The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find its volume. [Use `pi` = 3.14]

Answer

To find the volume of a frustum of a cone, we can use the formula:

\[ V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2) \]

where:
- \( V \) is the volume,
- \( h \) is the height of the frustum,
- \( r_1 \) is the radius of the lower base,
- \( r_2 \) is the radius of the upper base,
- \( \pi \) is a constant approximately equal to 3.14.

### Step-by-Step Solution

1. **Identify the diameters and calculate the radii:**
- The diameter of the lower end is 10 cm, so the radius \( r_1 \) is:
\[ r_1 = \frac{10}{2} = 5 \text{ cm} \]
- The diameter of the upper end is 30 cm, so the radius \( r_2 \) is:
\[ r_2 = \frac{30}{2} = 15 \text{ cm} \]

2. **Identify the height of the frustum:**
- The height \( h \) is given as 24 cm.

3. **Substitute the values into the volume formula:**
\[ V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2) \]
Substituting the known values:
\[ V = \frac{1}{3} \times 3.14 \times 24 \times (5^2 + 15^2 + 5 \times 15) \]

4. **Calculate \( r_1^2 \), \( r_2^2 \), and \( r_1 r_2 \):**
- \( r_1^2 = 5^2 = 25 \)
- \( r_2^2 = 15^2 = 225 \)
- \( r_1 r_2 = 5 \times 15 = 75 \)

5. **Add these values together:**
\[ r_1^2 + r_2^2 + r_1 r_2 = 25 + 225 + 75 = 325 \]

6. **Substitute back into the volume formula:**
\[ V = \frac{1}{3} \times 3.14 \times 24 \times 325 \]

7. **Calculate \( \frac{1}{3} \times 3.14 \):**
\[ \frac{1}{3} \times 3.14 \approx 1.047 \]

8. **Multiply by the height and the sum calculated:**
\[ V \approx 1.047 \times 24 \times 325 \]

9. **Calculate \( 1.047 \times 24 \):**
\[ 1.047 \times 24 \approx 25.128 \]

10. **Finally, multiply by 325:**
\[ V \approx 25.128 \times 325 \approx 8164 \text{ cm}^3 \]

### Final Answer
The volume of the bucket in the form of a frustum of a cone is approximately \( 8164 \text{ cm}^3 \).