Solving the Cone and Hemisphere Volume Ratio Problem

This problem asks us to find the ratio of the radius of a right circular cone to the radius of a hemisphere, given that their volumes are equal and there's a specific relationship between the cone's radius and its altitude.

Understanding the Formulas

First, let's recall the formulas for the volume of a right circular cone and a hemisphere:

• Volume of a right circular cone: \(V_{\text{cone}} = \frac{1}{3} \pi r_c^2 h_c\)
• Volume of a hemisphere: \(V_{\text{hemi}} = \frac{2}{3} \pi r_h^3\)

where \(r_c\) is the radius of the cone's base, \(h_c\) is the altitude (height) of the cone, and \(r_h\) is the radius of the hemisphere.

Using the Given Information

We are given two key pieces of information:

1. The radius of the cone's base is half of its altitude. This can be written as \(r_c = \frac{1}{2} h_c\). This relationship can also be expressed as \(h_c = 2 r_c\).
2. The volume of the cone and the volume of the hemisphere are equal: \(V_{\text{cone}} = V_{\text{hemi}}\).

Setting up the Equation

We need to substitute the given relationship for the cone's dimensions into the cone's volume formula. Since \(h_c = 2 r_c\), we substitute \(2 r_c\) for \(h_c\) in the cone volume formula:

\[V_{\text{cone}} = \frac{1}{3} \pi r_c^2 (2 r_c)\]

Simplifying this expression gives:

\[V_{\text{cone}} = \frac{2}{3} \pi r_c^3\]

Now, we use the second piece of information: \(V_{\text{cone}} = V_{\text{hemi}}\).

Substitute the formulas for the volumes:

\[\frac{2}{3} \pi r_c^3 = \frac{2}{3} \pi r_h^3\]

Solving for the Ratio

We need to solve this equation to find the ratio \(r_c : r_h\). We can cancel out the common terms on both sides of the equation.

\[\frac{2}{3} \pi r_c^3 = \frac{2}{3} \pi r_h^3\]

Divide both sides by \(\frac{2}{3} \pi\):

\[\frac{\frac{2}{3} \pi r_c^3}{\frac{2}{3} \pi} = \frac{\frac{2}{3} \pi r_h^3}{\frac{2}{3} \pi}\]

This simplifies to:

\[r_c^3 = r_h^3\]

To find the relationship between \(r_c\) and \(r_h\), we take the cube root of both sides:

\[\sqrt[3]{r_c^3} = \sqrt[3]{r_h^3}\]

This gives us:

\[r_c = r_h\]

The problem asks for the ratio of the radius of the cone to the radius of the hemisphere, which is \(r_c : r_h\).

Since \(r_c = r_h\), their ratio is:

\[\frac{r_c}{r_h} = \frac{1}{1}\]

So, the ratio is \(1:1\).

Summary of Steps

1. Write down the volume formulas for a cone and a hemisphere.
2. Use the given relationship \(r_c = \frac{1}{2} h_c\) to express the cone's height in terms of its radius (\(h_c = 2 r_c\)).
3. Substitute the expression for \(h_c\) into the cone volume formula.
4. Equate the volume of the cone and the hemisphere.
5. Solve the resulting equation for the ratio \(r_c : r_h\).

Conclusion

By setting the volumes of the cone and the hemisphere equal and using the relationship between the cone's radius and altitude, we found that the radius of the cone is equal to the radius of the hemisphere. Therefore, the ratio of their radii is \(1:1\).

Revision Table: Cone and Hemisphere Volumes

Additional Information: Solid Geometry Volumes

Calculating volumes of different 3D shapes is a fundamental part of solid geometry. Understanding the basic formulas is crucial. Let's look at a few more related concepts:

• Cylinder Volume: The volume of a cylinder is given by \(V = \pi r^2 h\), where \(r\) is the base radius and \(h\) is the height. Notice the cone volume is exactly one-third of a cylinder with the same base radius and height.
• Pyramid Volume: The volume of a pyramid is \(V = \frac{1}{3} \times (\text{Area of Base}) \times h\), where \(h\) is the height. A cone is essentially a pyramid with a circular base.
• Ratio Problems: Many geometry problems involve ratios. The key is often to set up an equation based on volumes, surface areas, or other properties and then simplify to find the required ratio. Always ensure the units are consistent when dealing with volume or area calculations.
• Units: Volume is measured in cubic units (e.g., \(cm^3\), \(m^3\), \(in^3\)). Radii and altitudes are measured in linear units (e.g., cm, m, in). The \(\pi\) constant is dimensionless.

Practicing problems involving equal volumes or surface areas of different shapes helps solidify the understanding of these formulas and relationships.