A TV tower has a height of `150 m`. The area of the region covered by the TV broadcast is (Radius of earth `= 6.4 xx 10^(6) m` )

Options

A. `9.6pixx10^(8)km^(2)`
B. `19.2pixx10^(8)km^(2)`
C. `19.2pixx10^(8)km^(2)`
D. `1.92pixx10^(8)km^(2)`

Answer

To find the area of the region covered by the TV broadcast from a tower of height 150 m, we can use the formula for the distance \( d \) from the top of the tower to the horizon, which is given by:

\[ d = \sqrt{2hR} \]

where:
- \( h \) is the height of the tower,
- \( R \) is the radius of the Earth.

### Step 1: Calculate the distance \( d \)

Given:
- Height of the tower \( h = 150 \, \text{m} \)
- Radius of the Earth \( R = 6.4 \times 10^6 \, \text{m} \)

Substituting the values into the formula:

\[ d = \sqrt{2 \times 150 \times 6.4 \times 10^6} \]

### Step 2: Simplify the expression

Calculating \( 2 \times 150 \):

\[ 2 \times 150 = 300 \]

Now substituting back into the equation:

\[ d = \sqrt{300 \times 6.4 \times 10^6} \]

### Step 3: Calculate \( 300 \times 6.4 \)

\[ 300 \times 6.4 = 1920 \]

So now we have:

\[ d = \sqrt{1920 \times 10^6} \]

### Step 4: Calculate the square root

\[ d = \sqrt{1920} \times \sqrt{10^6} \]

Since \( \sqrt{10^6} = 1000 \):

\[ d = \sqrt{1920} \times 1000 \]

### Step 5: Calculate \( \sqrt{1920} \)

To find \( \sqrt{1920} \):

\[ \sqrt{1920} \approx 43.82 \quad (\text{using a calculator}) \]

### Step 6: Calculate \( d \)

Now substituting back:

\[ d \approx 43.82 \times 1000 \approx 43820 \, \text{m} \]

### Step 7: Calculate the area \( A \)

The area \( A \) covered by the broadcast is given by:

\[ A = \pi d^2 \]

Substituting \( d \):

\[ A = \pi (43820)^2 \]

Calculating \( (43820)^2 \):

\[ (43820)^2 \approx 1920000000 \quad (\text{using a calculator}) \]

So,

\[ A \approx \pi \times 1920000000 \]

### Step 8: Convert to square kilometers

To convert from square meters to square kilometers, we divide by \( 10^6 \):

\[ A \approx \frac{\pi \times 1920000000}{10^6} = \pi \times 1920 \, \text{km}^2 \]

### Final Answer

Thus, the area of the region covered by the TV broadcast is approximately:

\[ A \approx 1920\pi \, \text{km}^2 \]