Question Number 78027 by ajfour last updated on 13/Jan/20
Commented by ajfour last updated on 13/Jan/20
$${Determine}\:{the}\:{radius}\:{in}\:{c}. \\ $$
Answered by MJS last updated on 15/Jan/20
$$\mathrm{easy}! \\ $$$$\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{center}\:\mathrm{divides} \\ $$$$\mathrm{the}\:\mathrm{height}\:\mathrm{2}:\mathrm{1}\:\Rightarrow\:{R}=\mathrm{2}{c}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{the}\:\mathrm{line}\:\mathrm{with}\:\mathrm{length}\:\mathrm{1}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{triangle}\right) \\ $$
Commented by ajfour last updated on 15/Jan/20
$${which}\:\bigtriangleup\:{is}\:{equilateral}\:{Sir}? \\ $$
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
$${radius}\:{is}\:{c}+{x}. \\ $$$${x}^{\mathrm{3}} −{x}=\mathrm{2}{c}.\:\:{The}\:{triangles}\:{cannot} \\ $$$${be}\:{equilateral}\:{for}\:{other}\:{values} \\ $$$${except}\:{a}\:{certain}\:{one},\:{Sir}. \\ $$
Commented by MJS last updated on 15/Jan/20
$$\mathrm{ok}.\:\mathrm{the}\:\mathrm{picture}\:\mathrm{looked}\:\mathrm{to}\:\mathrm{me}\:\mathrm{as}\:\mathrm{if}\:\mathrm{it}\:\mathrm{was}\:\mathrm{a} \\ $$$$\mathrm{regular}\:\mathrm{hexagon}... \\ $$