# Question-68503

Question Number 68503 by TawaTawa last updated on 12/Sep/19
Commented by Prithwish sen last updated on 12/Sep/19
$$\mathrm{2x}\frac{\pi\mathrm{6}^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{2}\left[\mathrm{36}\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)\right]=\mathrm{18}\left[\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{3}}\right]\:\backsim\:\mathrm{12}.\mathrm{33} \\$$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\$$
Commented by TawaTawa last updated on 12/Sep/19
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{answer},\:\mathrm{but}\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{more}.\:\mathrm{God}\:\mathrm{bless} \\$$$$\mathrm{you}\:\mathrm{sir} \\$$
Commented by mr W last updated on 12/Sep/19
$$\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{6}}×\mathrm{6}^{\mathrm{2}} −\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right)\right] \\$$$$=\mathrm{6}\left(\mathrm{3}\sqrt{\mathrm{3}}−\pi\right) \\$$
Commented by TawaTawa last updated on 13/Sep/19
$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{it}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{minus}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{quarter}\:\mathrm{circle}\:? \\$$
Commented by TawaTawa last updated on 13/Sep/19
$$\mathrm{Or}\:\mathrm{please}\:\mathrm{sir}\:\mathrm{help}\:\mathrm{me}\:\mathrm{label}\:\mathrm{the}\:\mathrm{diagram} \\$$
Commented by mr W last updated on 13/Sep/19
Commented by mr W last updated on 13/Sep/19
$${area}\:{of}\:\frac{\pi}{\mathrm{6}}−{sector}\:=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}\right) \\$$$${area}\:{of}\:{green}\:{segment}=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right) \\$$
Commented by TawaTawa last updated on 13/Sep/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\$$