# Question-76855

Question Number 76855 by aliesam last updated on 31/Dec/19
Commented by aliesam last updated on 31/Dec/19
$${AB}={BC}=……={Fg}={GA} \\$$$${prove}\:{that}\: \\$$$${the}\:{area}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{7}{tan}\left(\frac{\pi}{\mathrm{14}}\right)\right) \\$$
Commented by 67549972 last updated on 08/Feb/20
$$\\$$
Answered by mr W last updated on 31/Dec/19
Commented by mr W last updated on 31/Dec/19
$$\mathrm{2}\theta=\frac{\mathrm{2}\pi}{\mathrm{7}} \\$$$$\Rightarrow\theta=\frac{\pi}{\mathrm{7}} \\$$$${R}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\frac{{a}}{\mathrm{2}} \\$$$$\Rightarrow{R}=\frac{{a}}{\mathrm{2}\:\mathrm{cos}\:\frac{\pi}{\mathrm{14}}} \\$$$${A}_{{blue}} =\frac{{R}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}}{\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}} \\$$$${A}_{{green}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\theta−\mathrm{sin}\:\theta\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{7}}−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right) \\$$$$\\$$$${A}=\mathrm{7}\left({A}_{{blue}} +{A}_{{green}} \right) \\$$$$=\mathrm{7}\left[\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}}{\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{7}}−\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right)\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\pi+\frac{\mathrm{7}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}−\mathrm{7}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\pi+\frac{\mathrm{7}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\:\mathrm{cos}\:\frac{\pi}{\mathrm{7}}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}−\mathrm{7}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\pi+\mathrm{7}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\left(\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{7}}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}−\mathrm{1}\right)\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\pi+\mathrm{7}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}\left(\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{7}}−\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}\right)\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\pi−\frac{\mathrm{7}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\pi−\frac{\mathrm{14}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:\mathrm{cos}\:\frac{\pi}{\mathrm{14}}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}}\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\pi−\frac{\mathrm{7}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}}}{\mathrm{cos}\:\frac{\pi}{\mathrm{14}}}\right] \\$$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{7}\:\mathrm{tan}\:\frac{\pi}{\mathrm{14}}\right) \\$$
Commented by jagoll last updated on 31/Dec/19
$${waw}….{fantastic}\:{sir}\: \\$$
Commented by aliesam last updated on 31/Dec/19
$${briliant}\:{solution}\:{sir}\:.\:{thank}\:{you} \\$$
Commented by Tawa11 last updated on 29/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\$$