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 167690 by peter frank last updated on 22/Mar/22

Answered by som(math1967) last updated on 23/Mar/22

by solving x^2 +y^2 =4 and (x−2)^2 +y^2 =4  x=1 ,y=±(√3)  area of OABCO  =2×areaOMBAO  =2×2MBAM  =4∫_1 ^2 (√(4−x^2 ))dx  =4[((x(√(4−x^2 )))/2) +(4/2)sin^(−1) (x/2)]_1 ^2   =4[2sin^(−1) 1−((√3)/2) −2sin^(−1) (1/2)]  =4[π−(π/3)−((√3)/2)]  =(((8π)/3)−2(√3))sq unit

$${by}\:{solving}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:{and}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$$${x}=\mathrm{1}\:,{y}=\pm\sqrt{\mathrm{3}} \\ $$$${area}\:{of}\:{OABCO} \\ $$$$=\mathrm{2}×{areaOMBAO} \\ $$$$=\mathrm{2}×\mathrm{2}{MBAM} \\ $$$$=\mathrm{4}\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\left[\frac{{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}}\:+\frac{\mathrm{4}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\mathrm{4}\left[\mathrm{2sin}^{−\mathrm{1}} \mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\mathrm{2sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$=\mathrm{4}\left[\pi−\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right] \\ $$$$=\left(\frac{\mathrm{8}\pi}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}}\right){sq}\:{unit} \\ $$

Commented by som(math1967) last updated on 23/Mar/22

Terms of Service

Privacy Policy

Contact: info@tinkutara.com