Two triangles used in an art project are similar. The sides of the smaller triangle have lengths of 6 in., 7 in., and 12 in. The perimeter of the larger triangle is 75 in. What is the length of the shortest side of the larger triangle?
___ in.
Since the triangles are similar, the ratio of corresponding side lengths is the same. Let's say the ratio is $r$. We can set up the following proportion to find $r$: \begin{align*}
\frac{6}{x} & = \frac{7}{y} = \frac{12}{z} \\
\frac{6}{x} & = r \\
x & = \frac{6}{r} \\
y & = \frac{7}{r} \\
z & =\frac{12}{r}
\end{align*} where $x$, $y$, and $z$ are the side lengths of the smaller triangle, and $\frac{6}{x},\frac{7}{y},\frac{12}{z}$ are the corresponding side lengths of the larger triangle. We don't need to find the value of $r$; we just need to find the value of $\frac{6}{r}$. We can set up another proportion: \begin{align*}
\frac{6}{r} & = \frac{6}{x} \\
r & = \frac{x}{6} \\
75 & = \frac{6}{x} + \frac{7}{y} + \frac{12}{z} = r +\frac{7}{y} + \frac{12}{z} \\
75 & = \frac{x}{6} + \frac{7}{\frac{7}{r}} + \frac{12}{\frac{12}{r}} \\
75 & = \frac{x}{6} + r + r \\
75 & = \frac{6}{x} + \frac{6}{r} + \frac{6}{r} \\
75 & = 1 + 1 + \frac{6}{r} \\
73 & = \frac{6}{r} \\
r & = \frac{6}{73}
\end{align*} Since $r=\frac{6}{73}$, we have the equation $\frac{6}{x} = \frac{6}{73}$. Solving for $x$, we find $x=73$. The shortest side length of the large triangle is $\boxed{73}$ inches.