It's given that PQRSTU is a regular hexagon and O is the center of the hexagon. 

If we fold $$\triangle$$TSR ,$$\triangle$$PQR ,$$\triangle$$TUP along lines TR, PR, PT respectively then vertices S,Q,U will overlap each other exactly at center of the hexagon.

Hence we can say that Area of hexagon PQRSTU =  2($$\triangle$$PRT)

So Area of $$\triangle$$PRT = 0.5(Area of hexagon PQRSTU)

ALTERNATE EXPLANATION:-

Let the length of each side of the hexagon be a. 

Since it is a regular hexagon, $$PT=TR=PR=\sqrt{3}a$$. Thus, $$\triangle PTR$$ is an equilateral triangle, and the area of the triangle will be = $$\dfrac{\sqrt{3}}{4}\left(side\right)^2=\dfrac{\sqrt{3}}{4}\left(\sqrt{3}a\right)^2=\dfrac{3\sqrt{3}}{4}a^2$$.

Area of hexagon PQRSTU = $$\dfrac{3\sqrt{3}}{2}\left(side\right)^2=\dfrac{3\sqrt{3}}{2}a^2$$

Thus, the required ratio will be = $$\dfrac{\frac{3\sqrt{3}}{4}a^2}{\frac{3\sqrt{3}}{2}a^2}$$ = $$1:2$$