If the radii of the circular ends of a truncated conical bucket which is 45 cm high be 28 cm and 7 cm, then the capacity of the bucket (use `pi = (22)/(7)`)

Options

A. 48510
B. 45810
C. 48150
D. 48051

Answer

To find the capacity of the truncated conical bucket, we will use the formula for the volume of a frustum of a cone:

\[ V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) \]

where:
- \( V \) is the volume,
- \( h \) is the height of the frustum,
- \( R \) is the radius of the larger circular end,
- \( r \) is the radius of the smaller circular end,
- \( \pi \) is a constant (given as \( \frac{22}{7} \)).

### Step 1: Identify the values
From the problem:
- Height \( h = 45 \) cm
- Radius of the larger end \( R = 28 \) cm
- Radius of the smaller end \( r = 7 \) cm
- \( \pi = \frac{22}{7} \)

### Step 2: Substitute the values into the formula
Substituting the values into the volume formula:

\[ V = \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^2 + 7^2 + 28 \times 7) \]

### Step 3: Calculate \( R^2 \), \( r^2 \), and \( Rr \)
Calculating each term:
- \( R^2 = 28^2 = 784 \)
- \( r^2 = 7^2 = 49 \)
- \( Rr = 28 \times 7 = 196 \)

Now, add these values together:

\[ R^2 + r^2 + Rr = 784 + 49 + 196 = 1029 \]

### Step 4: Substitute back into the volume formula
Now substitute back into the volume formula:

\[ V = \frac{1}{3} \times \frac{22}{7} \times 45 \times 1029 \]

### Step 5: Simplify the expression
First, calculate \( \frac{1}{3} \times 45 = 15 \):

\[ V = 15 \times \frac{22}{7} \times 1029 \]

### Step 6: Calculate \( 15 \times 22 \)
Calculating \( 15 \times 22 = 330 \):

\[ V = \frac{330 \times 1029}{7} \]

### Step 7: Calculate \( 330 \times 1029 \)
Now, calculate \( 330 \times 1029 = 339570 \):

\[ V = \frac{339570}{7} \]

### Step 8: Divide by 7
Finally, divide \( 339570 \) by \( 7 \):

\[ V = 48510 \text{ cm}^3 \]

### Conclusion
The capacity of the bucket is \( 48510 \text{ cm}^3 \).

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