Let the widescreen width and traditional width be w and W respectively. Then the respective heights are

\frac{9 w}{16}

and

\frac{3 W}{4} .

As the areas are equal:

x \times \frac{9 w}{16} = W \times \frac{2 W}{4}

i.e.

w^{2} = \frac{4}{3} W^{2}

. Hence

w : W = 2 : \sqrt{3}