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Question Number 113219 by naka3546 last updated on 11/Sep/20

Answered by 1549442205PVT last updated on 11/Sep/20

Commented by Aina Samuel Temidayo last updated on 11/Sep/20

$$\mathrm{Please}\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{construct}\mathrm{these}$$$$\mathrm{things}?$$

Commented by 1549442205PVT last updated on 12/Sep/20

$$\mathrm{It}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}$$$$\stackrel{⌢}{\mathrm{FK}}=\stackrel{⌢}{\mathrm{KL}}=\stackrel{⌢}{\mathrm{LG}}=\stackrel{⌢}{\mathrm{GN}}=\stackrel{⌢}{\mathrm{NJ}}=\stackrel{⌢}{\mathrm{JE}}$$$$=\frac{\pi }{6}\Rightarrow \widehat{\mathrm{FEK}}=15°$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{sector}\mathrm{IKF}\mathrm{is}$$$${\mathrm{S}}_{\mathrm{sectorIKF}}=\frac{\pi {\mathrm{R}}^2}{12}=\frac{49\pi }{12}.\mathrm{The}\mathrm{are}\mathrm{of}\mathrm{the}$$$$\mathrm{equalateral}\mathrm{triangle}\mathrm{IKG}\mathrm{is}$$$${\mathrm{S}}_{\mathrm{\Delta }\mathrm{IKG}}=\frac{{\mathrm{R}}^2\sqrt{3}}{4}=\frac{49\sqrt{3}}{4}.\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{curve}$$$$\mathrm{triangle}\mathrm{DKF}\mathrm{equal}\mathrm{to}{\mathrm{S}}_{\mathrm{DGIF}}-2.{\mathrm{S}}_{\mathrm{sector}}$$$$-{\mathrm{S}}_{\mathrm{\Delta }\mathrm{IKG}}=49-\frac{49\pi }{6}-\frac{49\sqrt{3}}{4}\left(1\right)$$$$.\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{sector}$$$$\mathrm{ILM}\mathrm{is}{\mathrm{S}}_{\mathrm{sectorILM}}=\frac{\pi {\mathrm{R}}^2}{3}=\frac{49\pi }{3}(\mathrm{since}\stackrel{⌢}{\mathrm{LFM}}=120°)$$$${\mathrm{S}}_{\mathrm{\Delta }\mathrm{ILM}}={\mathrm{S}}_{\mathrm{\Delta }\mathrm{IKG}}=\frac{49\sqrt{3}}{4}.\mathrm{Hence},\mathrm{the}\mathrm{area}$$$$\mathrm{of}\mathrm{part}\mathrm{is}\mathrm{limited}\mathrm{by}\mathrm{the}\stackrel{⌢}{\mathrm{LIM}}\mathrm{and}$$$$\mathrm{the}\mathrm{arc}\stackrel{⌢}{\mathrm{LFM}}\mathrm{is}2({\mathrm{S}}_{\mathrm{sectorILM}}-{\mathrm{S}}_{\mathrm{\Delta }\mathrm{ILM}})$$$$=2(\frac{49\pi }{3}-\frac{49\sqrt{3}}{4})\left(2\right)$$$$\mathrm{The}\mathrm{are}\mathrm{of}\mathrm{eight}\mathrm{curve}\mathrm{white}\mathrm{triangle}\mathrm{are}$$$${\mathrm{S}}_1=8(49-\frac{49\pi }{6}-\frac{49\sqrt{3}}{4})\left(3\right).\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{two}$$$$\mathrm{big}\mathrm{white}\mathrm{region}\mathrm{are}{\mathrm{S}}_2=4(\frac{49\pi }{3}-\frac{49\sqrt{3}}{4})$$$$\mathrm{Consequently},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{is}\mathrm{shaded}$$$$\mathrm{region}\mathrm{equal}\mathrm{to}{\mathrm{S}}_{\mathrm{squareABCD}}-({\mathrm{S}}_1+{\mathrm{S}}_2)$$$$={14}^2-8(49-\frac{49\pi }{6}-\frac{49\sqrt{3}}{4})-4(\frac{49\pi }{3}-\frac{49\sqrt{3}}{4})$$$$=3.49\sqrt{3}-196\approx 58.61$$

Commented by 1549442205PVT last updated on 12/Sep/20

$$\mathrm{Using}\mathrm{Geogebra}\mathrm{software}.\mathrm{Drawing}$$$$\mathrm{in}\mathrm{tinkutara}\mathrm{also}\mathrm{did}\mathrm{this}\mathrm{thing}$$

Commented by Aina Samuel Temidayo last updated on 12/Sep/20

$$\mathrm{Ok}.\mathrm{Thanks}.$$

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