Let's solve this step-by-step to fill in the blanks with appropriate mathematical expressions and results.
1. Height of the pyramid:
Given that the height of the pyramid is 3 times longer than the base edge [tex]\( x \)[/tex]:
[tex]\[
\text{Height of the pyramid} = 3x
\][/tex]
2. Area of an equilateral triangle with side length [tex]\( x \)[/tex]:
The formula for the area of an equilateral triangle is:
[tex]\[
\text{Area of an equilateral triangle} = \frac{x^2 \sqrt{3}}{4} \text{ units}^2
\][/tex]
3. Area of the hexagon base:
Since the hexagon is composed of 6 equilateral triangles, its area is 6 times the area of one equilateral triangle:
[tex]\[
\text{Area of the hexagon base} = 6 \times \frac{x^2 \sqrt{3}}{4} = \frac{6x^2 \sqrt{3}}{4} = \frac{3x^2 \sqrt{3}}{2}
\][/tex]
4. Volume of the pyramid:
The formula for the volume of a pyramid is:
[tex]\[
\text{Volume of the pyramid} = \frac{1}{3} \times \text{area of the base} \times \text{height}
\][/tex]
Substituting the area of the hexagon and the height of the pyramid:
[tex]\[
\text{Volume of the pyramid} = \frac{1}{3} \times \frac{3x^2 \sqrt{3}}{2} \times 3x = \frac{1}{3} \times 3x^3 \sqrt{3} = x^3 \sqrt{3}
\][/tex]
So, let's summarize and fill in the blanks with these calculations:
Given the length of the base edge of a pyramid with a regular hexagon base is represented as [tex]\( x \)[/tex], the height of the pyramid is 3 times longer than the base edge.
- The height of the pyramid can be represented as [tex]\( 3x \)[/tex].
- The area of an equilateral triangle with side 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 one equilateral triangle.
- The volume of the pyramid is [tex]\( x^3 \sqrt{3} \)[/tex] cubic units.
