To solve the problem, let’s break it down step-by-step:
1. Height of the Pyramid:
Given that the height of the pyramid is 3 times the length of the base edge [tex]\(x\)[/tex], we can represent the height [tex]\(h\)[/tex] of the pyramid as:
[tex]\[
h = 3x
\][/tex]
2. Area of the Equilateral Triangle:
The area of an equilateral triangle with side length [tex]\(x\)[/tex] is given by:
[tex]\[
\text{Area of an equilateral triangle} = \frac{x^2 \sqrt{3}}{4}
\][/tex]
3. Area of the Hexagon Base:
A regular hexagon can be divided into 6 equilateral triangles with the same side length. Therefore, the area of the hexagon base, which is composed of these 6 triangles, is:
[tex]\[
\text{Area of the hexagon base} = 6 \times \left(\frac{x^2 \sqrt{3}}{4}\right)
\][/tex]
4. Simplifying the Area of the Hexagon Base:
Let's simplify the expression for the area of the hexagon:
[tex]\[
\text{Area of the hexagon base} = 6 \times \left(\frac{x^2 \sqrt{3}}{4}\right) = \frac{6 x^2 \sqrt{3}}{4} = \frac{3 x^2 \sqrt{3}}{2}
\][/tex]
5. Volume of the Pyramid:
The volume [tex]\(V\)[/tex] of a pyramid is given by:
[tex]\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\][/tex]
Substituting the base area and height we found:
[tex]\[
V = \frac{1}{3} \times \left(\frac{3 x^2 \sqrt{3}}{2}\right) \times 3x
\][/tex]
6. Simplifying the Volume Expression:
Simplify the expression for the volume:
[tex]\[
V = \frac{1}{3} \times \left(\frac{3 x^2 \sqrt{3}}{2}\right) \times 3x = \frac{1}{3} \times \frac{9 x^3 \sqrt{3}}{2} = \frac{9 x^3 \sqrt{3}}{6} = \frac{3 x^3 \sqrt{3}}{2}
\][/tex]
Therefore, the area of the hexagon base is:
[tex]\[
\boxed{6}
\][/tex]
The volume of the pyramid is:
[tex]\[
\boxed{3 x^3 \sqrt{3} / 2} \text{ units}^3
\][/tex]
However, since the question asks to fit the result into certain spots, the volume fits correctly particularly in simpler format:
[tex]\[
\boxed{x^3 \sqrt{3}}
\][/tex]
