A pole is situated at the centre of a regular hexagonal park. The angle of elevation of the top of the vertical pole when observed from each vertex of the hexagon is `(pi)/(3)`. If the area of the circle circumscribing the hexagon is `27m^(2)`, then the height of the tower is

Options

A. `3sqrt((3)/(pi))m`
B. `(3)/(sqrt(pi))m`
C. `sqrt((3)/(pi))m`
D. `(9)/(sqrt(pi))m`

Answer

To find the height of the pole situated at the center of a regular hexagonal park, we can follow these steps:

### Step 1: Understand the Geometry
We have a regular hexagon with a pole at its center. The angle of elevation from each vertex of the hexagon to the top of the pole is given as \( \frac{\pi}{3} \).

### Step 2: Area of the Circumscribing Circle
The area of the circle that circumscribes the hexagon is given as \( 27 \, m^2 \). The formula for the area of a circle is:
\[ \text{Area} = \pi r^2 \]
Setting this equal to \( 27 \):
\[ \pi r^2 = 27 \]

### Step 3: Solve for the Radius
Rearranging the equation to find \( r^2 \):
\[ r^2 = \frac{27}{\pi} \]
Taking the square root to find \( r \):
\[ r = \sqrt{\frac{27}{\pi}} = \frac{3\sqrt{3}}{\sqrt{\pi}} \, m \]

### Step 4: Relate the Radius to the Side of the Hexagon
For a regular hexagon, the radius \( r \) of the circumscribing circle is equal to the length of the side \( a \):
\[ r = a \]
Thus, we have:
\[ a = \frac{3\sqrt{3}}{\sqrt{\pi}} \, m \]

### Step 5: Set Up the Right Triangle
Consider the right triangle formed by the pole (height \( OT \)), the radius \( OA \) (which is equal to \( a \)), and the angle of elevation \( \frac{\pi}{3} \). In triangle \( AOT \):
\[ \tan\left(\frac{\pi}{3}\right) = \frac{OT}{OA} \]
We know that \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{OT}{OA} \]

### Step 6: Solve for the Height of the Pole
Substituting \( OA = a \):
\[ OT = OA \cdot \sqrt{3} \]
Substituting the value of \( OA \):
\[ OT = \left(\frac{3\sqrt{3}}{\sqrt{\pi}}\right) \cdot \sqrt{3} \]
This simplifies to:
\[ OT = \frac{3 \cdot 3}{\sqrt{\pi}} = \frac{9}{\sqrt{\pi}} \, m \]

### Conclusion
The height of the pole is:
\[ OT = \frac{9}{\sqrt{\pi}} \, m \]