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The length of the base edge of a pyramid with a regular hexagon base is represented as \( x \). The height of the pyramid is 3 times longer than the base edge.

The height of the pyramid can be represented as \( 3x \).

The area of an equilateral triangle with length \( x \) is \(\frac{x^2 \sqrt{3}}{4}\) 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 :

Let's break down the problem step-by-step:

1. Height Representation:
- The height of the pyramid is 3 times the length of its base edge, \(x\).
- Therefore, the height of the pyramid can be represented as \(3x\).

2. Area of the Equilateral Triangle:
- The area of an equilateral triangle with side length \(x\) is given by:
[tex]\[ \text{Area}_{\text{equilateral triangle}} = \frac{x^2 \sqrt{3}}{4} \][/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 one equilateral triangle:
[tex]\[ \text{Area}_{\text{hexagon base}} = 6 \times \left(\frac{x^2 \sqrt{3}}{4}\right) = \frac{3 x^2 \sqrt{3}}{2} \][/tex]

4. Volume of the Pyramid:
- The volume of a pyramid is calculated using the formula:
[tex]\[ \text{Volume}_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \][/tex]
- Plugging in the hexagon base area and the height, we get:
[tex]\[ \text{Volume}_{\text{pyramid}} = \frac{1}{3} \times \left(\frac{3 x^2 \sqrt{3}}{2}\right) \times (3x) \][/tex]
- Simplifying the above expression:
[tex]\[ \text{Volume}_{\text{pyramid}} = \frac{3 x^2 \sqrt{3}}{2} \times x = \frac{3}{2} \times x^3 \times \sqrt{3} = 1.5 \cdot x^3 \cdot \sqrt{3} \][/tex]

Now placing the results in the given question format:

The height of the pyramid can be represented as \(3x\).

The area of an equilateral triangle with length \(x\) is \(\frac{x^2 \sqrt{3}}{4}\) units\(^2\).

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

The volume of the pyramid is [tex]\(1.5 \cdot x^3 \cdot \sqrt{3}\)[/tex] units[tex]\(^3\)[/tex].