There is a pyramid on a base which is a regular hexagon of side 2a. If every slant edge of this pyramid is of length `((5a))/(2)` then the volume of this pyramid must be :

Options

A. a. `3a^(3)`
B. b. `3a^(3) sqrt(2) `
C. c. `3a^(3) sqrt(3) `
D. d. `6 a^(3) `

Answer

To find the volume of the pyramid with a regular hexagonal base, we will follow these steps:

### Step 1: Calculate the Area of the Hexagonal Base
The area \( A \) of a regular hexagon with side length \( s \) is given by the formula:
\[ A = \frac{3\sqrt{3}}{2} s^2 \]
In this case, the side length \( s = 2a \). Therefore, the area of the hexagonal base is:
\[ A = \frac{3\sqrt{3}}{2} (2a)^2 = \frac{3\sqrt{3}}{2} \cdot 4a^2 = 6\sqrt{3} a^2 \]

### Step 2: Determine the Height of the Pyramid
We know the slant height \( l \) of the pyramid is given as \( \frac{5a}{2} \). To find the height \( h \) of the pyramid, we can use the Pythagorean theorem. The height, slant height, and half the side length of the base form a right triangle. The half side length is:
\[ \text{Half side length} = \frac{2a}{2} = a \]
Using the Pythagorean theorem:
\[ l^2 = h^2 + a^2 \]
Substituting the values:
\[ \left(\frac{5a}{2}\right)^2 = h^2 + a^2 \]
\[ \frac{25a^2}{4} = h^2 + a^2 \]
\[ h^2 = \frac{25a^2}{4} - a^2 = \frac{25a^2}{4} - \frac{4a^2}{4} = \frac{21a^2}{4} \]
Taking the square root:
\[ h = \sqrt{\frac{21a^2}{4}} = \frac{\sqrt{21}a}{2} \]

### Step 3: Calculate the Volume of the Pyramid
The volume \( V \) of a pyramid is given by the formula:
\[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \]
Substituting the area of the base and the height:
\[ V = \frac{1}{3} \times 6\sqrt{3} a^2 \times \frac{\sqrt{21}a}{2} \]
Calculating this:
\[ V = \frac{1}{3} \times 6\sqrt{3} a^2 \times \frac{\sqrt{21}a}{2} = \frac{6\sqrt{3}\sqrt{21}a^3}{6} = \sqrt{63} a^3 = 3\sqrt{7} a^3 \]

### Final Answer
Thus, the volume of the pyramid is:
\[ V = 3\sqrt{7} a^3 \]