To find the radius $r$ of a sphere inscribed in a cone, we can use the formula derived from the properties of similar triangles and the relationship between the dimensions of the cone and the sphere.

Given:

• Radius of the cone's base $R = 5$
• Height of the cone $h = 12$

The radius $r$ of the inscribed sphere can be found using the formula:

$r = \frac{R \cdot h}{\sqrt{R^2 + h^2} + h}$

Substituting the given values:

$r = \frac{5 \cdot 12}{\sqrt{5^2 + 12^2} + 12}$

First, calculate the square root term:

$\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Now, substitute this back into the formula:

$r = \frac{5 \cdot 12}{13 + 12} = \frac{60}{25} = 2.4$

Thus, the radius of the inscribed sphere is $r = 2.4$.

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How would the radius of the sphere change if the cone's radius doubled?
2. What is the volume of the inscribed sphere?
3. How does the surface area of the inscribed sphere compare to that of the cone's base?
4. What is the relationship between the cone's slant height and the radius of the sphere?
5. If the height of the cone were doubled, how would that affect the radius of the inscribed sphere?

Tip: For geometric problems involving cones and spheres, visualizing the cross-section can help in understanding the relationships between different elements.