Chapter 12.2, Problem 12CE

To determine

To find: The volume of the pyramid $V-XYZ$ .

Answer to Problem 12CE

The volume of the pyramid $V-XYZ$ is equal to $18\sqrt{2}$ .

Explanation of Solution

Given information:

Each edge of the pyramid $V-XYZ$ is $6\text{cm}$ .

  

Calculation:

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

  $\text{V}=\frac{1}{3}\text{Bh}$

The base of the pyramid is equilateral triangle with edge of triangle is equal to $6\text{cm}$ .

The formula for the area of the equilateral triangle is,

  $\text{B}=\frac{\sqrt{3}}{4}{a}^2$

Substitute $6$ for $a$ in the formula for area.

  $\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}{\left(6\right)}^2\\ =\frac{\sqrt{3}}{4}\left(36\right)\\ =9\sqrt{3}\end{array}$

Use the Pythagorean theorem in $\Delta VMZ$ for the slant height $l$ of the pyramid.

  $\begin{array}{c}{\left(VZ\right)}^2={\left(VM\right)}^2+{\left(MZ\right)}^2\\ 6^2=l^2+3^2\\ l=\sqrt{36-9}\\ l=3\sqrt{3}\end{array}$

Use the Pythagorean theorem in $\Delta VOM$ for the height $\text{h}$ of the pyramid.

  $\begin{array}{c}{\left(VM\right)}^2={\left(VO\right)}^2+{\left(OM\right)}^2\\ {\left(3\sqrt{3}\right)}^2={\text{h}}^2+{\left(\sqrt{3}\right)}^2\\ \text{h}=\sqrt{27-3}\\ \text{h}=2\sqrt{6}\end{array}$

Substitute $2\sqrt{6}$ for $\text{h}$ and $9\sqrt{3}$ for $\text{B}$ in the formula for volume of pyramid.

  $\begin{array}{c}\text{V}=\frac{1}{3}\text{Bh}\\ =\frac{1}{3}\left(9\sqrt{3}\right)\left(2\sqrt{6}\right)\\ =3\sqrt{3}\times 2\sqrt{6}\\ =18\sqrt{2}\end{array}$

Therefore, the volume of the pyramid $V-XYZ$ is equal to $18\sqrt{2}$ .