4. The base of a quadrilateral pyramid is a square $ABCD$, with a side length of 2, and the lateral edge $SA$ is perpendicular to the base plane and also equals 2. A plane is drawn through the lateral edge $SC$ and a point on the side $AB$, such that the resulting cross-section of the pyramid has the smallest perimeter. Find the area of the cross-section.
Answer: $\sqrt{6}$.

Solution. The length of the edge $S C=\sqrt{2^{2}+(2 \sqrt{2})^{2}}=2 \sqrt{3}$. Let the section be made through point $M$ on side $A B$. To minimize the perimeter of the section, it is necessary that the sum $S M+M C$ be minimized. Consider the unfolding of faces $S A B$ and $A B C D$ in the form of trapezoid $S A D C$ (point $A$ lies on the base $S D$). On it, $S M C$ is a broken line, the sum $S M+M C$ is its length, which is minimal if point $M$ lies on the line $S C$. Then $M$ is the midpoint of edge $A B$. Segments $S M$ and $M C$ are equal to $\sqrt{2^{2}+1^{2}}=\sqrt{5}$. The height of the isosceles triangle $S M C$, dropped to the base $S C$, is equal to $\sqrt{(\sqrt{5})^{2}-\left(\frac{2 \sqrt{3}}{2}\right)^{2}}=\sqrt{2}$, and the area is equal to $\frac{1}{2} \cdot 2 \sqrt{3} \cdot \sqrt{2}=\sqrt{6}$.

Comment. Other methods of solution are possible. For example, the function can be studied using the derivative. If the section with the smallest perimeter is not found - 0 points. If the section with the smallest perimeter is indicated (without proof), and then its area is correctly found - 2 points. If the function was studied, the root of the derivative was correctly found, but it was not proven that the root is the point of the minimum value, and then the area of the section was correctly found - 3 points. For algebraic errors in transformations with correct logic of the solution, the score is reduced by 1-2 points.