95. The base of the pyramid is a square $ABCD$ with side $a$, the lateral edge $SC$ is perpendicular to the base plane and equals $b$. $M$ is a point on the edge $AS$. Points $M, B$, and $D$ lie on the lateral surface of a right circular cone with vertex at point $A$, and point $C$ lies in the base plane of this cone. Determine the lateral surface area of the cone.
95. Let's take a point $K$ on the edge $A S$ such that $|A K|=a$. Then the points $B$; $D$ and $K$ belong to the section of the cone by a plane parallel to the base of the cone ( $|A B|=|A D|=|A K|)$. From the fact that $C$ lies in the plane of the base, it follows that the plane $B D K$ bisects the height of the cone. Thus, the surface area of our cone is four times the surface area of the cone whose base radius is equal to the radius of the circle circumscribed around $\triangle B D K$, with a generatrix equal to $a$.

$$
\text { Ans: } \frac{4 \pi \sqrt{2} a^{2}\left(\sqrt{b^{2}+2 a^{2}}-a\right)}{\sqrt[4]{b^{2}+2 a^{2}} \cdot \sqrt{3 \sqrt{b^{2}+2 a^{2}}-4 a}}
$$