5. As shown in Figure 1, given that the length of the rectangle is 8 and the width is 4, fold the rectangle along one of its diagonals and press it flat. Then the area of the overlapping part (the shaded triangle) is ( ).
(A) 10
(B) 12
(C) 14
(D) 16
5.A.

As shown in Figure 6, draw $E F \perp B D$ at $F$. It is easy to see that point $F$ is the midpoint of $B D$.

Also, points $A, B, F, E$ are concyclic, so
$$
D E \cdot D A=D F \cdot D B,
$$

which means $D E \cdot 8=2 \sqrt{5} \times 4 \sqrt{5} \Rightarrow D E=5$.
Thus, $S_{\triangle B D E}=\frac{1}{2} D E \cdot A B=10$.