Geometry problem 10 by Dhaval Furia

Let each side of the triangle be of length $a$ and that of the hexagon be $b$ . Then

$2\times \dfrac {b}{2}\sec 60\degree+b=a\implies b=\dfrac {a}{3}$

Area of the triangle is $\dfrac {\sqrt 3a^2}{4}$ and that of

the hexagon is $6\times \dfrac {\sqrt 3b^2}{4}=\dfrac {\sqrt 3a^2}{6}$

Hence the required ratio is $\dfrac {\sqrt 3a^2}{6}:\dfrac {\sqrt 3a^2}{4}=\boxed {2:3}$ .