Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 140139 by mathsuji last updated on 04/May/21

Answered by mr W last updated on 04/May/21

$$sayradiusofcurcumcircleisr$$$$\mathrm{\Sigma }{\sin }^{-1}\frac{a}{2r}={\sin }^{-1}\frac{6}{2r}+\sin \frac{3}{2r}+{\sin }^{-1}\frac{\sqrt{11}}{2r}+{\sin }^{-1}\frac{6}{2r}+{\sin }^{-1}\frac{\sqrt{2}}{2r}=\pi$$$$\Rightarrow 2r=7.09124208$$$$A=\frac{1}{4}\mathrm{\Sigma }a\sqrt{4r^2-a^2}$$$$\approx 23.81176$$

Commented by mr W last updated on 04/May/21

Commented by mathsuji last updated on 05/May/21

$$thankyouverymuchSir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com