10. As shown in Figure 3, in the right trapezoid $A B C D$, $\angle A B C$ $=\angle B C D=90^{\circ}, A B=$ $B C=10$, point $M$ is on side $B C$, such that $\triangle A D M$ is an equilateral triangle. Then the sum of the areas of $\triangle A B M$ and $\triangle D C M$ is $\qquad$ (require the exact value)
10. $150(2-\sqrt{3})$.

As shown in Figure 6, draw $A E \perp C D$, intersecting the extension of $C D$ at point $E$. Then
$$
\begin{array}{l}
A E=B C \\
=A B=10 .
\end{array}
$$

Since quadrilateral $A B C E$ is a square, and $A D=A M$, we have
$$
\begin{array}{l}
\mathrm{Rt} \triangle A E D \cong \mathrm{Rt} \triangle A B M \\
\Rightarrow E D=B M, C D=C M .
\end{array}
$$

Let $C D=x$. Then $B M=10-x$.
By the Pythagorean theorem, we get
$$
x^{2}+x^{2}=D M^{2}=A M^{2}=10^{2}+(10-x)^{2} \text {. }
$$

Solving for $x$, we get $x=10(\sqrt{3}-1)$.
$$
\begin{array}{l}
\text { Therefore, } S_{\triangle M C D}+S_{\triangle A B M} \\
=\frac{1}{2} x^{2}+\frac{1}{2} 10(10-x) \\
=150(2-\sqrt{3}) .
\end{array}
$$