5. Given the lengths of the four sides of a trapezoid are $1, 2, 3, 4$. Then the area of this trapezoid is ( ).
(A) 4
(B) 6
(C) $8 \sqrt{2}$
(D) $\frac{10}{3} \sqrt{2}$
5. (D).

Using $1, 2, 3, 4$ as sides to form a trapezoid, there are only six possible combinations:
(1) with 1, 2 as the bases; (2) with 1, 3 as the bases; (3) with 1, 4 as the bases; (4) with 2, 3 as the bases; (5) with 2, 4 as the bases; (6) with 3, 4 as the bases.

It is easy to see that only (3) can form a trapezoid, while the others cannot form a trapezoid.
As shown in Figure 2, in trapezoid
$A B C D$, $\dot{A} B=3, B C$
$=4, C D=2, D A=1$, draw
$A H \perp B C$ at $H$, draw
$A E / / D C$ intersecting $B C$ at $E$,
then $\triangle A B E$ is an isosceles triangle. By
$$
\frac{1}{2} A H \cdot B E=S_{\triangle A B E}=\frac{1}{2} A E \cdot \sqrt{A B^{2}-\left(\frac{A E}{2}\right)^{2}},
$$

we get $A H=\frac{2}{3} \sqrt{3^{2}-1^{2}}=\frac{4 \sqrt{2}}{3}$.
Then $S_{\text {trapezoid } A I C D}=\frac{1}{2}(1+4) \times \frac{4 \sqrt{2}}{3}=\frac{10 \sqrt{2}}{3}$.