10. In a square lawn with side length $a$, there are sprinkler devices installed at each of the four corners. The sprinkler devices can rotate to spray water at a $90^{\circ}$ angle, and each can spray water from one side of its corner to the other side, with an effective range of $a$. Then the proportion of the lawn that can be simultaneously covered by the four sprinkler devices is $\qquad$
10. $\frac{\pi}{3}+1-\sqrt{3}$.

As shown in Figure 4, we need to find the ratio of the area of the curved quadrilateral $E F G H$ to the area of the square $A B C D$.
Draw $E K \perp C D$ through point $E$.
By symmetry, we know $B E=C E$.
Thus, $\triangle B C E$ is an equilateral triangle.
Therefore, $\angle D C E=\angle B C F=\angle E C F=\frac{\pi}{6}$
Then, $S_{\text {circular segment } E F}=\frac{1}{2} a^{2} \cdot \frac{\pi}{6}-\frac{1}{2} a^{2} \sin \frac{\pi}{6}$
$$
=\left(\frac{\pi}{12}-\frac{1}{4}\right) a^{2} \text {, }
$$
$$
\begin{array}{l}
S_{\text {curved quadrilateral } E F G H}=4 S_{\text {circular segment } E F}+S_{\text {square } E F G H} \\
=\left(\frac{\pi}{3}-1\right) a^{2}+\left(2 a \sin \frac{\pi}{12}\right)^{2} \\
=\left(\frac{\pi}{3}+1-\sqrt{3}\right) a^{2} .
\end{array}
$$

Therefore, the required ratio is $\frac{\pi}{3}+1-\sqrt{3}$.