Задачи по геометрии, нужно свериться

Problem 3:

To find the volume of the cone, we'll use the given information about the area of its axial section and the area of its base.

Step 1: Find the radius of the base.

The area of the base of the cone is given by:
[
S{\text{base}} = \pi r^2 = 25\pi.
\]
Solving for \( r \):
\[
\pi r^2 = 25\pi \implies r^2 = 25 \implies r = 5.
\]

**Step 2: Find the height of the cone.**

The axial (central) section of the cone is a triangle with base \( 2r \) and height \( h \). The area of the axial section is:
\[
A{\text{axial}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2r \times h = r h = 30.
\]
Using \( r = 5 \):
[
5h = 30 \implies h = 6.
\]

Step 3: Calculate the volume of the cone.

The volume \( V \) of a cone is:
[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (5)^2 (6) = \frac{1}{3} \pi (25)(6) = 50\pi.
\]

Answer: Problem 3 Answer: The volume of the cone is 50 × π.

---

Problem 4:

To find the volume of the sphere inscribed in the cube, we'll use the volume of the cube to find its side length and then calculate the sphere's volume.

Step 1: Find the side length of the cube.

The volume of the cube is:
[
V{\text{cube}} = s^3 = 24.
\]
Solving for \( s \):
\[
s = \sqrt[3]{24}.
\]

**Step 2: Find the radius of the sphere.**

The diameter of the inscribed sphere is equal to the side length of the cube, so the radius \( r \) is:
\[
r = \frac{s}{2} = \frac{\sqrt[3]{24}}{2}.
\]

**Step 3: Calculate the volume of the sphere.**

The volume \( V{\text{sphere}} \) is:
[
V{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left( \frac{s}{2} \right)^3 = \frac{4}{3} \pi \left( \frac{s^3}{8} \right) = \frac{\pi s^3}{6}.
\]
Substituting \( s^3 = 24 \):
\[
V{\text{sphere}} = \frac{\pi \times 24}{6} = 4\pi.
\]

Answer: Problem 4 Answer: The volume of the sphere is 4 × π.

Problem 3 Answer: The volume of the cone is 50 × π.

Problem 4 Answer: The volume of the sphere is 4 × π.