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 intersecting the plane of the base $\triangle ABC$ with a sphere centered at the center of the inscribed sphere and with a radius of $2r$ is $\qquad$.
It is known that $r=\frac{\sqrt{2}}{6}$, then the radius of the circle formed by the intersection of the sphere and the base $r^{\prime}=\sqrt{3} r$ $=\frac{\sqrt{6}}{6}$, as shown in the figure. From $O_{1} D=\frac{\sqrt{3}}{6}$, we know that the central angle of the sector $O_{1} E F$ is a right angle, so $S_{\text {sector } O_{1} E F}=\frac{\pi}{24}$.
Thus, $S_{\text {segment } E F}=\frac{\pi}{24}-\frac{1}{12}$. Therefore, the area of the required figure $S=\frac{\pi}{6}-\frac{\pi-2}{8}=\frac{\pi+6}{24}$.