Let's solve this step-by-step.
### Height of the Pyramid
The height of the pyramid is given to be 3 times the base edge length. If the base edge length is represented by [tex]\( x \)[/tex], then the height [tex]\( h \)[/tex] can be expressed as:
[tex]\[
h = 3x
\][/tex]
So, the height of the pyramid can be represented as [tex]\( 3x \)[/tex].
### Area of an Equilateral Triangle
The area [tex]\( A \)[/tex] of an equilateral triangle with side length [tex]\( x \)[/tex] is given by the formula:
[tex]\[
A = \frac{x^2 \sqrt{3}}{4}
\][/tex]
Thus, the area of an equilateral triangle with length [tex]\( x \)[/tex] is [tex]\(\frac{x^2 \sqrt{3}}{4}\)[/tex] units[tex]\(^2\)[/tex].
### Area of the Hexagon Base
A regular hexagon can be divided into 6 equilateral triangles. Therefore, the area of the regular hexagon base is 6 times the area of one equilateral triangle. Since the area of one equilateral triangle is [tex]\(\frac{x^2 \sqrt{3}}{4}\)[/tex], the area [tex]\( A_\text{hex} \)[/tex] of the hexagon base is:
[tex]\[
A_\text{hex} = 6 \times \frac{x^2 \sqrt{3}}{4} = \frac{6}{4} x^2 \sqrt{3} = \frac{3}{2} x^2 \sqrt{3}
\][/tex]
So, the area of the hexagon base is [tex]\( 3 \sqrt{3} x^2/2 \)[/tex] units[tex]\(^2\)[/tex].
### Volume of the Pyramid
The volume [tex]\( V \)[/tex] of a pyramid is given by the formula:
[tex]\[
V = \frac{1}{3} \times \text{base area} \times \text{height}
\][/tex]
Substituting the base area [tex]\( A_\text{hex} = \frac{3}{2} x^2 \sqrt{3} \)[/tex] and the height [tex]\( h = 3x \)[/tex]:
[tex]\[
V = \frac{1}{3} \times \frac{3}{2} x^2 \sqrt{3} \times 3x = \frac{3}{2} x^2 \sqrt{3} \times x = \frac{3 \cdot 3}{2} x^3 \sqrt{3} = 3 \cdot \frac{3}{2} x^3 \sqrt{3} = \frac{9}{2} x^3 \sqrt{3} = \frac{18}{4} x^3 \sqrt{3} = 1.5 x^3 \sqrt{3}
\][/tex]
So, the volume of the pyramid is [tex]\( 1.5 x^3 \sqrt{3} \)[/tex] units[tex]\(^3\)[/tex].
Therefore, we fill in the blanks as follows:
The height of the pyramid can be represented as:
[tex]\[ \boxed{3x} \][/tex]
The [tex]\(\boxed{\text{area}}\)[/tex] of an equilateral triangle with length [tex]\( x \)[/tex] is [tex]\(\frac{x^2 \sqrt{3}}{4}\)[/tex] units[tex]\(^2\)[/tex].
The area of the hexagon base is [tex]\(\boxed{6}\)[/tex] times the area of the equilateral triangle.
The volume of the pyramid is [tex]\( \boxed{1.5\, x^3 \sqrt{3}} \)[/tex] units[tex]\(^3\)[/tex].
