Question-212719

Question Number 212719 by Spillover last updated on 22/Oct/24

Answered by A5T last updated on 22/Oct/24

Commented by A5T last updated on 22/Oct/24

a=2((π/6)−((sin60°)/2))=2((π/6)−((√3)/4))=(π/3)−((√3)/2) A=(π/2)−((2π)/3)+(√3)=(√3)−(π/6) [Non-shaded]=6A+6a=3(√3)+π [Hexagon]=6×2×((√3)/2)=6(√3) [Shaded]=6(√3)−3(√3)−π=3(√3)−π

$${a}=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{{sin}\mathrm{60}°}{\mathrm{2}}\right)=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${A}=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}\pi}{\mathrm{3}}+\sqrt{\mathrm{3}}=\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{6}} \\ $$$$\left[{Non}-{shaded}\right]=\mathrm{6}{A}+\mathrm{6}{a}=\mathrm{3}\sqrt{\mathrm{3}}+\pi \\ $$$$\left[{Hexagon}\right]=\mathrm{6}×\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{6}\sqrt{\mathrm{3}} \\ $$$$\left[{Shaded}\right]=\mathrm{6}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}−\pi=\mathrm{3}\sqrt{\mathrm{3}}−\pi \\ $$

Commented by Spillover last updated on 23/Oct/24