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 177596 by cherokeesay last updated on 07/Oct/22

Answered by mr W last updated on 07/Oct/22

Commented by mr W last updated on 07/Oct/22

$$\sqrt{2}{r}_1=R=4$$$$\Rightarrow r_1=2\sqrt{2}$$$${(r_1+r_2)}^2-{(r_1-r_2)}^2={(R-r_2)}^2-r_2^2$$$$4r_1{r}_2=R(R-2r_2)$$$$\sqrt{2}{r}_2=2-r_2$$$$\Rightarrow r_2=\frac{2}{\sqrt{2}+1}=2(\sqrt{2}-1)$$$$A_{green}=\frac{\pi R^2}{4}-\frac{\pi r_1^2}{4}-\pi r_2^2$$$$=\frac{\pi \times 4^2}{4}-\frac{\pi {\left(2\sqrt{2}\right)}^2}{4}-\pi {\left(2(\sqrt{2}-1)\right)}^2$$$$=2(4\sqrt{2}-5)\pi \approx 4.127$$

Commented by Ar Brandon last updated on 07/Oct/22

$$\mathrm{Sir},\mathrm{why}\sqrt{2}{r}_1=R?$$

Commented by mr W last updated on 07/Oct/22

Commented by Tawa11 last updated on 07/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by Ar Brandon last updated on 07/Oct/22

Thanks Sir

Terms of Service

Privacy Policy

Contact: info@tinkutara.com