4. In a convex quadrilateral $ABCD$, $E, F, G, H$ are the midpoints of $AB, BC, CD, DA$ respectively, and $EG$ intersects $FH$ at point $O$. Let the areas of quadrilaterals $AEOH, BFOE, CGOF$ be $3, 4, 5$ respectively. Then the area of quadrilateral $DHOG$ is ( ).
(A) $\frac{\sqrt{15}}{2}$
(B) $\frac{15}{4}$
(C) 4
(D) 6
4.C.

As shown in Figure 3, connect $O A, O B, O C, O D$. Then
$$
\begin{array}{l}
S_{\triangle A E O}=S_{\triangle B E O}, \\
S_{\triangle B F O}=S_{\triangle C F O}, \\
S_{\triangle C O O}=S_{\triangle D C O}, \\
S_{\triangle D H O}=S_{\triangle A H O} .
\end{array}
$$

Therefore, $S_{\text {quadrilateral }}$ EOH $+S_{\text {quadrilateral }}$ OCC
$$
=S_{\text {quadrilateral } B F O E}+S_{\text {quadrilateralDHOC }} \text {. }
$$

Thus, $S_{\text {quadrilateralDHOc }}=3+5-4=4$.