2015 AMC 10B Problem 13

1. Find the corners. The line meets the $x$-axis where $y = 0$: $12x = 60$, so $x = 5$. It meets the $y$-axis where $x = 0$: $5y = 60$, so $y = 12$. The triangle has corners $(0,0)$, $(5,0)$ and $(0,12)$.
2. It is a right triangle. The two axes meet at a right angle, so the legs are $5$ and $12$. The long side is $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$.
3. Two altitudes come free. An altitude is a segment from a corner that meets the opposite side at a right angle. In a right triangle, each leg meets the other leg at a right angle, so the legs $5$ and $12$ are two of the three altitudes.
4. The third altitude. Call the altitude to the long side $h$. Work out the area two ways. Using the legs: $\frac12 \cdot 5 \cdot 12 = 30$. Using the long side as the base: $\frac12 \cdot 13 \cdot h$. So
$$\frac{13h}{2} = 30, \qquad h = \frac{60}{13}.$$
5. Add.
$$5 + 12 + \frac{60}{13} = \frac{221}{13} + \frac{60}{13} = \frac{281}{13}.$$

The answer is $\boxed{\textbf{(E)}\ \frac{281}{13}}$.

A line like $12x + 5y = 60$ always makes a right triangle with the axes, with the right angle at the origin. In a right triangle the legs are also altitudes, and the altitude to the long side comes from the area: leg times leg equals long side times that altitude. Spotting the $5$-$12$-$13$ triangle saves working out the square root.

The altitude to the long side is the distance from the origin to the line $12x + 5y = 60$. The distance formula from a point to a line gives $\dfrac{|12 \cdot 0 + 5 \cdot 0 - 60|}{\sqrt{12^2 + 5^2}} = \dfrac{60}{13}$. Then add $5 + 12$ as before.