Question Number 197196 by sonukgindia last updated on 10/Sep/23
Answered by mahdipoor last updated on 10/Sep/23
$${blue}\equiv \\ $$$$\pi{r}^{\mathrm{2}} \left(\frac{\alpha_{\mathrm{1}} }{\mathrm{360}}+\frac{\alpha_{\mathrm{2}} }{\mathrm{360}}+\frac{\alpha_{\mathrm{3}} }{\mathrm{360}}...+\frac{\alpha_{{n}} }{\mathrm{360}}\right) \\ $$$${orange}\equiv \\ $$$$\pi{r}^{\mathrm{2}} \left(\frac{\mathrm{360}−\alpha_{\mathrm{1}} }{\mathrm{360}}+\frac{\mathrm{360}−\alpha_{\mathrm{2}} }{\mathrm{360}}+\frac{\mathrm{360}−\alpha_{\mathrm{3}} }{\mathrm{360}}...+\frac{\mathrm{360}−\alpha_{{n}} }{\mathrm{360}}\right) \\ $$$${O}−{B}=\pi{r}^{\mathrm{2}} \left({n}−\mathrm{2}\frac{\Sigma\alpha}{\mathrm{360}}\right)=\pi{r}^{\mathrm{2}} \left({n}−\mathrm{2}\frac{\mathrm{180}×\left({n}−\mathrm{2}\right)}{\mathrm{360}}\right) \\ $$$$=\mathrm{2}\pi{r}^{\mathrm{2}} \\ $$$$ \\ $$