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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

$${\mathrm{A}}_{\mathrm{seg}}={\mathrm{A}}_{\sec }-{\mathrm{A}}_{\mathrm{tri}}$$$${\mathrm{A}}_{\mathrm{seg}}=\frac{120°}{180°}\pi \cdot {\mathrm{R}}^2\cdot \frac{1}{2}-\frac{\mathrm{R}}{2}\cdot \frac{\mathrm{R}\sqrt{3}}{2}=\frac{{\mathrm{R}}^2\pi }{3}-\frac{{\mathrm{R}}^2\sqrt{3}}{4}$$$$=\frac{{\mathrm{R}}^2(4\pi -3\sqrt{3})}{12}$$$${\mathrm{A}}_{\mathrm{elipse}}=3\pi {\mathrm{R}}^2$$$$\mathrm{Yellow}=3\pi {\mathrm{R}}^2-2[\pi {\mathrm{R}}^2-\frac{{\mathrm{R}}^2(4\pi -3\sqrt{3})}{12}]+[\frac{{\mathrm{R}}^2(4\pi -3\sqrt{3})}{6}-\frac{\pi {\mathrm{R}}^2}{4}]$$$$=3\pi {\mathrm{R}}^2-2\pi {\mathrm{R}}^2+\frac{{\mathrm{R}}^2(4\pi -3\sqrt{3})}{3}-\frac{\pi {\mathrm{R}}^2}{4}$$$$=\frac{3\pi {\mathrm{R}}^2}{4}+\frac{{\mathrm{R}}^2(4\pi -3\sqrt{3})}{3}=\frac{25\pi {\mathrm{R}}^2-{12\mathrm{R}}^2\sqrt{3}}{12}$$$$=\frac{{\mathrm{R}}^2(25\pi -12\sqrt{3})}{12}$$$$\mathrm{if}\mathrm{R}=7,\mathrm{Yellow}=49\times \frac{(25\pi -12\sqrt{3})}{12}\approx 235.{8\mathrm{u}}^2$$

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