Chapter 12.2, Problem 14CE

To determine

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

Answer to Problem 14CE

The lateral area of the pyramid $V-XYZ$ is equal to $27\sqrt{3}$ .

Explanation of Solution

Given information:

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

  

Calculation:

The lateral area of the pyramid is the product of number of faces and the area of the each face.

  $\text{L}\text{.A}\text{.}=n\left(\text{B}\right)$

The area of each face is equal to the area of the base.

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.

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

Substitute $9\sqrt{3}$ for $\text{B}$ and $3$ for $n$ in the formula for lateral area.

  $\begin{array}{c}\text{L}\text{.A}\text{.}=n\left(\text{B}\right)\\ =3\left(9\sqrt{3}\right)\\ =27\sqrt{3}\end{array}$

Therefore, the lateral area of the pyramid $V-XYZ$ is equal to $27\sqrt{3}$ .