The base of a solid right pyramid is a regular hexagon with a radius of [tex]2x[/tex] units and an apothem of [tex]x \sqrt{3}[/tex] units.

Which expression represents the area of the base of the pyramid?

A. [tex]x^2 \sqrt{3} \, \text{units}^2[/tex]
B. [tex]3x^2 \sqrt{3} \, \text{units}^2[/tex]
C. [tex]4x^2 \sqrt{3} \, \text{units}^2[/tex]
D. [tex]6x^2 \sqrt{3} \, \text{units}^2[/tex]



Answer :

To find the area of the base of the right pyramid, which is a regular hexagon, we need to follow these steps:

1. Identify the given parameters:
- The radius (distance from the center to a vertex) of the hexagon is [tex]\(2x\)[/tex] units.
- The apothem (perpendicular distance from the center to the midpoint of a side) of the hexagon is [tex]\(x \sqrt{3}\)[/tex] units.

2. Determine the side length of the hexagon:
- The side length [tex]\(s\)[/tex] of the hexagon can be derived using the apothem. For a regular hexagon, the relationship between the side length [tex]\(s\)[/tex] and the apothem (a) is given by [tex]\( a = \frac{s \sqrt{3}}{2} \)[/tex].
- Given the apothem is [tex]\(x \sqrt{3}\)[/tex], we can set up the equation:
[tex]\[ x \sqrt{3} = \frac{s \sqrt{3}}{2} \][/tex]
- Solving for [tex]\(s\)[/tex]:
[tex]\[ s \sqrt{3} = 2x \sqrt{3} \][/tex]
[tex]\[ s = 2x \][/tex]

3. Calculate the area of one equilateral triangle:
- The hexagon can be divided into 6 equilateral triangles with side length [tex]\(s = 2x\)[/tex].
- The area [tex]\(A\)[/tex] of one equilateral triangle with side length [tex]\(s\)[/tex] is given by:
[tex]\[ A = \frac{\sqrt{3}}{4} s^2 \][/tex]
- Substituting [tex]\(s = 2x\)[/tex]:
[tex]\[ A = \frac{\sqrt{3}}{4} (2x)^2 \][/tex]
[tex]\[ A = \frac{\sqrt{3}}{4} \cdot 4x^2 \][/tex]
[tex]\[ A = x^2 \sqrt{3} \][/tex]

4. Calculate the area of the hexagon:
- Since the hexagon is composed of 6 equilateral triangles, the total area of the hexagon [tex]\(A_{hex}\)[/tex] is:
[tex]\[ A_{hex} = 6 \times (x^2 \sqrt{3}) \][/tex]
[tex]\[ A_{hex} = 6x^2 \sqrt{3} \][/tex]

Therefore, the expression that represents the area of the base of the pyramid is [tex]\(6 x^2 \sqrt{3}\)[/tex] units[tex]\(^2\)[/tex].