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'Television Transition' printed from https://nrich.maths.org/
Let the widescreen width and traditional width be $w$ and $W$
respectively. Then the respective heights are $\frac{9w}{16}$ and
$\frac{3W}{4}. $ As the areas are equal: $$x \times \frac{9w}{16} =
W \times \frac{2W}{4}$$ i.e. $$w^2 = \frac {4}{3}W^2$$ Hence $w : W
= 2 : \sqrt{3}$.