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.
Similar Questions
The volume of a hemisphere is 155232 cm 3. What is the radius of the hemisphere?
The radius and height of a right circular cone are in the ratio 3 : 7. If the volume of the cone is 528 cm 3, then what is the height of the cone? \(\left( {{\rm{Take}}\,\,{\rm{\pi }}\,{\rm{ = }}\frac{{22}}{7}} \right)\)
The length, breadth and height of a cuboid are in the ratio 27 : 8 : 1. The cuboid is melted and recast into a cube. If p is the surface area of the cuboid and q is the surface area of the cube, then what is p/q equal to?
A square sheet of side length 44 cm is rolled along one of its sides to form a cylinder by making opposite edges just to touch each other. What is the volume of the cylinder ? (Take π = 22/7)
A lamp shade is in the shape of a part of a cone and its top and bottom ends are circles whose circumferences are respectively 30 cm and 40 cm. The perpendicular distance between the ends is 6 cm. If the cone were to be completed, then how far would its vertex be from the top end?
Three solid lead spheres of radius 6 cm, 8 cm and 10 cm are melted together and recast as a solid sphere. What is the percentage diminution of the surface area as compared to the sum of the surface areas of the three spheres ?
A solid sphere of radius 3 cm is melted to form a hollow cylinder of height 4 cm and external diameter 10 cm. What is the thickness of the cylinder?
A cylindrical pipe has inner diameter of 14 cm. Water flows through it at a rate of 154 litres per minute. What is the speed of water in km/hr? \(\left( {{\rm{Take}}\,\,{\rm{\pi }}\,{\rm{ = }}\frac{{22}}{7}} \right)\)
What is the radius of the base of the cone ?
A cone of height 16 cm and diameter 14 cm is mounted on a hemisphere of same diameter. What is the volume of the solid thus formed? (take π = 22/7)
Important Questions from Solid Figures
A metallic solid cuboid of dimensions 36 cm × 18 cm × 12 cm is melted and recast in the form of cubes of side 6 cm. Find the number of cubes so formed.
A solid cylinder has a radius of 9 cm and a height of 25 cm. What is the ratio of its total surface area to its curved surface area?
If the surface area of a sphere is 64 π cm 2, then the volume of the sphere is:
Find the surface area of a sphere of diameter 21 cm. (Use π = \(\frac{{22}}{7}\) )
A cuboid of dimension 24 cm, 9 cm and 8 cm is melted and smaller cubes of side 3 cm are formed. How many such cubes can be formed?
