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Question Number 35750 by ajfour last updated on 23/May/18

Commented by ajfour last updated on 23/May/18

$$Findproportionofblue,brown,$$$$yellow,andredcolouredareas.$$$$Given:chordmakesangle\theta with$$$$thediametershownandthatthe$$$$pointofintersectionofthechord$$$$andthediameterisatdistance$$$$\mathit{\lambda }\mathit{r}fromcentreofcircle.$$

Commented by ajfour last updated on 23/May/18

Commented by ajfour last updated on 23/May/18

$$CE=\lambda r$$

Commented by ajfour last updated on 23/May/18

$$AB=2\sqrt{r^2-{\lambda }^2{r}^2\sin {}^2\theta }$$$$AE=\frac{AB}{2}+CE\cos \theta$$$$=r\sqrt{1-{\lambda }^2\sin {}^2\theta }+\lambda r\cos \theta$$$$BE=r\sqrt{1-{\lambda }^2\sin {}^2\theta }-\lambda r\cos \theta$$$$Area(△ACE)=\frac{1}{2}\left(AE\right)\left(CE\right)\sin \theta$$$$=\frac{\lambda r^2\sin \theta }{2}(\sqrt{1-{\lambda }^2\sin {}^2\theta }+\lambda \cos \theta )$$$$similarly$$$$Area(△BCE)=\frac{1}{2}\left(BE\right)\left(CE\right)\sin \theta$$$$=\frac{\lambda r^2\sin \theta }{2}(\sqrt{1-{\lambda }^2\sin {}^2\theta }-\lambda \cos \theta )$$$$let\sin \mathrm{\angle }CAE=\sin \mathrm{\angle }CBE=\sin \alpha$$$$\sin \alpha =\frac{\lambda r\sin \theta }{r}=\lambda \sin \theta$$$$\Rightarrow \alpha ={\sin }^{-1}\left(\lambda \sin \theta \right)$$$$\mathrm{\angle }ACD=\alpha +\theta$$$$Area\left(sectorCAD\right)=\frac{r^2}{2}(\theta +\alpha )$$$$---------------$$$$Area\left(blue\right)=Area(△ACE)+$$$$Area\left(sectorCAD\right)$$$$=\frac{\lambda r^2\sin \theta }{2}(\sqrt{1-{\lambda }^2\sin {}^2\theta }+\lambda \cos \theta )$$$$+\frac{r^2}{2}[\theta +{\sin }^{-1}\left(\lambda \sin \theta \right)]$$$$---------------$$$$Area\left(yellow\right)=Area\left(sectorACF\right)$$$$-Area(△ACE)$$$$=\frac{r^2}{2}[\pi -\theta -{\sin }^{-1}\left(\lambda \sin \theta \right)]$$$$-\frac{\lambda r^2\sin \theta }{2}(\sqrt{1-{\lambda }^2\sin {}^2\theta }+\lambda \cos \theta )$$$$---------------$$$$Area(△BCE)$$$$=\frac{\lambda r^2\sin \theta }{2}(\sqrt{1-{\lambda }^2\sin {}^2\theta }-\lambda \cos \theta )$$$$Area\left(brown\right)=\frac{r^2}{2}(\pi -\theta +\alpha )$$$$+Area(△BCE)$$$$=\frac{r^2}{2}[\pi -\theta +{\sin }^{-1}\left(\lambda \sin \theta \right)]$$$$+\frac{\lambda r^2\sin \theta }{2}(\sqrt{1-{\lambda }^2\sin {}^2\theta }-\lambda \cos \theta )$$$$---------------$$$$Area\left(red\right)=\frac{r^2}{2}[\theta -{\sin }^{-1}\left(\lambda \sin \theta \right)]$$$$-\frac{\lambda r^2\sin \theta }{2}(\sqrt{1-{\lambda }^2\sin {}^2\theta }-\lambda \cos \theta )$$$$---------------.$$

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