The ratio of the width to the height of a rectangular screen is $3: 2$. If the length of a diagonal of the screen is $65 \mathrm{~cm}$, what is the area of the screen, in $\mathrm{cm}^{2}$ ?
Since the ratio of the width to the height is $3: 2$, we let the width be $3 x \mathrm{~cm}$ and the height be $2 x \mathrm{~cm}$, for some $x>0$.

In terms of $x$, the area of the screen is $(3 x \mathrm{~cm})(2 x \mathrm{~cm})=6 x^{2} \mathrm{~cm}^{2}$.

Since the diagonal has length $65 \mathrm{~cm}$, then by the Pythagorean Theorem, we obtain $(3 x)^{2}+(2 x)^{2}=65^{2}$.

Simplifying, we obtain $9 x^{2}+4 x^{2}=65^{2}$ or $13 x^{2}=65 \cdot 65$.

Thus, $x^{2}=5 \cdot 65$ and so $6 x^{2}=6 \cdot 5 \cdot 65=30 \cdot 65=1950$.

Therefore, the area of the screen is $1950 \mathrm{~cm}^{2}$.

ANSWER: $1950 \mathrm{~cm}^{2}$