7. Given a regular tetrahedron $P-ABC$ with a base edge length of 1 and a height of $\sqrt{2}$, the radius of the inscribed sphere is $r$. Then, the area of the figure obtained by the intersection of the sphere with the center at the center of the inscribed sphere and a radius of $2r$ with the base $\triangle ABC$ is . $\qquad$
7. $\frac{1}{4}+\frac{\pi}{24}$.

As shown in Figure 2, the slant height of the regular tetrahedron $P-ABC$ is $\frac{5 \sqrt{3}}{6}$. Then $\frac{r}{\frac{\sqrt{3}}{6}}=\frac{\sqrt{2}-r}{\frac{5 \sqrt{3}}{6}} \Rightarrow r=\frac{\sqrt{2}}{6}$.
As shown in Figure 3, according to the problem,
$$
O M=\sqrt{(2 r)^{2}-r^{2}}=\frac{\sqrt{6}}{6} \text {. }
$$

Also, $O D=\frac{\sqrt{3}}{6}$, then
$$
\angle M O N=90^{\circ}, \angle M O F=30^{\circ} \text {. }
$$

Therefore, the area of the required figure is
$$
3 \times \frac{1}{2} \times\left(\frac{\sqrt{6}}{6}\right)^{2}+\frac{\pi}{4} \times\left(\frac{\sqrt{6}}{6}\right)^{2}=\frac{1}{4}+\frac{\pi}{24} \text {. }
$$