58. The area of a trapezoid formed by line segments of lengths $2,3,4,5$ is $\qquad$
Answer: $\frac{14 \sqrt{5}}{3}$
Solution: Draw the following diagram,
Take two segments from the line segments of lengths $2,3,4,5$ as the legs of the trapezoid, then the difference between the two legs is at least 1, so in $\triangle \mathrm{CDE}$, $\mathrm{DE}>1$, meaning the difference between the upper and lower bases of the trapezoid is at least 2.

Taking 2 and 4 as the bases of the trapezoid, then in $\triangle \mathrm{CDE}$, the lengths of the three sides are 2,3,5, which is not valid.
Similarly, a trapezoid with bases 3 and 5 does not exist.
Therefore, the bases of the trapezoid can only be 2 and 5, then in $\triangle \mathrm{CDE}$, $\mathrm{CE}=\mathrm{DE}=3, \mathrm{CD}=4$, by the Pythagorean theorem, the height on side $\mathrm{CD}$ is $\sqrt{3^{2}-\left(\frac{4}{2}\right)^{2}}=\sqrt{5}$,

Then, by the formula for the area of a triangle, the height of the trapezoid is $\frac{4 \sqrt{5}}{3}$,

So the area of the trapezoid is
$$
(2+5) \times \frac{4 \sqrt{5}}{3} \div 2=\frac{14 \sqrt{5}}{3} \text {. }
$$