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 \( \frac{\pi}{3} \) (which is equivalent to 60 degrees).

### Step 2: Area of the Circumcircle
We know the area of the circle circumscribing the hexagon is given as \( 27 \, m^2 \). The formula for the area of a circle is:
\[ \text{Area} = \pi r^2 \]
where \( r \) is the radius of the circle. Setting this equal to \( 27 \):
\[ \pi r^2 = 27 \]
From this, we can solve for \( r^2 \):
\[ r^2 = \frac{27}{\pi} \]
Taking the square root gives us the radius \( r \):
\[ r = \sqrt{\frac{27}{\pi}} = \frac{3\sqrt{3}}{\sqrt{\pi}} \]

### Step 3: Set Up the Triangle
Let \( O \) be the center of the hexagon and \( T \) be the top of the pole. The distance from the center \( O \) to any vertex \( A \) of the hexagon is \( OA = r \). The height of the pole is denoted as \( h = OT \).

### Step 4: Use Trigonometry
In triangle \( OAT \), we have:
\[ \tan\left(\frac{\pi}{3}\right) = \frac{OT}{OA} \]
Substituting the known values:
\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
Thus, we have:
\[ \sqrt{3} = \frac{h}{r} \]
Rearranging gives:
\[ h = r \cdot \sqrt{3} \]

### Step 5: Substitute for \( r \)
Substituting the value of \( r \):
\[ h = \left(\frac{3\sqrt{3}}{\sqrt{\pi}}\right) \cdot \sqrt{3} \]
This simplifies to:
\[ h = \frac{3 \cdot 3}{\sqrt{\pi}} = \frac{9}{\sqrt{\pi}} \]

### Conclusion
The height of the pole is:
\[ \boxed{\frac{9}{\sqrt{\pi}}} \]