The perimeter of a rectangle $A B C D$ is $20 \mathrm{~m}$. The smallest length, in meters, that the diagonal $A C$ can have is:
(a) 0
(b) $\sqrt{50}$
(c) 10
(d) $\sqrt{200}$
(e) $20 \sqrt{5}$
Let $a$ and $b$ denote the lengths of the sides of the rectangle, so $2a + 2b = 20$, $\log a + b = 10$. On the other hand, the square of the length of the diagonal can be calculated using the Pythagorean theorem, thus $d^2 = a^2 + b^2$. Therefore,

$$
\begin{aligned}
2d^2 & = 2a^2 + 2b^2 = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) \\
& = (a + b)^2 + (a - b)^2 \\
& = 100 + (a - b)^2
\end{aligned}
$$

we have that the length of the diagonal is minimized when $a = b$, and in this case $2d^2 = 100$ and $d = \sqrt{50}$. The correct option is (b).