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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=2a sin (π/(12)) A_(shade) =(((√3)p^2 )/2)+(2−1)×(a^2 /2)((π/6)−sin (π/6)) A_(shade) =((4a^2 (√3))/2)sin^2 (π/(12))+(a^2 /2)((π/6)−sin (π/6)) A_(shade) =a^2 (√3)(1−cos (π/6))+(a^2 /2)((π/6)−sin (π/6)) A_(shade) =((a^2 (2(√3)−3))/2)+((a^2 (π−3))/(12)) A_(shade) =((a^2 (π+12(√3)−21))/(12))≈0.24385a^2

$${p}=\mathrm{2}{a}\:\mathrm{sin}\:\frac{\pi}{\mathrm{12}} \\ $$$${A}_{{shade}} =\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} }{\mathrm{2}}+\left(\mathrm{2}−\mathrm{1}\right)×\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${A}_{{shade}} =\frac{\mathrm{4}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{12}}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${A}_{{shade}} ={a}^{\mathrm{2}} \sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\right)+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${A}_{{shade}} =\frac{{a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right)}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} \left(\pi−\mathrm{3}\right)}{\mathrm{12}} \\ $$$${A}_{{shade}} =\frac{{a}^{\mathrm{2}} \left(\pi+\mathrm{12}\sqrt{\mathrm{3}}−\mathrm{21}\right)}{\mathrm{12}}\approx\mathrm{0}.\mathrm{24385}{a}^{\mathrm{2}} \\ $$

Commented by $@ty@m123 last updated on 17/May/20

Thanks a lot.

$${Thanks}\:{a}\:{lot}. \\ $$

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