Question

A camping tent in hemispherical shape of radius \(1.4\) m, has a door opening of area \(0.50\) \(\text{m}^2\). Outer surface area of the tent is

• A. \(11.78\) \(\text{m}^2\)
• B. \(12.32\) \(\text{m}^2\)
• C. \(11.82\) \(\text{m}^2\)
• D. \(12.86\) \(\text{m}^2\)

Hint:

Remember to use \(\pi = 22/7\) when radius is a multiple of \(7\) (like \(1.4\)) as it simplifies the calculations significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The outer surface area of a hemispherical tent is its Curved Surface Area (CSA). However, since there is a door opening, that specific area must be subtracted from the total CSA to find the actual outer surface area of the material used.
Step 2: Key Formula or Approach:
\[ \text{Net Outer Surface Area} = (\text{CSA of Hemisphere}) - (\text{Area of Door}) \]
\[ \text{CSA of Hemisphere} = 2 \pi r^2 \]
Step 3: Detailed Explanation:
Given:
Radius \(r = 1.4\) m
Area of door = \(0.50\) \(\text{m}^2\)
Using \(\pi = \frac{22}{7}\):
\[ \text{CSA} = 2 \times \frac{22}{7} \times (1.4) \times (1.4) \]
\[ \text{CSA} = 44 \times 0.2 \times 1.4 \]
\[ \text{CSA} = 8.8 \times 1.4 = 12.32 \text{ m}^2 \]
Now, Net Surface Area:
\[ \text{Area} = 12.32 - 0.50 = 11.82 \text{ m}^2 \]
Step 4: Final Answer:
The outer surface area of the tent is \(11.82\) \(\text{m}^2\).