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:

w \times \frac{9 w}{16} = W \times \frac{3 W}{4}

i.e.

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

Hence $w$ : $W$ = 2 : $\sqrt{3}$