5. Square $A B C D$, square $B E F G$, and square $P K R F$ are positioned as shown in Figure 2, with point $G$ on line segment $D K$, and the side length of square $B E F G$ is 2. Then the area of $\triangle D E K$ is $(\quad)$.
(A) 4
(B) 3
(C) 2
(D) $\sqrt{2}$
5. A.

As shown in Figure 8, connect $D B$, $G E$, and $F K$.
Then $D B / / G E / / F K$
$$
\begin{array}{l} 
\Rightarrow S_{\triangle D E K}=S_{\triangle D G E}+S_{\triangle E G K}=S_{\triangle E B G}+S_{\triangle F E G} \\
\quad=S_{\text {square BEFG }}=4 .
\end{array}
$$