Expressing the Volume of a Pyramid with Variables

The length of the base edge of a pyramid with a regular hexagon base is represented as [tex][tex]$x$[/tex][/tex]. The height of the pyramid is 3 times longer than the base edge.

1. The height of the pyramid can be represented as [tex][tex]$3x$[/tex][/tex].
2. The area of an equilateral triangle with length [tex][tex]$x$[/tex][/tex] is [tex][tex]$\frac{x^2 \sqrt{3}}{4}$[/tex][/tex] square units.
3. The area of the hexagon base is 6 times the area of the equilateral triangle.
4. The volume of the pyramid is [tex][tex]$\frac{3 \sqrt{3}}{2} x^3$[/tex][/tex] cubic units.



Answer :

Let's follow the details step-by-step to determine the required expressions for the volume of the pyramid:

1. Expressing the Height of the Pyramid:
The height of the pyramid [tex]\( h \)[/tex] is given to be 3 times the length of the base edge [tex]\( x \)[/tex]. Therefore,
[tex]\[ h = 3x \][/tex]
Thus, the height of the pyramid can be represented as:
[tex]\[ 3x \][/tex]

2. Area of an Equilateral Triangle:
The area [tex]\( A_{\triangle} \)[/tex] of an equilateral triangle with side length [tex]\( x \)[/tex] is given by:
[tex]\[ A_{\triangle} = \frac{x^2 \sqrt{3}}{4} \][/tex]
Thus, the area of the equilateral triangle with length [tex]\( x \)[/tex] is:
[tex]\[ \frac{x^2 \sqrt{3}}{4} \text{ units}^2 \][/tex]

3. Area of the Hexagon Base:
A regular hexagon can be divided into 6 equilateral triangles. Therefore, the area of the hexagon base [tex]\( A_{\text{hex}} \)[/tex] is 6 times the area of one equilateral triangle:
[tex]\[ A_{\text{hex}} = 6 \times \frac{x^2 \sqrt{3}}{4} = \frac{6x^2 \sqrt{3}}{4} = \frac{3x^2 \sqrt{3}}{2} \][/tex]
Thus, the area of the hexagon base is:
[tex]\[ 3 \times \frac{x^2 \sqrt{3}}{2} \][/tex]
times the area of the equilateral triangle.

4. Volume of the Pyramid:
The volume [tex]\( V \)[/tex] of a pyramid is calculated using the formula:
[tex]\[ V = \frac{1}{3} \times \text{Base area} \times \text{Height} \][/tex]
Substituting the respective values, we have:
[tex]\[ V = \frac{1}{3} \times \frac{3x^2 \sqrt{3}}{2} \times 3x = \frac{1}{3} \times \frac{9x^3 \sqrt{3}}{2} = \frac{9x^3 \sqrt{3}}{6} = \frac{3x^3 \sqrt{3}}{2} = 1.5 \cdot \left( x^3 \sqrt{3} \right) \][/tex]
Thus, the volume of the pyramid is:
[tex]\[ 1.5 \cdot x^3 \sqrt{3} \text{ units}^3 \][/tex]

So, filling in the blanks, the detailed solution is:

- 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]\( \frac{x^2 \sqrt{3}}{4} \text{ units}^2 \)[/tex].
- The area of the hexagon base is [tex]\( 3 \)[/tex] times the area of the equilateral triangle.
- The volume of the pyramid is [tex]\( 1.5 \cdot x^3 \sqrt{3} \text{ units}^3 \)[/tex].