A right circular cone is 9.8 cm high with the radius of its base 4 cm. It is melted and recast into a right circular cone with the radius of its base 3.5 cm. Find its height (use π = \(\frac{22}{7}\))
12.8 cm
Understanding Cone Volume and Recasting Problems
This problem involves melting a right circular cone and recasting it into a different right circular cone. The key principle here is that when a solid is melted and reshaped, its volume remains constant. We need to find the height of the new cone using the volume formula for a cone.
Formula for Volume of a Cone
The volume (\(V\)) of a right circular cone with radius (\(r\)) and height (\(h\)) is given by the formula:
\(V = \frac{1}{3} \pi r^2 h\)
Information Given for the First Cone
- Height (\(h_1\)) = 9.8 cm
- Radius (\(r_1\)) = 4 cm
- Using \(\pi = \frac{22}{7}\)
Calculating the Volume of the First Cone
Using the volume formula:
\(V_1 = \frac{1}{3} \times \pi \times r_1^2 \times h_1\)
\(V_1 = \frac{1}{3} \times \frac{22}{7} \times (4 \, \text{cm})^2 \times 9.8 \, \text{cm}\)
\(V_1 = \frac{1}{3} \times \frac{22}{7} \times 16 \, \text{cm}^2 \times 9.8 \, \text{cm}\)
Let's simplify the calculation. \(9.8\) divided by \(7\) is \(1.4\).
\(V_1 = \frac{1}{3} \times 22 \times 16 \times 1.4 \, \text{cm}^3\)
\(V_1 = \frac{22 \times 16 \times 1.4}{3} \, \text{cm}^3\)
\(V_1 = \frac{352 \times 1.4}{3} \, \text{cm}^3\)
\(V_1 = \frac{492.8}{3} \, \text{cm}^3\)
Information Given for the Second Cone
- Radius (\(r_2\)) = 3.5 cm
- Height (\(h_2\)) = ? (what we need to find)
- Using \(\pi = \frac{22}{7}\)
Setting Volumes Equal and Solving for Height
Since the first cone is melted and recast into the second cone, their volumes are equal:
\(V_1 = V_2\)
Using the volume formula for the second cone:
\(V_2 = \frac{1}{3} \pi r_2^2 h_2\)
\(V_2 = \frac{1}{3} \times \frac{22}{7} \times (3.5 \, \text{cm})^2 \times h_2\)
\(V_2 = \frac{1}{3} \times \frac{22}{7} \times (3.5 \times 3.5) \, \text{cm}^2 \times h_2\)
\(V_2 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \, \text{cm}^2 \times h_2\)
Now, set \(V_1 = V_2\):
\(\frac{1}{3} \times \frac{22}{7} \times 16 \times 9.8 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times h_2\)
We can cancel the common terms \(\frac{1}{3}\) and \(\frac{22}{7}\) from both sides:
\(16 \times 9.8 = 12.25 \times h_2\)
Calculate the left side:
\(16 \times 9.8 = 156.8\)
The equation is now:
\(156.8 = 12.25 \times h_2\)
Solve for \(h_2\):
\(h_2 = \frac{156.8}{12.25}\)
To perform the division, it might be easier to remove the decimals. Multiply both numerator and denominator by 100:
\(h_2 = \frac{156.8 \times 100}{12.25 \times 100} = \frac{15680}{1225}\)
Performing the division:
\(h_2 = 12.8\)
So, the height of the new cone is 12.8 cm.
Summary of Calculations
| Item | First Cone | Second Cone |
|---|---|---|
| Height (\(h\)) | 9.8 cm | \(h_2\) |
| Radius (\(r\)) | 4 cm | 3.5 cm |
| Volume (\(V\)) | \(V_1 = \frac{1}{3} \times \frac{22}{7} \times 4^2 \times 9.8\) | \(V_2 = \frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times h_2\) |
| Volume Relation | \(V_1 = V_2\) | |
| Equation | \(16 \times 9.8 = 3.5^2 \times h_2\) | |
| Result | \(h_2 = 12.8\) cm | |
The height of the new cone is 12.8 cm.
Revision Table: Geometry Formulas
| Shape | Volume Formula | Key Variables |
|---|---|---|
| Cone | \(\frac{1}{3} \pi r^2 h\) | \(r\) = radius, \(h\) = height |
| Cylinder | \(\pi r^2 h\) | \(r\) = radius, \(h\) = height |
| Sphere | \(\frac{4}{3} \pi r^3\) | \(r\) = radius |
| Cube | \(s^3\) | \(s\) = side length |
| Cuboid | \(l \times w \times h\) | \(l\) = length, \(w\) = width, \(h\) = height |
Additional Information: Conservation of Volume
The principle of conservation of volume is fundamental in problems involving melting and recasting solids. It states that when a solid is transformed from one shape to another without any loss or addition of material, its total volume remains unchanged. This applies whether a single solid is reshaped or multiple smaller solids are combined to form a larger one, or vice versa.
- This principle assumes the density of the material does not change during the process.
- It is widely used in geometry and mensuration problems.
- Knowing the volumes of different 3D shapes is essential for solving such problems.
Important Questions from Solid Figures
The area of the floor of a cubical room is 192 m 2. The length of the longest rod that can be kept in that room is :
A solid metallic rectangular block of dimensions 112 cm × 44 cm × 25 cm is melted and recast into a cylinder of radius 35 cm. The curved surface area (in cm 2) of the cylinder is: (Take π = 22/7)
If the volume of a cube is 175616 cm 3, what is its side?
The volume of a right circular cone is 1232 cm 3. If the height of the cone is 24 cm, then what will be the radius of its base?
A right triangle contains the right angle between the sides 5 cm and 7 cm. A cone is generated by revolving about the side 5 cm. The volume of this cone is:
