A cone is inscribed in a regular quadrilateral pyramid. Find the ratio of the area of the total surface of the cone to the area of its lateral surface, if the side of the base of the pyramid is 4, and the angle between the height of the pyramid and the plane of the lateral face is $30^{\circ}$.
Let the circle of the base of a cone inscribed in a given regular quadrilateral pyramid $P A B C D$ with vertex $P$ touch the side $A B$ of the base $A B C D$ at point $M, O$ - the center of the base. Then $M$ is the midpoint of $A B, P M$ is the apothem of the pyramid, $M R O$ is the angle between the height $P O$ of the pyramid and the plane of the lateral face $A P B$. According to the problem, $\angle M P O=30^{\circ}$. Therefore,

$$
P M=2 \cdot O M=2 \cdot A D / 2=2 \cdot 2=4
$$

If $r$ is the radius of the base of the cone, $l$ is the slant height, $S_{1}$ is the lateral surface area of the cone, and $S_{2}$ is the total surface area of the cone, then

$$
\begin{gathered}
S_{1}=\pi r l=\pi \cdot O M \cdot P M=\pi \cdot 2 \cdot 4=8 \pi \\
S_{2}=\pi r^{2}+\pi r l=\pi \cdot O M^{2}+\pi \cdot O M \cdot P M=4 \pi+8 \pi=12 \pi
\end{gathered}
$$

Therefore, $S_{2} / S_{1}=12 \pi /(8 \pi)=3 / 2$.

## Answer

$$
\text { 1.5. }
$$

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