The outer and inner surfaces areas of a hemispherical bowl are `1152 picm^(2) and 648 pi cm^(2)` , respectively .What is the total surface area of the bowl ?

Options

A. `2052 pi cm^(2)`
B. `2362 pi cm^(2)`
C. `2325 pi cm^(2)`
D. `2000 pi cm^(2)`

Answer

To find the total surface area of the hemispherical bowl, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Information:**
- Outer surface area of the hemispherical bowl: \( 1152 \pi \, \text{cm}^2 \)
- Inner surface area of the hemispherical bowl: \( 648 \pi \, \text{cm}^2 \)

2. **Use the Formula for Surface Area of a Hemisphere:**
The surface area \( A \) of a hemisphere (including the base) is given by:
\[ A = 2\pi r^2 + \pi r^2 = 3\pi r^2 \]
where \( r \) is the radius of the hemisphere.

3. **Calculate the Radius of the Inner Hemisphere:**
For the inner hemisphere, we have:
\[ 3\pi r^2 = 648\pi \]
Dividing both sides by \( \pi \):
\[ 3r^2 = 648 \]
Dividing by 3:
\[ r^2 = 216 \]
Taking the square root:
\[ r = \sqrt{216} = 14.7 \, \text{cm} \, (\text{approximately}) \]

4. **Calculate the Radius of the Outer Hemisphere:**
For the outer hemisphere, we have:
\[ 3\pi R^2 = 1152\pi \]
Dividing both sides by \( \pi \):
\[ 3R^2 = 1152 \]
Dividing by 3:
\[ R^2 = 384 \]
Taking the square root:
\[ R = \sqrt{384} = 19.6 \, \text{cm} \, (\text{approximately}) \]

5. **Calculate the Area of Thickness:**
The area of thickness is the difference between the outer and inner surface areas:
\[ \text{Area of thickness} = \text{Outer surface area} - \text{Inner surface area} \]
\[ = 1152\pi - 648\pi = 504\pi \, \text{cm}^2 \]

6. **Calculate the Total Surface Area:**
The total surface area of the bowl is the sum of the inner surface area, outer surface area, and the area of thickness:
\[ \text{Total Surface Area} = \text{Inner Surface Area} + \text{Outer Surface Area} + \text{Area of Thickness} \]
\[ = 648\pi + 1152\pi + 504\pi = 2304\pi \, \text{cm}^2 \]

### Final Answer:
The total surface area of the hemispherical bowl is \( 2304\pi \, \text{cm}^2 \).