All the edges of a regular triangular pyramid are x units long. Find the volume of the pyramid in terms of x.

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

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 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$\frac{\sqrt{2}}{12}{x}^3$ .