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Question Number 78027 by ajfour last updated on 13/Jan/20

Commented by ajfour last updated on 13/Jan/20

$$Determinetheradiusinc.$$

Answered by MJS last updated on 15/Jan/20

$$\mathrm{easy}!$$$$\mathrm{in}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{the}\mathrm{center}\mathrm{divides}$$$$\mathrm{the}\mathrm{height}2:1\Rightarrow R=2c=2\sqrt{3}$$$$(\mathrm{the}\mathrm{line}\mathrm{with}\mathrm{length}1\mathrm{is}\frac{1}{6}\mathrm{of}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}$$$$\mathrm{triangle})$$

Commented by ajfour last updated on 15/Jan/20

$$which△isequilateralSir?$$

Commented by MJS last updated on 15/Jan/20

$$\mathrm{both}\mathrm{triangles}\mathrm{inscribed}\mathrm{in}\mathrm{the}\mathrm{hexagon}$$

Commented by ajfour last updated on 15/Jan/20

$$radiusisc+x.$$$$x^3-x=2c.Thetrianglescannot$$$$beequilateralforothervalues$$$$exceptacertainone,Sir.$$

Commented by MJS last updated on 15/Jan/20

$$\mathrm{ok}.\mathrm{the}\mathrm{picture}\mathrm{looked}\mathrm{to}\mathrm{me}\mathrm{as}\mathrm{if}\mathrm{it}\mathrm{was}\mathrm{a}$$$$\mathrm{regular}\mathrm{hexagon}...$$

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