10. As shown in Figure 2, the diameter $A B=a \mathrm{~cm}$ of the large circle $\odot O$, circles $\odot O_{1}$ and $\odot O_{2}$ are drawn with $O A$ and $O B$ as diameters, respectively, and two equal circles $\odot O_{3}$ and $\odot O_{4}$ are drawn in the space between $\odot O$ and $\odot O_{1}$ and $\odot O_{2}$, such that these circles are internally or externally tangent to each other. Then the area of quadrilateral $O_{1} O_{4} O_{2} O_{3}$ is $\mathrm{cm}^{2}$.
10. $\frac{1}{6} a^{2}$.

Connecting $\mathrm{O}_{3} \mathrm{O}_{4}$ must pass through point $\mathrm{O}$, then $\mathrm{O}_{3} \mathrm{O}_{4} \perp \mathrm{AB}$. Let the radii of $\odot \mathrm{O}_{3}$ and $\odot \mathrm{O}_{4}$ be $x \mathrm{~cm}$. Since the radii of $\odot \mathrm{O}_{1}$ and $\odot \mathrm{O}_{2}$ are $\frac{a}{4} \mathrm{~cm}$, in the right triangle $\triangle O_{3} O_{1} O$, we have $\left(\frac{a}{4}\right)^{2}+\left(\frac{a}{2}-x\right)^{2}=\left(\frac{a}{4}+x\right)^{2}$.
Solving for $x$ gives $x=\frac{a}{6}$. Therefore, $O_{3} O=\frac{a}{2}-\frac{a}{6}=\frac{1}{3} a$.
It is easy to see that quadrilateral $\mathrm{O}_{1} \mathrm{O}_{4} \mathrm{O}_{2} \mathrm{O}_{3}$ is a rhombus, so $S_{\text {shape } O_{1} O_{2} O_{3} O_{4}}=\frac{a}{2} \times \frac{a}{3}=\frac{a^{2}}{6}\left(\mathrm{~cm}^{2}\right)$.