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 23179 by Joel577 last updated on 27/Oct/17

Commented by Joel577 last updated on 27/Oct/17

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{part}$$

Answered by ajfour last updated on 27/Oct/17

$$let10cm=a$$$$eq.ofcirclewithcentreastop$$$$leftcornerisy=a-\sqrt{a^2-x^2}$$$$Areaofbottomunshaded$$$$region=2{\int }_0^{a/2}[a-\sqrt{a^2-x^2}]dx$$$$=2{[ax-\frac{x}{2}\sqrt{a^2-x^2}-\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)]}_0^{a/2}$$$$=2[\frac{a^2}{2}-\frac{a^2\sqrt{3}}{8}-\frac{\pi a^2}{12}]$$$$Shadedarea=a^2-8[\frac{a^2}{2}-\frac{a^2\sqrt{3}}{8}-\frac{\pi a^2}{12}]$$$$=-3a^2+a^2\sqrt{3}+\frac{2\pi }{3}{a}^2$$$$witha=10cm$$$$\mathit{S}haded\mathit{A}rea=100(\frac{2\pi }{3}+\sqrt{3}-3){cm}^2$$$$\approx 100(2.095+1.732-3){cm}^2$$$$\approx 82.7{cm}^2.$$

Commented by Joel577 last updated on 27/Oct/17

$$Idontunderstandhowcanugety=a-\sqrt{a^2-x^2}$$

Commented by Rasheed.Sindhi last updated on 27/Oct/17

$$\mathrm{Can}\mathrm{this}\mathrm{be}\mathrm{solved}\mathrm{without}\mathrm{calculus}?$$

Commented by Joel577 last updated on 27/Oct/17

$$thankyouverymuch$$

Answered by mrW1 last updated on 27/Oct/17

$$\mathrm{let}\mathrm{a}=10$$$$\mathrm{b}=2\mathrm{a}\times \sin \frac{30°}{2}=2\mathrm{a}\times \sin 15°=\mathrm{a}\sqrt{2-\sqrt{3}}$$$${\mathrm{A}}_{\mathrm{shade}}=8\times ⌢+4\times \mathrm{\Delta }+1\times ◻$$$$=8\times \frac{{\mathrm{a}}^2}{2}(\frac{\pi }{6}-\sin {30}^{°})+4\times \frac{\mathrm{b}}{2}\times \frac{\sqrt{3}\mathrm{b}}{2}+{\mathrm{b}}^2$$$$={4\mathrm{a}}^2(\frac{\pi }{6}-\frac{1}{2})+(\sqrt{3}+1)(2-\sqrt{3}){\mathrm{a}}^2$$$$=(\frac{2\pi }{3}+\sqrt{3}-3){\mathrm{a}}^2$$$$\approx 82.64{\mathrm{cm}}^2$$

Commented by Rasheed.Sindhi last updated on 27/Oct/17

$$\mathrm{Thanks}\mathrm{Sir}!\mathrm{Can}\mathrm{it}\mathrm{be}\mathrm{solved}\mathrm{even}$$$$\mathrm{without}\mathrm{trigonomeric}\mathrm{ratios}.$$

Commented by mrW1 last updated on 27/Oct/17

Commented by Joel577 last updated on 27/Oct/17

$$thankyouverymuch$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com