II. Answer Questions (Total 56 points)
9. (16 points) Given a tetrahedron $P-ABC$ with each edge of the base being 1 and the height being $\sqrt{2}$, the center of its inscribed sphere is $O$, and the radius is $r$. Find the area of the planar region within the base $ABC$ consisting of points that are no more than $2r$ away from point $O$.
9. First, find the radius $r$ of the inscribed sphere.
As shown in Figure 3, let the projections of the sphere center $O$ on the plane $ABC$ and the plane $ABP$ be $H$ and $K$, respectively, and let the midpoint of $AB$ be $M$. Then, $P, K, M$ and $P, O, H$ are collinear, respectively.
Thus,
$$
\begin{array}{l}
\angle PHM = \angle PKO \\
= \frac{\pi}{2},
\end{array}
$$

and $OH = OK = r$,
$$
\begin{array}{c}
PO = PH - OH = \sqrt{2} - r, \\
MH = \frac{\sqrt{3}}{6} AB = \frac{\sqrt{3}}{6}, \\
PM = \sqrt{MH^2 + PH^2} = \frac{5 \sqrt{3}}{6} . \\
\text{Therefore, } \frac{r}{\sqrt{2} - r} = \frac{OK}{PO} = \frac{MH}{PM} = \frac{1}{5} .
\end{array}
$$

Thus, $\frac{r}{\sqrt{2} - r} = \frac{OK}{PO} = \frac{MH}{PM} = \frac{1}{5}$.
Solving for $r$, we get $r = \frac{\sqrt{2}}{6}$.
Let $T$ be any point on the base $ABC$. Then
$$
TO \leqslant 2r \Leftrightarrow TH = \sqrt{TO^2 - r^2} \leqslant \sqrt{3} r = \frac{\sqrt{6}}{6} \text{. }
$$

Construct a circle $\odot H$ with radius $r_1 = \frac{\sqrt{6}}{6}$. The plane region to be considered is the intersection of $\triangle ABC$ and $\odot H$.
As shown in Figure 4, let $\odot H$ intersect $AB$ at points $U, V$, $BC$ at points $W, X$, and $CA$ at points $Y, Z$.
Notice that,
$$
MH = \frac{\sqrt{3}}{6} = \frac{\sqrt{2}}{2} r_1 .
$$

Thus, $\angle UHM = \frac{\pi}{4}$.
Therefore, $\triangle HUV, \triangle HWX, \triangle HYZ$ are all isosceles right triangles with leg length $r_1$, and the regions $HVW, HXY, HZU$ are all sectors with radius $r_1$ and central angle $\frac{\pi}{6}$. Hence, the area $S$ of the plane region is the sum of the areas of these three triangles and three sectors, i.e.,
$$
\begin{array}{l}
S = 3 \times \frac{1}{2} r_1^2 + 3 \times \frac{1}{2} \times \frac{\pi}{6} r_1^2 \\
= \frac{3}{2} \left( \frac{\sqrt{6}}{6} \right)^2 \left( 1 + \frac{\pi}{6} \right) = \frac{1}{4} + \frac{\pi}{24} .
\end{array}
$$