21. As shown in Figure 4, the angles of hexagon $A B C D E F$ are all equal, the areas of square $A B J$ and square $F E H G$ are $18$ and $32$ respectively, $\triangle J B K$ is an equilateral triangle, and $E F = B C$. Then the area of $\triangle K B C$ is ( ).
(A) $6 \sqrt{2}$
(B) 9
(C) 12
(D) $9 \sqrt{2}$
(E) 32
21. C.

From the problem, we have
$$
\begin{array}{l}
B J=B K=\sqrt{18}=3 \sqrt{2}, \\
B C=E F=\sqrt{32}=4 \sqrt{2} .
\end{array}
$$

Notice that all the angles of the hexagon $A B C D E F$ are equal.
Thus, $\angle A B C=\frac{180^{\circ}(6-2)}{6}=120^{\circ}$.
Then, $\angle K B C=360^{\circ}-\angle J B K-\angle J B A-\angle A B C$
$$
=360^{\circ}-90^{\circ}-90^{\circ}-120^{\circ}=90^{\circ} \text {. }
$$

Therefore, $\triangle B K C$ is a right triangle.
Hence, $S_{\triangle K B C}=\frac{1}{2} B K \cdot B C=12$.