2002 AMC 10A Problem 13

Since $15^2 + 20^2 = 225 + 400 = 625 = 25^2$, the triangle is right-angled with legs $15$ and $20$. Its area is $\frac{1}{2} \cdot 15 \cdot 20 = 150$.

For a fixed area, the altitude to a side is $\frac{2 \cdot \text{area}}{\text{side}}$, so the shortest altitude is the one to the longest side, the hypotenuse:
$$h = \frac{2 \cdot 150}{25} = 12.$$

The answer is $\boxed{\textbf{(B)}\ 12}$.

Recognizing a scaled $3$-$4$-$5$ triangle makes the area immediate. Then the single formula $\text{area} = \frac12 bh$, read three ways, converts the area into any altitude; larger base means smaller altitude.

• Answering 15, the shorter leg, forgetting that the altitude to the hypotenuse is shorter than either leg.
• Not noticing the right angle and attempting Heron's formula, which wastes time but still gives area $150$.