21. In trapezoid $ABCD$, it is known that $AB \parallel CD, AB=33$, $CD=21$, the lengths of the two non-parallel sides are 10 and 14, and $\angle A, \angle B$ are both acute angles. Then the length of the shorter diagonal of the trapezoid is ( ).
(A) $10 \sqrt{6}$
(B) 25
(C) $8 \sqrt{10}$
(D) $18 \sqrt{2}$
(E) 26
21. B.

As shown in Figure 7, draw $DE \parallel BC$, intersecting $AB$ at point $E$.
$$
\begin{array}{l}
\text{Then } DE=10, \\
AE=12, \\
AD=14 .
\end{array}
$$

By the cosine rule,
$$
\cos A=\frac{14^{2}+12^{2}-10^{2}}{2 \times 14 \times 12}=\frac{5}{7} \text{. }
$$

In $\triangle ABD$,
$$
BD^{2}=AB^{2}+AD^{2}-2 AB \cdot AD \cos A=625 \text{. }
$$

Thus, $BD=25$.
Similarly, $AC=\sqrt{1057}>25$.