If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$, in inches, is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$
For rectangle $ABCD$ with perimeter 20, the diagonal $AC$ is given by:
\[AC = \sqrt{l^2 + w^2}\]
To minimize $AC$, $l$ and $w$ should be equal (i.e., the rectangle is a square). Thus, $l = w = 5$.
So, the minimum $AC$ is:
\[AC = \sqrt{5^2 + 5^2} = \boxed{\textbf{(B) } \sqrt{50}}\]
~ proloto