Chapter 12.2, Problem 24WE

To determine

To find: The slant height, lateral area and volume of the pyramid.

Answer to Problem 24WE

The slant height, lateral area and volume of the pyramid is equal to $\sqrt{13}$ , $6\sqrt{39}$ and $12\sqrt{3}$ , respectively.

Explanation of Solution

Given information:

The length of $VA$ is equal to $5$ and length of $h$ is equal to $3$ in the given pyramid:

  

Calculation:

Use the Pythagoras theorem in $\Delta AOV$ for the length of side $AO$ .

  $\begin{array}{c}{\left(AV\right)}^2={\left(AO\right)}^2+{\left(VO\right)}^2\\ 5^2={\left(AO\right)}^2+3^2\\ AO=\sqrt{25-9}\\ AO=4\end{array}$

So, the length of the side $AO$ is equal to $4$ .

The length of the side $OM$ is half the length of the side $AO$ . So, the length of the side $OM$ is equal to $2$ .

The length of the total side $AM$ is the sum of the lengths of sides $AO$ and $OM$ . So, the length of the side $AM$ is equal to $4+2=6$ .

As $AM$ is the altitude of the equilateral triangle and the relationship between the side of equilateral triangle and altitude is given below simplify it for the length of side of equilateral triangle.

  $\begin{array}{c}BC=\frac{2}{\sqrt{3}}\left(AM\right)\\ =\frac{2}{\sqrt{3}}\left(6\right)\\ =4\sqrt{3}\end{array}$

So, the length of each side of equilateral triangle is equal to $4\sqrt{3}$ .

The length of the side $MC$ is the half of the length of the side $BC$ as $M$ is the midpoint of the side. So, the length of the side $MC$ is equal to $2\sqrt{3}$ .

The length of each lateral edge of pyramid is equal. So, the length of side $VC$ is also equal to $5$ .

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

  $\begin{array}{c}{\left(VC\right)}^2={\left(VM\right)}^2+{\left(MC\right)}^2\\ 5^2=l^2+{\left(2\sqrt{3}\right)}^2\\ l=\sqrt{25-12}\\ l=\sqrt{13}\end{array}$

So, the slant height $l$ of the pyramid is equal to $\sqrt{13}$ .

The formula for the lateral area of the pyramid is half the product of perimeter of the base and slant height.

  $\text{A}=\frac{1}{2}pl$

The length of each side of equilateral triangle is equal to $2\sqrt{3}$ . So, the perimeter of the equilateral triangle is equal to $3\left(4\sqrt{3}\right)=12\sqrt{3}$ .

Substitute $6\sqrt{3}$ for $p$ and $\sqrt{13}$ for $l$ in the formula for lateral area.

  $\begin{array}{c}\text{A}=\frac{1}{2}pl\\ =\frac{1}{2}\left(12\sqrt{3}\right)\left(\sqrt{13}\right)\\ =6\sqrt{39}\end{array}$

So, the lateral area of the pyramid is equal to $6\sqrt{39}$ .

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 $4\sqrt{3}$ .

  $\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}{\left(4\sqrt{3}\right)}^2\\ =\frac{\sqrt{3}}{4}\left(48\right)\\ =12\sqrt{3}\end{array}$

Substitute $12\sqrt{3}$ for $\text{B}$ and $3$ for $h$ in the formula for volume of pyramid.

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

So, the volume of the pyramid is equal to $12\sqrt{3}$ .

Therefore, the slant height, lateral area and volume of the pyramid is equal to $\sqrt{13}$ , $6\sqrt{39}$ and $12\sqrt{3}$ , respectively.