Chapter 12.2, Problem 29WE

To determine

To find: The volume of the pyramid in terms of $x$ .

Answer to Problem 29WE

The volume of the pyramid in terms of $x$ is equal to $\frac{\sqrt{2}}{12}{x}^3$ .

Explanation of Solution

Given information:

The length of all the edges of a regular triangular pyramid is equal to $x$ units.

Calculation:

Consider the regular triangular pyramid:

  

The length of each base edge of pyramid is equal. So, the length of the side $BC$ is equal to the length of the side $AB$ equal to $x$ .

The length of the side $BM$ is the half of the length of the side $BC$ as $M$ is the midpoint of the side. So, the length of the side $BM$ is equal to $\frac{x}{2}$ .

Consider the triangle $BOM$ and the angle made by side $BO$ and $OM$ is equal to $30°$ .

  $\begin{array}{c}\cos 30°=\frac{BM}{BO}\\ BO=\frac{BM}{\cos 30°}\\ BO=\frac{\frac{x}{2}}{\frac{\sqrt{3}}{2}}\\ BO=\frac{x}{\sqrt{3}}\end{array}$ .

So, the length of the side $BO$ is equal to $\frac{x}{\sqrt{3}}$ .

Use the Pythagoras theorem in $\Delta BOV$ for the height of pyramid.

  $\begin{array}{c}{\left(VB\right)}^2={\left(VO\right)}^2+{\left(BO\right)}^2\\ {\left(h\right)}^2=x^2-{\left(\frac{x}{\sqrt{3}}\right)}^2\\ h=\sqrt{x^2-\frac{x^2}{3}}\\ h=\sqrt{\frac{2}{3}}x\end{array}$

So, the height $h$ of the pyramid is equal to $\sqrt{\frac{2}{3}}x$ .

The formula for the volume of pyramid is the one third of the product of base area and height.

  $\text{V}=\frac{1}{3}\text{B}h$

Simplify the area of equilateral triangle with side $x$ .

  $\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}{x}^2\end{array}$

Substitute $\frac{\sqrt{3}}{4}{x}^2$ for $\text{B}$ and $\sqrt{\frac{2}{3}}x$ for $h$ in the formula for volume of pyramid.

  $\begin{array}{c}\text{V}=\frac{1}{3}\text{B}h\\ =\frac{1}{3}\left(\frac{\sqrt{3}}{4}{x}^2\right)\left(\sqrt{\frac{2}{3}}x\right)\\ =\frac{\sqrt{2}}{12}{x}^3\end{array}$

Therefore, the volume of the pyramid in terms of $x$ is equal to $\frac{\sqrt{2}}{12}{x}^3$ .