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Question Number 195012 by sonukgindia last updated on 22/Jul/23

Answered by a.lgnaoui last updated on 23/Jul/23

Commented by a.lgnaoui last updated on 23/Jul/23

Answered by mr W last updated on 23/Jul/23

Commented by mr W last updated on 23/Jul/23

$$r=radiusofsmallcircle$$$$R=radiusofsemicircle=2r$$$$\theta =60°$$$$a=\sqrt{7}+\sqrt{3}$$$$OD=\frac{r}{\tan \frac{\theta }{2}}$$$$AD=2r-\frac{r}{\tan \frac{\theta }{2}}$$$$DB=2r+\frac{r}{\tan \frac{\theta }{2}}$$$${AC}^2={(2r-\frac{r}{\tan \frac{\theta }{2}})}^2+a^2+2a(2r-\frac{r}{\tan \frac{\theta }{2}})\cos \theta$$$${BC}^2={(2r+\frac{r}{\tan \frac{\theta }{2}})}^2+a^2-2a(2r+\frac{r}{\tan \frac{\theta }{2}})\cos \theta$$$${AC}^2+{BC}^2={AB}^2$$$${(2r-\frac{r}{\tan \frac{\theta }{2}})}^2+a^2+2a(2r-\frac{r}{\tan \frac{\theta }{2}})\cos \theta +{(2r+\frac{r}{\tan \frac{\theta }{2}})}^2+a^2-2a(2r+\frac{r}{\tan \frac{\theta }{2}})\cos \theta ={\left(4r\right)}^2$$$$r^2(4-\frac{1}{{\tan }^2\frac{\theta }{2}})=a^2$$$$\Rightarrow r^2=\frac{a^2}{4-{\tan }^2\frac{\theta }{2}}\Rightarrow r=\frac{a}{\sqrt{4-{\tan }^2\frac{\theta }{2}}}$$$$areaofsmallcircle=\pi r^2$$$$=\frac{\pi a^2}{4-{\tan }^2\frac{\theta }{2}}=\frac{\pi {(\sqrt{7}+\sqrt{3})}^2}{4-{\left(\frac{1}{\sqrt{3}}\right)}^2}=\frac{6\pi (5+\sqrt{21})}{11}$$

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