## Task B-3.1.

The intersection of a right circular cone by a plane containing its axis of symmetry is a right triangle. Calculate the surface area of the sphere inscribed in the cone if the height of the cone is $2 \mathrm{~cm}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-17.jpg?height=522&width=1402&top_left_y=961&top_left_x=248)

The cross-section of the cone and the inscribed sphere is an isosceles right triangle with an inscribed circle of radius $R$. The generators of the cone are the legs, and the diameter of the base of the cone is the hypotenuse of this right triangle. (sketch)

The radius $R$ of the inscribed circle in a right triangle is calculated using the formula $R=\frac{1}{2}(a+b-c)$, where $a$ and $b$ are the legs, and $c$ is the hypotenuse

(or using the area: $\left.P=R \cdot \frac{1}{2}(a+b+c)=\frac{1}{2} c h\right)$.

Since $h=2 \mathrm{~cm}$, it follows that $a=b=s=2 \sqrt{2} \mathrm{~cm}$ and $c=2 r=4 \mathrm{~cm}$, so

$$
R=\frac{1}{2}(2 \sqrt{2}+2 \sqrt{2}-4)=2(\sqrt{2}-1) \mathrm{cm}
$$

Substituting $R$ into the formula for the surface area of the sphere, $O=4 r^{2} \pi$, we get

$$
O=16 \pi(3-2 \sqrt{2}) \mathrm{cm}^{2}
$$

Note: The student can also find the radius of the sphere using the equality $h=R+R \sqrt{2}$ (2 points), from which $R=\frac{h}{1+\sqrt{2}}=\frac{2}{1+\sqrt{2}}(1$ point).