Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 86453 by TawaTawa1 last updated on 28/Mar/20

Commented by john santu last updated on 28/Mar/20

$$2\mathrm{a}=\mathrm{distance}\mathrm{the}\mathrm{centre}\mathrm{point}\mathrm{of}$$$$\mathrm{two}\mathrm{circle}?$$

Commented by TawaTawa1 last updated on 28/Mar/20

$$\mathrm{Yes}\mathrm{sir}$$

Commented by jagoll last updated on 29/Mar/20

$$\mathrm{what}\mathrm{this}\mathrm{answer}\mathrm{miss}\mathrm{tawa}?$$

Answered by jagoll last updated on 29/Mar/20

Commented by jagoll last updated on 29/Mar/20

$$\cos \theta =\frac{\mathrm{a}}{\mathrm{R}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{\mathrm{a}}{\mathrm{R}}\right)$$$$\mathrm{the}\mathrm{area}\mathrm{of}1\mathrm{shaded}\mathrm{area}\mathrm{is}$$$$=\frac{{2\cos }^{-1}\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}{2\pi }\times \pi {\mathrm{R}}^2-\frac{1}{2}{\mathrm{R}}^2\sin 2\theta$$$$={\mathrm{R}}^2{\cos }^{-1}\left(\frac{\mathrm{a}}{\mathrm{R}}\right)-{\mathrm{R}}^2\left(\frac{\mathrm{a}\sqrt{{\mathrm{R}}^2-{\mathrm{a}}^2}}{{\mathrm{R}}^2}\right)$$$$={\mathrm{R}}^2{\cos }^{-1}\left(\frac{\mathrm{a}}{\mathrm{R}}\right)-\mathrm{a}\sqrt{{\mathrm{R}}^2-{\mathrm{a}}^2}$$$$\mathrm{now}\mathrm{the}\mathrm{total}\mathrm{area}\mathrm{your}\mathrm{request}$$$$={2\mathrm{R}}^2{\cos }^{-1}\left(\frac{\mathrm{a}}{\mathrm{R}}\right)-2\mathrm{a}\sqrt{{\mathrm{R}}^2-{\mathrm{a}}^2}$$

Commented by TawaTawa1 last updated on 29/Mar/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.\mathrm{I}\mathrm{will}\mathrm{use}\mathrm{your}\mathrm{answer}.$$$$\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{answer}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com