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A cone is inscribed in a regular tetrahedron with edge length s

s. The vertex of the cone is at the centroid of the tetrahedron, and the base of the cone is a circle inscribed in one of the faces of the tetrahedron. The equation of the circle can be written in the form (x−h)2+(y−k)2=r2

(x−h)2

+(y−k)2

=r2

, where (h,k)

(h,k) is the center of the circle and r

r is the radius. Find the volume of the cone in terms of s

s.

To find the volume of the cone inscribed in the regular tetrahedron, we need to break down the problem into a series of steps that incorporate geometric relationships in the tetrahedron.

### Step 1: Understanding the Tetrahedron and Inscribed Circle

Given:
- The edge length of the tetrahedron is \( s \).
- The cone is inscribed in the tetrahedron, with its vertex at the centroid and its base being a circle inscribed in one of the tetrahedron's faces.

#### Geometry of a Regular Tetrahedron:
- A regular tetrahedron has 4 equilateral triangular faces.
- Each triangular face has side lengths \( s \).
- The centroid of the tetrahedron is equidistant from all four vertices and lies inside the tetrahedron.

#### Geometry of the Inscribed Circle:
- The circle inscribed in one triangular face is tangent to all three sides of the equilateral triangle. The center of this circle is the centroid of the triangular face, and the radius \( r \) is the inradius of the triangle.

For an equilateral triangle, the inradius \( r \) is given by the formula:
\[
r = \frac{s \sqrt{3}}{6}
\]
where \( s \) is the side length of the equilateral triangle.

### Step 2: Height of the Cone

The height of the cone is the distance from the centroid of the tetrahedron (the vertex of the cone) to the centroid of the triangular face (the center of the inscribed circle).

For a regular tetrahedron, the distance from the centroid of the tetrahedron to the centroid of one of its faces (the height of the cone) is given by:
\[
h = \frac{s \sqrt{6}}{4}
\]

### Step 3: Volume of the Cone

The volume of a cone is given by the formula:
\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h
\]

Substituting the values of \( r \) and \( h \) derived above:

1. \( r = \frac{s \sqrt{3}}{6} \)
2. \( h = \frac{s \sqrt{6}}{4} \)

Thus, the volume of the cone becomes:
\[
V_{\text{cone}} = \frac{1}{3} \pi \left( \frac{s \sqrt{3}}{6} \right)^2 \times \frac{s \sqrt{6}}{4}
\]

### Step 4: Simplification

First, simplify \( r^2 \):
\[
r^2 = \left( \frac{s \sqrt{3}}{6} \right)^2 = \frac{s^2 \cdot 3}{36} = \frac{s^2}{12}
\]

Now, substitute this into the volume formula:
\[
V_{\text{cone}} = \frac{1}{3} \pi \cdot \frac{s^2}{12} \cdot \frac{s \sqrt{6}}{4}
\]

Simplify the expression:
\[
V_{\text{cone}} = \frac{1}{3} \cdot \frac{\pi s^3 \sqrt{6}}{48}
\]

Thus, the volume of the cone is:
\[
V_{\text{cone}} = \frac{\pi s^3 \sqrt{6}}{144}
\]

### Final Answer:
The volume of the cone in terms of \( s \) is:
\[
\boxed{ \frac{\pi s^3 \sqrt{6}}{144} }
\]

---

Would you like further details or have any questions about the steps?

Here are some related questions:
1. How do you find the centroid of a tetrahedron?
2. What is the significance of the inradius in geometry problems?
3. How do you compute the surface area of a regular tetrahedron?
4. Can the volume of the inscribed cone be generalized for other polyhedra?
5. How do you derive the inradius of an equilateral triangle?

**Tip:** When dealing with 3D geometry, breaking the problem into smaller 2D sections (like triangular faces) often simplifies the solution.

A cone is inscribed in a regular tetrahedron with edge length s. The vertex of the cone is at the centroid of the tetrahedron, and the base of the cone is a circle inscribed in one of the faces of the tetrahedron. Find the volume of the cone in terms of s.

Discover how to calculate the volume of a cone inscribed in a regular tetrahedron with edge length s. This problem covers key geometry concepts, including the inradius of an equilateral triangle and the centroid of a tetrahedron. Suitable for students in Grades 10-12.

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