11.141. The area of the section of a regular tetrahedron that has the shape of a square is $m^{2}$. Find the surface area of the tetrahedron.
Solution.

Let the square $D P Q F$ be the section of the regular tetrahedron $M A B C$ (Fig. 11.37), as mentioned in the problem statement. Then $P Q = P D = m$. Since $P Q$ is parallel to $D F$, the line $P Q$ is parallel to the plane $A B C$. Therefore, the plane $A M C$, passing through the line $P Q$, intersects the plane $A B C$ along the line $A C$, which is parallel to $P Q$. Consequently, $\triangle P M Q \sim \triangle A M C$ and $\triangle P M Q$ is equilateral, with $P M = P Q = m$. Similarly, $P D \parallel B M, A P = P D = m$. Therefore, the edge of the tetrahedron $a = 2 m$. The surface area of the tetrahedron $S = 4 S_{\triangle A B C} = 4 \cdot \frac{a^{2}}{4} \sqrt{3} = a^{2} \sqrt{3} = 4 m^{2} \sqrt{3}$.

Answer: $4 m^{2} \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0850.jpg?height=608&width=508&top_left_y=798&top_left_x=737)
Fig. 11.39

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0851.jpg?height=607&width=569&top_left_y=193&top_left_x=91)

Fig. 11.40

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0851.jpg?height=761&width=570&top_left_y=31&top_left_x=692)