Question Number 189465 by normans last updated on 17/Mar/23
Answered by HeferH last updated on 17/Mar/23
Commented by HeferH last updated on 17/Mar/23
![A_(seg) = A_(sec) −A_(tri) A_(seg) = ((120°)/(180°))π∙ R^2 ∙(1/2)− (R/2)∙((R(√3))/2) = ((R^2 π)/3)−((R^2 (√3))/4) = ((R^2 (4π−3(√3)))/(12)) A_(elipse) = 3πR^2 Yellow = 3πR^2 −2[πR^2 −((R^2 (4π−3(√3)))/(12))]+ [((R^2 (4π−3(√3)))/6) −((πR^2 )/4)] = 3πR^2 −2πR^2 + ((R^2 (4π−3(√3)))/3)−((πR^2 )/4) = ((3πR^2 )/4) + ((R^2 (4π−3(√3)))/3) = ((25πR^2 −12R^2 (√3))/(12)) = ((R^2 (25π−12(√3)))/(12)) if R = 7, Yellow = 49 × (((25π−12(√3)))/(12))≈ 235.8u^2](https://www.tinkutara.com/question/Q189480.png)
$$\:\mathrm{A}_{\mathrm{seg}} \:=\:\mathrm{A}_{\mathrm{sec}} −\mathrm{A}_{\mathrm{tri}} \\ $$$$\:\mathrm{A}_{\mathrm{seg}} \:=\:\frac{\mathrm{120}°}{\mathrm{180}°}\pi\centerdot\:\mathrm{R}^{\mathrm{2}} \:\centerdot\frac{\mathrm{1}}{\mathrm{2}}−\:\frac{\mathrm{R}}{\mathrm{2}}\centerdot\frac{\mathrm{R}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\frac{\mathrm{R}^{\mathrm{2}} \pi}{\mathrm{3}}−\frac{\mathrm{R}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{R}^{\mathrm{2}} \left(\mathrm{4}\pi−\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{12}} \\ $$$$\:\mathrm{A}_{\mathrm{elipse}} =\:\mathrm{3}\pi\mathrm{R}^{\mathrm{2}} \\ $$$$\:\mathrm{Yellow}\:=\:\mathrm{3}\pi\mathrm{R}^{\mathrm{2}} −\mathrm{2}\left[\pi\mathrm{R}^{\mathrm{2}} −\frac{\mathrm{R}^{\mathrm{2}} \left(\mathrm{4}\pi−\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{12}}\right]+\:\left[\frac{\mathrm{R}^{\mathrm{2}} \left(\mathrm{4}\pi−\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{6}}\:−\frac{\pi\mathrm{R}^{\mathrm{2}} }{\mathrm{4}}\right] \\ $$$$\:=\:\mathrm{3}\pi\mathrm{R}^{\mathrm{2}} −\mathrm{2}\pi\mathrm{R}^{\mathrm{2}} +\:\frac{\mathrm{R}^{\mathrm{2}} \left(\mathrm{4}\pi−\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{3}}−\frac{\pi\mathrm{R}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:=\:\frac{\mathrm{3}\pi\mathrm{R}^{\mathrm{2}} }{\mathrm{4}}\:+\:\frac{\mathrm{R}^{\mathrm{2}} \left(\mathrm{4}\pi−\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{3}}\:=\:\:\frac{\mathrm{25}\pi\mathrm{R}^{\mathrm{2}} −\mathrm{12R}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{12}}\: \\ $$$$\:=\:\frac{\mathrm{R}^{\mathrm{2}} \left(\mathrm{25}\pi−\mathrm{12}\sqrt{\mathrm{3}}\right)}{\mathrm{12}} \\ $$$$\:\mathrm{if}\:\mathrm{R}\:=\:\mathrm{7},\:\mathrm{Yellow}\:=\:\mathrm{49}\:×\:\frac{\left(\mathrm{25}\pi−\mathrm{12}\sqrt{\mathrm{3}}\right)}{\mathrm{12}}\approx\:\mathrm{235}.\mathrm{8u}^{\mathrm{2}} \\ $$