A cone and a hemisphere have equal bases and volumes. What is the ratio of the height of the cone to the radius of the hemisphere?
2 ∶ 1
Solving the Cone and Hemisphere Volume Ratio Problem
This problem asks us to find the ratio of the height of a cone to the radius of a hemisphere, given that they have equal bases and equal volumes.
Understanding the Problem Conditions
- Equal Bases: The base of a cone is a circle, and the base of a hemisphere is also a circle (the great circle). Having equal bases means their base areas are equal. The area of a circle is given by \(\pi r^2\). If the radius of the cone's base is \(r_c\) and the radius of the hemisphere's base is \(r_h\), then \(\pi r_c^2 = \pi r_h^2\). This implies \(r_c^2 = r_h^2\), and since radii are positive, we have \(r_c = r_h\). Let's denote this common radius by \(r\). So, the radius of the cone's base is \(r\), and the radius of the hemisphere is also \(r\).
- Equal Volumes: The volume of the cone is equal to the volume of the hemisphere.
Formulas for Volume
We need the formulas for the volume of a cone and a hemisphere:
- Volume of a cone (\(V_c\)) with radius \(r_c\) and height \(h_c\): \(V_c = \frac{1}{3} \pi r_c^2 h_c\)
- Volume of a hemisphere (\(V_h\)) with radius \(r_h\): \(V_h = \frac{2}{3} \pi r_h^3\)
Setting Up the Equation
We are given that the volumes are equal, \(V_c = V_h\). Using the common radius \(r\) (since \(r_c = r_h = r\)), we can write:
\(\frac{1}{3} \pi r^2 h_c = \frac{2}{3} \pi r^3\)
Solving for the Ratio of Height to Radius
Now, we need to solve this equation for the ratio \(\frac{h_c}{r}\). Let's simplify the equation:
- Cancel out the common factors on both sides. Both sides have \(\frac{1}{3}\pi\) and \(r^2\) (assuming \(r \neq 0\), which is true for shapes with volume).
Dividing both sides by \(\frac{1}{3}\pi r^2\):
\(\frac{\frac{1}{3} \pi r^2 h_c}{\frac{1}{3} \pi r^2} = \frac{\frac{2}{3} \pi r^3}{\frac{1}{3} \pi r^2}\)
\(h_c = \frac{2 r^3}{r^2}\)
\(h_c = 2r\)
The question asks for the ratio of the height of the cone (\(h_c\)) to the radius of the hemisphere. The radius of the hemisphere is \(r_h\), which we established is equal to \(r\). So, we need to find the ratio \(\frac{h_c}{r_h} = \frac{h_c}{r}\).
From our equation \(h_c = 2r\), we can rearrange it to find the ratio:
\(\frac{h_c}{r} = 2\)
This can be written as a ratio \(2:1\).
Conclusion
The ratio of the height of the cone to the radius of the hemisphere, when they have equal bases and equal volumes, is \(2:1\).
| Shape | Radius (Base/Hemisphere) | Height (Cone) | Volume Formula |
|---|---|---|---|
| Cone | \(r_c = r\) | \(h_c\) | \(V_c = \frac{1}{3} \pi r_c^2 h_c = \frac{1}{3} \pi r^2 h_c\) |
| Hemisphere | \(r_h = r\) | N/A | \(V_h = \frac{2}{3} \pi r_h^3 = \frac{2}{3} \pi r^3\) |
Given \(V_c = V_h\):
\(\frac{1}{3} \pi r^2 h_c = \frac{2}{3} \pi r^3\)
\(h_c = 2r\)
Ratio \(\frac{h_c}{r_h} = \frac{h_c}{r} = \frac{2r}{r} = \frac{2}{1}\)
Revision Table: Cone and Hemisphere Properties
| Property | Cone | Hemisphere |
|---|---|---|
| Base Shape | Circle | Circle |
| Base Area (radius \(r\)) | \(\pi r^2\) | \(\pi r^2\) |
| Volume (radius \(r\), cone height \(h\)) | \(\frac{1}{3} \pi r^2 h\) | \(\frac{2}{3} \pi r^3\) |
| Equal Bases Condition | Base radii are equal (\(r_c = r_h\)) | |
| Equal Volumes Condition | \(V_{cone} = V_{hemisphere}\) | |
Additional Information: Solids of Revolution and Volumes
The cone and hemisphere are examples of common three-dimensional geometric shapes. Their volumes are derived using calculus (integration) by considering them as solids of revolution or through other geometric methods.
- A cone can be formed by rotating a right-angled triangle about one of its perpendicular sides (the height).
- A sphere (of which a hemisphere is half) can be formed by rotating a circle about its diameter.
Understanding these fundamental shapes and their volume formulas is crucial for solving many geometry and mensuration problems. The principle of equating volumes or surface areas based on given conditions is a common technique in these types of questions.
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