10. In quadrilateral $A B C D$, $A B=B C=C D=$ $26, A D=30 \sqrt{3}, A C$ intersects $B D$ at point $O, \angle A O B=$ $60^{\circ}$. Then $S_{\text {quadrilateral } A B C D}=$ $\qquad$
10. $506 \sqrt{3}$.

Let $O A=a, O B=b, O C=c, O D=d$. Then
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} O A \cdot O B \sin 60^{\circ}=\frac{\sqrt{3}}{4} a b, \\
S_{\triangle O B C}=\frac{1}{2} O B \cdot O C \sin 120^{\circ}=\frac{\sqrt{3}}{4} b c, \\
S_{\triangle O C D}=\frac{1}{2} O C \cdot O D \sin 60^{\circ}=\frac{\sqrt{3}}{4} c d, \\
S_{\triangle O D A}=\frac{1}{2} O D \cdot O A \sin 120^{\circ}=\frac{\sqrt{3}}{4} d a .
\end{array}
$$

By the cosine rule, we have
$$
\begin{array}{l}
A B^{2}=a^{2}+b^{2}-a b, B C^{2}=b^{2}+c^{2}+b c, \\
C D^{2}=c^{2}+d^{2}-c d, D A^{2}=d^{2}+a^{2}+d a . \\
\text { Hence } B C^{2}+D A^{2}-C D^{2}-A B^{2} \\
=a b+b c+c d+d a=2024 \\
\Rightarrow S_{\text {quadrilateral } A B C D}=\frac{\sqrt{3}}{4}(a b+b c+c d+d a) \\
=506 \sqrt{3} .
\end{array}
$$