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Question Number 65980 by Tony Lin last updated on 07/Aug/19

Commented by Tanmay chaudhury last updated on 07/Aug/19

Answered by Tanmay chaudhury last updated on 07/Aug/19

$$eqnofcirclecentreAis{x}^2+y^2={10}^2$$$$eqnofcirclecentremidpointofAD{x}^2+{(y-5)}^2=5^2$$$$eqnofcirclecentremidpointofCD{(x-5)}^2+{(y-5)}^2=5^2$$$$pointP...$$$${(x-5)}^2+{(y-5)}^2=5^2$$$$x^2+{(y-5)}^2=5^2$$$$substruct$$$$-10x+25=0x_P=\frac{5}{2}$$$${(y-5)}^2=25-\frac{25}{4}=\frac{75}{4}y=5\pm \frac{5\sqrt{3}}{2}\to y=\frac{5(1+\sqrt{3})}{2}$$$$negetivevalueofynotfeasible$$$$P(\frac{5}{2},\frac{5(1+\sqrt{3})}{2})$$$$forpointQsolve$$$$x^2+y^2={10}^2{(x-5)}^2+{(y-5)}^2=5^2$$$$100-10x-10y+50=25$$$$x+y=\frac{125}{10}=\frac{25}{2}$$$$x^2+\frac{625}{4}-25x+x^2=100$$$$8x^2-100x+625-400=0$$$$8x^2-100x+225=0$$$$x_Q=\frac{100\pm \sqrt{10000-32\times 225}}{16}$$$$x_Q=\frac{100\pm 20\sqrt{7}}{16}\to x_Q=\frac{25+5\sqrt{7}}{4}[-veignored]$$$$y=\frac{25}{2}-\frac{25+5\sqrt{7}}{4}=\frac{25-5\sqrt{7}}{4}$$$$soreqiuiredarea$$$${\int }_0^{x_Q}\sqrt{100-x^2}dx-\frac{\pi {\left(\frac{5}{2}\right)}^2}{4}-{\int }_{x_P}^{x_Q}5+\sqrt{25-{(x-5)}^2}dx$$$$itisreflectionofthought...$$

Commented by Kunal12588 last updated on 07/Aug/19

$$ithinkyougotalittlewrongtheir$$$$eqnofcirclecentremidpointofCD{(x-5)}^2+{(y-5)}^2=5^2$$$$shouldbeeqnofcirclecentremidpointofCD{(x-5)}^2+{(y-10)}^2=5^2$$$$bysolvingP(5,5)andQ(8,6)$$$${\int }_0^{x_Q}\sqrt{100-x^2}dx-\frac{\pi {\left(\frac{5}{2}\right)}^2}{4}-{\int }_{x_P}^{x_Q}5+\sqrt{25-{(x-5)}^2}dx$$$$alsoi{don}^{\prime }tgetsubtractingblueterm$$$$shouldbe{\int }_0^{x_Q}\sqrt{100-x^2}dx-{\int }_0^{x_P}5+\sqrt{25-x^2}dx-{\int }_{x_P}^{x_Q}10+\sqrt{25-{(x-5)}^2}dx$$

Commented by Tony Lin last updated on 08/Aug/19

$$thankssir$$

Answered by MJS last updated on 07/Aug/19

$$A=\left(\begin{array}{c}-5\\ -5\end{array}\right)B=\left(\begin{array}{c}5\\ -5\end{array}\right)C=\left(\begin{array}{c}5\\ 5\end{array}\right)$$$$D=\left(\begin{array}{c}-5\\ 5\end{array}\right)P=\left(\begin{array}{c}0\\ 0\end{array}\right)Q=\left(\begin{array}{c}3\\ 1\end{array}\right)$$$$\mathrm{red}\mathrm{area}$$$$\underset{-5}{\stackrel{3}{\int }}(-5+\sqrt{100-{(x+5)}^2})dx-\underset{-5}{\stackrel{0}{\int }}\sqrt{25-{(x+5)}^2}dx-\underset{0}{\stackrel{3}{\int }}(5-\sqrt{25-x^2})dx=$$$$=\frac{25}{4}(8\arcsin \frac{4}{5}+2\arcsin \frac{3}{5}-\pi -4)=$$$$=\frac{25}{2}(\pi -2-\frac{1}{2}\arctan \frac{10296}{11753})\approx 9.77357$$$$\approx 9.77357$$

Commented by Tony Lin last updated on 08/Aug/19

$$thankssir$$

Answered by mr W last updated on 07/Aug/19

Commented by mr W last updated on 07/Aug/19

$$\phi +\theta =\pi$$$$10\sin \frac{\phi }{2}=5\sin \frac{\theta }{2}$$$$2\sin \frac{\phi }{2}=\sin (\frac{\pi }{2}-\frac{\phi }{2})=\cos \frac{\phi }{2}$$$$\Rightarrow \tan \frac{\phi }{2}=\frac{1}{2}$$$$\sin \phi =\frac{2\times \frac{1}{2}}{1+{\left(\frac{1}{2}\right)}^2}=\frac{4}{5}$$$$\Rightarrow \phi =2{\tan }^{-1}\frac{1}{2}$$$$\Rightarrow \theta =\pi -2{\tan }^{-1}\frac{1}{2}$$$$A_{green}=2\times \frac{5^2}{2}(\frac{\pi }{2}-1)=\frac{25\pi }{2}-25$$$$A_{green}+A_{red}=\frac{{10}^2}{2}(\phi -\sin \phi )+\frac{5^2}{2}(\theta -\sin \theta )$$$$=50(2{\tan }^{-1}\frac{1}{2}-\frac{4}{5})+\frac{25}{2}(\pi -2{\tan }^{-1}\frac{1}{2}-\frac{4}{5})$$$$=75{\tan }^{-1}\frac{1}{2}-50+\frac{25\pi }{2}$$$$\Rightarrow A_{red}=75{\tan }^{-1}\frac{1}{2}-50+\frac{25\pi }{2}-\frac{25\pi }{2}+25$$$$\Rightarrow A_{red}=25(3{\tan }^{-1}\frac{1}{2}-1)=9.7736$$

Commented by mr W last updated on 07/Aug/19

Commented by Tony Lin last updated on 08/Aug/19

$$thankssir$$

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