Solution:
$$
135 / 28
$$

Note that the sides of $S$ are parallel to the sides of $T$, so $S$ is a triangle similar to $T$. The semiperimeter of $T$ is $s=\frac{1}{2}(26+51+73)=75$. By Heron's formula, the area of $T$ is $\sqrt{75 \cdot 49 \cdot 24 \cdot 2}=420$. If $r$ is the inradius of $T$, then the area of $T$ is $r s$, so $r=420 / 75=28 / 5$. It follows that the inradius of $S$ is $r-5=3 / 5$, and the ratio of similitude between $S$ and $T$ is $3 / 28$. Therefore, the area of $S$ is $420 \cdot(3 / 28)^{2}=135 / 28$.
8. Let $T$ be a triangle with side lengths 26,51 , and 73 . Let $S$ be the set of points inside $T$ which do not lie within a distance of 5 of any side of $T$. Find the area of $S$.