$$$\alpha =\frac{\pi }{2}-\beta =\frac{\pi }{3}-{\cos }^{-1}\frac{\sqrt{3}\lambda }{2}$$$$A=2(\frac{\alpha R^2}{2}+\frac{Ra\sin \beta }{2})$$$$A=R^2(\alpha +\lambda \sin \beta )$$$$areaoftriangle=\sqrt{3}{a}^2=\sqrt{3}{\lambda }^2{R}^2=A+S..\left(i\right)$$$$areaofsemicircle=\frac{\pi R^2}{2}=A+2S..\left(ii\right)$$$$2\times \left(i\right)-\left(ii\right):$$$$\Rightarrow 2\sqrt{3}{\lambda }^2{R}^2-\frac{\pi R^2}{2}=A=R^2(\alpha +\lambda \sin \beta )$$$$\Rightarrow 2\sqrt{3}{\lambda }^2-\frac{\pi }{2}=\alpha +\lambda \sin \beta$$$$\Rightarrow 2\sqrt{3}{\lambda }^2-\frac{\pi }{2}=\frac{\pi }{3}-{\cos }^{-1}\frac{\sqrt{3}\lambda }{2}+\frac{\lambda \sqrt{3}(\lambda +\sqrt{4-3{\lambda }^2})}{4}$$$$\Rightarrow 7\sqrt{3}{\lambda }^2-\lambda \sqrt{3(4-3{\lambda }^2)}+4{\cos }^{-1}\frac{\sqrt{3}\lambda }{2}=\frac{10\pi }{3}$$$$\Rightarrow \lambda \approx 0.894034$$$$$$$$from\left(ii\right):$$$$S=\frac{1}{2}(\frac{\pi R^2}{2}-A)=\frac{\pi R^2}{4}-\frac{A}{2}$$$$\frac{S}{A}=\frac{\pi R^2}{4\times (2\sqrt{3}{\lambda }^2{R}^2-\frac{\pi R^2}{2})}-\frac{1}{2}$$$$\Rightarrow \frac{S}{A}=\frac{\pi }{8\sqrt{3}{\lambda }^2-2\pi }-\frac{1}{2}\approx 0.155564$$

Commented by mr W last updated on 23/Jan/20

$$anexactsolutionisnotpossible.$$

Commented by M±th+et£s last updated on 23/Jan/20

$$godblessyousir.nicejob$$

Commented by M±th+et£s last updated on 23/Jan/20

Commented by M±th+et£s last updated on 23/Jan/20

$$howcanyoufindsin\left(B\right)fromBvaluep$$

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