Question Number 94201 by $@ty@m123 last updated on 17/May/20
Commented by $@ty@m123 last updated on 17/May/20
Find the area of shaded region if side of square is a and radius of arc is a.
Answered by mr W last updated on 17/May/20
Commented by mr W last updated on 17/May/20
$${p}=\mathrm{2}{a}\:\mathrm{sin}\:\frac{\pi}{\mathrm{12}} \\ $$$${A}_{{shade}} =\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} }{\mathrm{2}}+\left(\mathrm{2}−\mathrm{1}\right)×\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${A}_{{shade}} =\frac{\mathrm{4}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{12}}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${A}_{{shade}} ={a}^{\mathrm{2}} \sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\right)+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${A}_{{shade}} =\frac{{a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right)}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} \left(\pi−\mathrm{3}\right)}{\mathrm{12}} \\ $$$${A}_{{shade}} =\frac{{a}^{\mathrm{2}} \left(\pi+\mathrm{12}\sqrt{\mathrm{3}}−\mathrm{21}\right)}{\mathrm{12}}\approx\mathrm{0}.\mathrm{24385}{a}^{\mathrm{2}} \\ $$
Commented by $@ty@m123 last updated on 17/May/20
$${Thanks}\:{a}\:{lot}. \\ $$