A geometry problem by DEBABRATA MUKHERJEE

Let both's Perimeter be 6a. Now Side of triangle = 2a And side of hexagon = a Area of Eq. Triangle = $\frac{\sqrt{3}}{4} \times a^{2}$ & Area of reg. Hex. = $\frac{3\sqrt{3}}{2} \times a^{2}$ Hence find the ratio

$3S_t = 6S_h$

$S_t = 2S_h$

$A_t = \frac{(2S_h)^2\sin{60}}{2}$

$A_h = 6(\frac{1}{2})(\frac{S_h}{2\sin{30}})^2\sin{60}$

$\frac{A_t}{A_h} = \boxed{\frac{2}{3}}$

i will not solve it but give the most important hint for this problem.......................a regular hexagon can be divided into 6 equilateral triangles....thats it..............