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Question Number 70497 by mr W last updated on 04/Oct/19

Commented by mr W last updated on 04/Oct/19

$$shadedarea=?$$

Answered by ajfour last updated on 04/Oct/19

Commented by ajfour last updated on 04/Oct/19

$$2a\sin \beta =a\sin \alpha ....\left(i\right)$$$$2a\cos \beta -a\cos \alpha =a\sqrt{2}...\left(ii\right)$$$$\Rightarrow 4(1-\sin {}^2\beta )=2+\cos {}^2\alpha +2\sqrt{2}\cos \alpha$$$$4-(1-\cos {}^2\alpha )=2+\cos {}^2\alpha +2\sqrt{2}\cos \alpha$$$$\Rightarrow \cos \alpha =\frac{1}{2\sqrt{2}},\sin \alpha =\frac{\sqrt{7}}{2\sqrt{2}}$$$$\sin \beta =\frac{\sqrt{7}}{4\sqrt{2}}$$$$Shadedarea=A$$$$\frac{A}{4}=\frac{a^2\alpha }{2}-\frac{4a^2\beta }{2}+\frac{1}{2}\left(a\sqrt{2}\right)\left(a\sin \alpha \right)$$$$A=2a^2\{\alpha -4\beta +\sqrt{2}\sin \alpha )$$$$A=50\{{\sin }^{-1}\frac{\sqrt{7}}{2\sqrt{2}}-{4\sin }^{-1}\frac{\sqrt{7}}{4\sqrt{2}}+\frac{\sqrt{7}}{2}\}{cm}^2$$$$\approx 29.27625{cm}^2.$$

Commented by mr W last updated on 05/Oct/19

$$thankssir!$$$$wecanalsoget\beta inthisway:$$$$a^2={\left(2a\right)}^2+{\left(\sqrt{2}a\right)}^2-2\times 2a\times \sqrt{2}a\cos \beta$$$$\Rightarrow \cos \beta =\frac{5}{4\sqrt{2}}$$$$\Rightarrow \sin \beta =\frac{\sqrt{7}}{4\sqrt{2}}$$

Commented by ajfour last updated on 05/Oct/19

$$Tworoadsdivergedinawood,$$$$Ichosetheonelesstraveledby..$$

Commented by mr W last updated on 05/Oct/19

$$truely,allroadsleadtoRome!$$

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