Question

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 ( x ). the height of the pyramid is 3 times longer than the base edge.
the height of the pyramid can be represented as (\boxed{quad}) (dropdown).
the (\boxed{quad}) (dropdown) of an equilateral triangle with length ( x ) is (\frac{x^2sqrt{3}}{4}) units(^2).
the area of the hexagon base is (\boxed{quad}) (dropdown) times the area of the equilateral triangle.
the volume of the pyramid is (\boxed{quad}x^3sqrt{3}) units(^3) (dropdown).

Explanation:

Step1: Determine the height of the pyramid

The base edge is \( x \), and the height is 3 times the base edge. So the height \( h = 3x \).

Step2: Identify the term for the formula

The formula \( \frac{x^2\sqrt{3}}{4} \) is the area of an equilateral triangle with side length \( x \). So the missing term is "area".

Step3: Find the number of equilateral triangles in a regular hexagon

A regular hexagon can be divided into 6 equilateral triangles with side length equal to the side length of the hexagon. So the area of the hexagon base is 6 times the area of one equilateral triangle.

Step4: Calculate the volume of the pyramid

The volume of a pyramid is \( V=\frac{1}{3}Bh \), where \( B \) is the base area and \( h \) is the height.
The base area \( B = 6\times\frac{x^2\sqrt{3}}{4}=\frac{3x^2\sqrt{3}}{2} \).
The height \( h = 3x \).
Substitute into the volume formula:
\( V=\frac{1}{3}\times\frac{3x^2\sqrt{3}}{2}\times3x \)
Simplify:
\( V=\frac{1}{3}\times\frac{9x^3\sqrt{3}}{2}=\frac{3x^3\sqrt{3}}{2} \)? Wait, no, wait. Wait, let's re - calculate. Wait, the base area: a regular hexagon with side length \( x \) has area \( 6\times\frac{\sqrt{3}}{4}x^2=\frac{3\sqrt{3}}{2}x^2 \). The height \( h = 3x \). Then volume \( V=\frac{1}{3}\times B\times h=\frac{1}{3}\times\frac{3\sqrt{3}}{2}x^2\times3x \).
First, \( \frac{1}{3}\times\frac{3\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \), then \( \frac{\sqrt{3}}{2}\times3x^3=\frac{3}{2}x^3\sqrt{3} \)? Wait, but the problem says the volume is \( \square x^3\sqrt{3} \). So we need to find the coefficient. Let's do it again:

\( B = 6\times\frac{x^2\sqrt{3}}{4}=\frac{3x^2\sqrt{3}}{2} \)

\( h = 3x \)

\( V=\frac{1}{3}\times B\times h=\frac{1}{3}\times\frac{3x^2\sqrt{3}}{2}\times3x \)

\( \frac{1}{3}\times\frac{3}{2}\times3=\frac{3}{2} \)? Wait, no, \( \frac{1}{3}\times3 = 1 \), then \( 1\times\frac{3}{2}\times x^3\sqrt{3}=\frac{3}{2}x^3\sqrt{3} \)? Wait, maybe I made a mistake. Wait, the area of an equilateral triangle is \( \frac{\sqrt{3}}{4}x^2 \), so 6 of them is \( 6\times\frac{\sqrt{3}}{4}x^2=\frac{3\sqrt{3}}{2}x^2 \). Then volume is \( \frac{1}{3}\times\frac{3\sqrt{3}}{2}x^2\times3x \). The \( \frac{1}{3} \) and 3 cancel out, leaving \( \frac{3\sqrt{3}}{2}x^3 \)? Wait, no, \( \frac{1}{3}\times3 = 1 \), so \( \frac{3\sqrt{3}}{2}x^3 \) is the same as \( \frac{3}{2}x^3\sqrt{3} \). But maybe the problem has a different approach. Wait, let's check the steps again.

Wait, the first blank: height is \( 3x \).

Second blank: "area" (since the formula is for the area of an equilateral triangle).

Third blank: 6 (since a regular hexagon has 6 equilateral triangles as its components).

Fourth blank: Let's recalculate the volume.

Volume of pyramid \( V=\frac{1}{3}Bh \), where \( B \) is base area, \( h \) is height.

\( B = 6\times\frac{x^2\sqrt{3}}{4}=\frac{3x^2\sqrt{3}}{2} \)

\( h = 3x \)

\( V=\frac{1}{3}\times\frac{3x^2\sqrt{3}}{2}\times3x=\frac{1}{3}\times3\times\frac{3x^3\sqrt{3}}{2}=\frac{3x^3\sqrt{3}}{2} \)? Wait, no, \( \frac{1}{3}\times\frac{3x^2\sqrt{3}}{2}=\frac{x^2\sqrt{3}}{2} \), then multiply by \( 3x \): \( \frac{x^2\sqrt{3}}{2}\times3x=\frac{3x^3\sqrt{3}}{2} \). But the problem states the volume is \( \square x^3\sqrt{3} \). So the coefficient is \( \frac{3}{2} \)? Wait, maybe there is a miscalculation. Wait, let's check the area of the regular hexagon again. A regular hexagon with side length \( s \) has area \( \frac{3\sqrt{3}}{2}s^2 \). If \( s = x \), then \( B=\frac{3\sqrt{3}}{2}x^2 \). Height \( h = 3x \). Then \( V=\frac{1}{3}\times\frac{3\sqrt{3}}{2}x^2\times3x=\frac{3\sqrt{3}}{2}x^3 \), which is \( \frac{3}{2}x^3\sqrt{3} \). So the coefficient is \( \frac{3}{2} \).

Answers (filling the blanks):

1. The height of the pyramid can be represented as \( \boldsymbol{3x} \).
2. The \(\boldsymbol{\text{area}}\) of an equilateral triangle with length \( x \) is \( \frac{x^2\sqrt{3}}{4} \) units\(^2\).
3. The area of the hexagon base is \(\boldsymbol{6}\) times the area of the equilateral triangle.
4. The volume of the pyramid is \(\boldsymbol{\frac{3}{2}}\) \( x^3\sqrt{3} \) units\(^3\).

Answer:

s (filling the blanks):

1. The height of the pyramid can be represented as \( \boldsymbol{3x} \).
2. The \(\boldsymbol{\text{area}}\) of an equilateral triangle with length \( x \) is \( \frac{x^2\sqrt{3}}{4} \) units\(^2\).
3. The area of the hexagon base is \(\boldsymbol{6}\) times the area of the equilateral triangle.
4. The volume of the pyramid is \(\boldsymbol{\frac{3}{2}}\) \( x^3\sqrt{3} \) units\(^3\).