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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 the length of 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] square units.

The area of the hexagon base is 6 times the area of the equilateral triangle.

The volume of the pyramid is [tex]\(\frac{1}{2} x^3 \sqrt{3}\)[/tex] cubic units.

Sagot :

To solve this question, we'll break it down into smaller parts and understand each step involving geometrical relationships and formulas. Let's begin:

1. Height of the Pyramid:
- The length of the base edge of the pyramid is denoted as [tex]\( x \)[/tex].
- We're given that the height of the pyramid is 3 times the length of the base edge.
- Therefore, 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 side length [tex]\( x \)[/tex] can be calculated using the formula:
[tex]\[ \frac{x^2 \sqrt{3}}{4} \][/tex]
- Thus, the area of an equilateral triangle with side length [tex]\( x \)[/tex] is:
[tex]\[ \frac{x^2 \sqrt{3}}{4} \text{ square units} \][/tex]

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

4. Volume of the Pyramid:
- The volume of a pyramid is calculated using the formula:
[tex]\[ \text{Volume of a pyramid} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \][/tex]
- Plugging in the base area [tex]\( \frac{3 \sqrt{3}}{2}x^2 \)[/tex] and height [tex]\( 3x \)[/tex]:
[tex]\[ \text{Volume} = \frac{1}{3} \times \frac{3 \sqrt{3}}{2}x^2 \times 3x \][/tex]
- Simplifying the equation:
[tex]\[ = \frac{1}{3} \times \frac{9 \sqrt{3}}{2} x^3 \][/tex]
[tex]\[ = \frac{9 \sqrt{3}}{6} x^3 \][/tex]
[tex]\[ = 1.5 \sqrt{3} x^3 \][/tex]
- So, the volume of the pyramid is:
[tex]\[ 1.5 \sqrt{3} x^3 \text{ cubic units} \][/tex]

Thus, the final answers for the blanks are:
- The height of the pyramid can be represented as:
[tex]\[ 3x \][/tex]

- The [tex]\(\square\)[/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]\( 6 \)[/tex] 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].