Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 177578 by BaliramKumar last updated on 07/Oct/22

Commented by BaliramKumar last updated on 07/Oct/22

area of green colour

Commented by som(math1967) last updated on 07/Oct/22

$$\frac{\pi \sqrt{2}}{2}-1?$$

Commented by BaliramKumar last updated on 07/Oct/22

$$yessir$$

Commented by som(math1967) last updated on 07/Oct/22

$$Ar.ofcircle=2\pi$$$$letR=radofquadrant$$$$OB=R-\sqrt{2}$$$$OABCsquare\therefore OB=\sqrt{2}\times \sqrt{2}=2$$$$\therefore R=2+\sqrt{2}$$$$ar.ofquadrant=\frac{1}{4}\pi {(2+\sqrt{2})}^2$$$$ar.ofquadrantformwithOABC$$$$\frac{1}{4}\times \pi \times {\left(\sqrt{2}\right)}^2=\frac{\pi }{2}$$$$A_{green}=\frac{1}{2}[\frac{1}{4}\pi {(2+\sqrt{2})}^2-2\pi -(2-\frac{\pi }{2})]$$$$=\frac{1}{2}[\frac{1}{4}\pi \times 2(3+\sqrt{2})-\frac{3\pi }{2}-2]$$$$=\frac{\pi \sqrt{2}}{2}-1squnit$$$$$$

Commented by som(math1967) last updated on 07/Oct/22

Commented by Tawa11 last updated on 07/Oct/22

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

Answered by cherokeesay last updated on 07/Oct/22

Commented by BaliramKumar last updated on 07/Oct/22

$$greatsir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com