Given the information, let's fill in the blanks step-by-step:
1. Height of the Pyramid:
The height of the pyramid is 3 times the length of the base edge. Given the base edge length is [tex]\( x \)[/tex], the height of the pyramid can be represented as [tex]\( 3x \)[/tex].
2. Area of an Equilateral Triangle:
The area of an equilateral triangle with edge length [tex]\( x \)[/tex] is given by:
[tex]\[
\text{Area} = \frac{x^2 \sqrt{3}}{4} \text{ units}^2.
\][/tex]
3. Area of the Hexagon Base:
A regular hexagon can be divided into 6 equilateral triangles. Thus, the area of the hexagon base is 6 times the area of the equilateral triangle. Therefore:
[tex]\[
\text{Area of Hexagon} = 6 \times \frac{x^2 \sqrt{3}}{4} = \frac{3 x^2 \sqrt{3}}{2}.
\][/tex]
4. 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 the height, we get:
[tex]\[
\text{Volume} = \frac{1}{3} \times \frac{3 x^2 \sqrt{3}}{2} \times 3x = \frac{9 x^3 \sqrt{3}}{6} = 1.5 \sqrt{3} x^3 \text{ units}^3.
\][/tex]
Now, we can fill in the blanks with the derived values:
- The height of the pyramid can be represented as [tex]\( 3x \)[/tex].
- The area 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 6 times the area of the equilateral triangle.
- The volume of the pyramid is [tex]\( 1.5 \sqrt{3} x^3 \)[/tex] units[tex]\(^3\)[/tex].
So the final filled statements would be:
1. The height of the pyramid can be represented as [tex]\( 3x \)[/tex].
2. The area of an equilateral triangle with length [tex]\( x \)[/tex] is [tex]\( \frac{x^2 \sqrt{3}}{4} \)[/tex] units[tex]\(^2\)[/tex].
3. The area of the hexagon base is 6 times the area of the equilateral triangle.
4. The volume of the pyramid is [tex]\( 1.5 \sqrt{3} x^3 \)[/tex] units[tex]\(^3\)[/tex].
