Chapter 12.2, Problem 25WE

To determine

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

Answer to Problem 25WE

The lateral area and volume of the pyramid is equal to $144$ and $24\sqrt{39}$ , respectively.

Explanation of Solution

Given information:

The length of $VA$ is equal to $10$ and length of $AB$ is equal to $12$ in the given pyramid:

  

Calculation:

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 $12$ .

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 $6$ .

The length of each lateral side of pyramid is also equal. So, the length of the side $VB$ is equal to $10$ .

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

  $\begin{array}{c}{\left(VB\right)}^2={\left(BM\right)}^2+{\left(VM\right)}^2\\ {10}^2={\left(6\right)}^2+l^2\\ l=\sqrt{100-36}\\ l=8\end{array}$

So, the slant height of the pyramid is equal to $8$ .

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.

  $\begin{array}{c}AM=\frac{\sqrt{3}}{2}\left(BC\right)\\ =\frac{\sqrt{3}}{2}\left(12\right)\\ =6\sqrt{3}\end{array}$

So, the length of side $AM$ is equal to $6\sqrt{3}$ .

The length of the side $OM$ is one third of the length of side $AM$ . The length of the side $OM$ is equal to $2\sqrt{3}$ .

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

  $\begin{array}{c}{\left(VM\right)}^2={\left(VO\right)}^2+{\left(OM\right)}^2\\ 8^2=h^2+{\left(2\sqrt{3}\right)}^2\\ h=\sqrt{64-12}\\ h=\sqrt{52}\end{array}$

So, the height $h$ of the pyramid is equal to $\sqrt{52}$ .

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 $12$ . So, the perimeter of the equilateral triangle is equal to $3\left(12\right)=36$ .

Substitute $36$ for $p$ and $8$ for $l$ in the formula for lateral area.

  $\begin{array}{c}\text{A}=\frac{1}{2}pl\\ =\frac{1}{2}\left(36\right)\left(8\right)\\ =144\end{array}$

So, the lateral area of the pyramid is equal to $144$ .

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 $12$ .

  $\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{a}^2\\ =\frac{\sqrt{3}}{4}{\left(12\right)}^2\\ =\frac{\sqrt{3}}{4}\left(144\right)\\ =36\sqrt{3}\end{array}$

Substitute $36\sqrt{3}$ for $\text{B}$ and $\sqrt{52}$ for $h$ in the formula for volume of pyramid.

  $\begin{array}{c}\text{V}=\frac{1}{3}\text{B}h\\ =\frac{1}{3}\left(36\sqrt{3}\right)\left(\sqrt{52}\right)\\ =12\sqrt{3}\times 2\sqrt{13}\\ =24\sqrt{39}\end{array}$

So, the volume of the pyramid is equal to $24\sqrt{39}$ .

Therefore, the lateral area and volume of the pyramid is equal to $144$ and $24\sqrt{39}$ respectively.