The perimeter of a rectangle $A B C D$ measures $20 \mathrm{~m}$. The smallest length that the diagonal $A C$ can have, in meters, is:
(a) 0 ;
(b) $\sqrt{50}$;
(c) 10 ;
(d) $\sqrt{200}$;
(e) $20 \sqrt{5}$.
The correct option is (b).

Denoting by $a$ and $b$ the lengths of the sides of the rectangle, we have $2a + 2b = 20$, so $a + b = 10$. The square of the length of the diagonal, given by the Pythagorean Theorem, is $d^2 = a^2 + b^2$. But,

$$
(a+b)^2 + (a-b)^2 = \left(a^2 + 2ab + b^2\right) + \left(a^2 - 2ab + b^2\right) = 2a^2 + 2b^2 = 2d^2
$$

and $(a+b)^2 = 100$, therefore $d^2 = 50 - \frac{1}{2}(a+b)^2$. Thus, we see that the minimum length of the diagonal occurs when $a = b$, in which case $d = \sqrt{50}$.