The base of the pyramid is a square $ABCD$ with a side length of 1.5, the lateral edge $SC$ is perpendicular to the base plane and equals 1.75. Points $S, B$, and $D$ lie on the lateral surface of a cone with the vertex at point $A$, and point $C$ lies in the base plane of this cone. Find the lateral surface area of the cone.
Let rays $A B, A D$ and $A S$ intersect the circumference of the base of the cone at points $B 1, D 1$ and $S 1$ respectively. Then segments $A B 1, A D 1$ and $A S 1$ are the generators of the cone. The plane of the base of the cone intersects the plane $A B C D$ along the line $B 1 D 1$, and since point $C$, lying in the plane $A B C D$, also lies in the plane of the base of the cone, this point lies on the segment $B 1 D 1$. Moreover, since $A B 1 = A D 1$ and $A B = A D$, the line $B 1 D 1$ is parallel to the diagonal $B D$ of the square $A B C D$. The opposite sides of the quadrilateral $B D C B 1$ are pairwise parallel, so this quadrilateral is a parallelogram, hence $B B 1 = C D = A B = \frac{\frac{3}{2}}{2}$ and $A B 1 = 3$. Similarly, $A D 1 = 3$. Then $B 1 D 1 = 3 \sqrt{2}$. From the right triangle $B C S$ we find that

$$
B S = \sqrt{B C^{2} + S C^{2}} = \sqrt{\left(\frac{3}{2}\right)^{2} + \left(\frac{7}{4}\right)^{2}} = \frac{\sqrt{85}}{4}
$$

Let $\angle S A B = \alpha, \angle S 1 B 1 D 1 = \angle S 1 D 1 B 1 = \beta$. From the theorem of three perpendiculars, it follows that triangle $A B S$ is a right triangle, $\angle A B S = 90^{\circ}$. Then

$$
\operatorname{tg} \alpha = \frac{B S}{A B} = \frac{\frac{\sqrt{93}}{1}}{\frac{3}{2}} = \frac{\sqrt{85}}{6}, \cos \alpha = \frac{1}{\sqrt{1 + \lg^{2} 2_{2}}} = \frac{1}{\sqrt{1 + \frac{9}{3}}} = \frac{6}{11}
$$

By the cosine theorem

$$
B 1 S 1 = \sqrt{A B_{1}^{2} + A S_{1}^{2} - 2 A B_{1} \cdot A S_{1} \cos \alpha} = \sqrt{3^{2} + 3^{2} - 2 \cdot 3 \cdot 3 \cdot \frac{6}{11}} = 3 \sqrt{\frac{10}{11}}
$$

From the right triangle $B 1 C S 1$ we find that

$$
\cos \beta = \frac{C B_{1}}{B_{1} S_{1}} = \frac{\frac{3 \sqrt{2}}{2}}{3 \sqrt{\frac{10}{11}}} = \frac{\sqrt{11}}{2 \sqrt{5}}
$$

then

$$
\sin \beta = \sqrt{1 - \cos^{2} \beta} = \sqrt{1 - \left(\frac{\sqrt{11}}{2 \sqrt{5}}\right)^{2}} = \frac{3}{2 \sqrt{5}}
$$

Let $R$ be the radius of the circle circumscribed around the isosceles triangle $B 1 S 1 D 1$. Points $B 1, S 1$ and $D 1$ lie on the circumference of the base of the cone, so $R$ is the radius of the base of the cone. By the sine theorem

$$
R = \frac{B_{1} S_{1}}{2 \sin \beta} = \frac{3 \sqrt{\frac{10}{10}}}{2 \cdot \frac{3}{2 \sqrt{\beta}}} = \frac{5 \sqrt{2}}{\sqrt{11}}
$$

Let $S_{\text {bok. }}$ be the area of the lateral surface of the cone. Then

$$
S_{\text {bok. }} = \pi R \cdot A B 1 = \pi \cdot \sqrt{\frac{5 \sqrt{2}}{11}} \cdot 3 = \frac{30 \pi}{\sqrt{22}}
$$

## Answer

$30 \pi$

$\sqrt{22}$