The height of the cone is 20, and the radius of the base is 25. Find the area of the section passing through the vertex if its distance from the center of the base of the cone is 12.

#
Let $A$ be the vertex of the cone, $O$ the center of the base, $BC$ a chord where the secant plane intersects the base of the cone, $M$ the midpoint of $BC$, and $OK$ the height of triangle $AOM$. Since $BC \perp OM$ and $BC \perp AO$, $BC$ is perpendicular to the plane $AOM$. Therefore, $AM \perp BC$ and $OK \perp BC$. Thus, the line $OK$ is perpendicular to two intersecting lines $BC$ and $AM$ in the plane of the section. Hence, $OK$ is perpendicular to the plane of the section. Given $OK=12$. Let $\angle OAM = \angle KOM = \alpha$. Then

$$
\begin{gathered}
\sin \alpha = \frac{OK}{AO} = \frac{12}{20} = \frac{3}{5}, \cos \alpha = \frac{4}{5} \\
AM = \frac{AO}{\cos \alpha} = \frac{20}{\frac{4}{5}} = 25, OM = \frac{OK}{\sin \alpha} = \frac{12}{\frac{3}{5}} = 15
\end{gathered}
$$

From the right triangle $BOM$, we find that

$$
BM = \sqrt{OB^2 - OM^2} = \sqrt{25^2 - 15^2} = \sqrt{10 \cdot 40} = 20
$$

Therefore,

$$
S_{\triangle ABC} = \frac{1}{2} BC \cdot AM = BM \cdot AM = 20 \cdot 25 = 500
$$

## Answer

500.00

## Solution

Since the lateral faces of the given pyramid with vertex $D$ form equal dihedral angles with the base plane, and the height $DO$ passes inside the pyramid, the point $O$ is the center of the circle inscribed in the base triangle. Clearly, the circle of the base of the cone of maximum volume with vertex $D$, located inside the pyramid, is the circle inscribed in the triangle $ABC$ of the base. Let $r$ be the radius of this circle, $p$ the semiperimeter of the triangle, $S$ its area, $V1$ the volume of the cone, and $V$ the volume of the pyramid. Then

$$
p = \frac{13 + 14 + 15}{2} = 21
$$

$$
S = \sqrt{p(p-13)(p-14)(p-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 7 \cdot 3 \cdot 4 = 84
$$

$$
r = \frac{S}{p} = \frac{84}{21} = 4, V = \frac{1}{3} S \cdot OD = \frac{1}{3} \cdot 84 \cdot 24 = 84 \cdot 8 = 672
$$

$$
V1 = \frac{1}{3} \pi r^2 \cdot OD = \frac{1}{3} \cdot 16 \cdot 24 \pi = 16 \cdot 8 \pi = 128 \pi
$$

Therefore,

$$
V - V1 = 672 - 128 \pi = 32(21 - 4 \pi)
$$

## Answer

$32(21 - 4 \pi)$.