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

To find the outer surface area of a hemispherical tent, we can use the formula for the curved surface area of a hemisphere. The formula is:

\(CSA = 2\pi r^2\)

where \(r\) is the radius of the hemisphere.

In this problem, the given radius \(r\) is \(1.4 \, \text{m}\).

Let's calculate the curved surface area:

\(CSA = 2\pi (1.4)^2\)

First calculate \((1.4)^2\):

\((1.4)^2 = 1.96\)

Substituting back, we get:

\(CSA = 2 \times \pi \times 1.96\)

Using the value of \(\pi \approx 3.14\), the calculation becomes:

\(CSA = 2 \times 3.14 \times 1.96 = 12.2992\)

This is approximately equal to \(12.30 \, \text{m}^2\), which is the total curved surface area, including the door opening.

Since the door opening area is given as \(0.50 \, \text{m}^2\), the net outer surface area of the tent, excluding the door area, will be:

\(Outer \, Surface \, Area = 12.30 - 0.50 = 11.80 \, \text{m}^2\)

However, the closest correct option available is:

\(11.82 \, \text{m}^2\)

Thus, the correct answer is \(11.82 \, \text{m}^2\). This minor difference is likely due to approximation of the value of \(\pi\) or rounding during intermediate steps, which is common in exam scenarios.