The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is: 
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$
To begin, let's notice that the inscribed circle of the right triangle is its [incircle](https://artofproblemsolving.com/wiki/index.php/Incircle), and that the radius of the incircle is the right triangle's [inradius](https://artofproblemsolving.com/wiki/index.php/Inradius). In this case, the hypotenuse is 10, and the inradius is 1. The formula for the inradius of a right triangle is $r = (a+b-c)/2$, where $r$ is the length of the inradius of the triangle, $c$ is the length hyptotenuse the the right triangle, and $a$ and $b$ are the lengths of the legs of the right triangle. From this, we can plug in values to notice that:
\begin{align*} r & = \frac{a+b-c}{2}\\\\ 1 & = \frac{a+b-10}{2}\\\\ 2 & = a+b-10\\\\ 12 &= a+b\\\\ \end{align*}
From this, we arrive at $a+b+c = 12+10 = 22$. The answer is clearly $\boxed{\textbf{(B)}}$.