Chapter 12.2, Problem 23WE

(a)

To determine

To find: The length of the sides $OM$ , $OA$ and $BC$ .

Answer to Problem 23WE

The length of the sides $OM$ , $OA$ and $BC$ is equal to $3$ , $6$ and $6\sqrt{3}$ , respectively.

Explanation of Solution

Given information:

The length of $h$ is equal to $4$ and the length of $l$ is equal to $5$ in the given pyramid:

  

Calculation:

Use the Pythagoras theorem in $\Delta VOM$ for the length of the side $OM$ .

  $\begin{array}{c}{\left(VM\right)}^2={\left(VO\right)}^2+{\left(OM\right)}^2\\ l^2=h^2+{\left(OM\right)}^2\\ {\left(OM\right)}^2=5^2-4^2\\ {\left(OM\right)}^2=25-16\end{array}$

Further simplify,

  $\begin{array}{c}{\left(OM\right)}^2=25-16\\ OM=\sqrt{9}\\ OM=3\end{array}$

So, the length of the side $OM$ is equal to $3$ .

The length of the side $AO$ is always the double the length of the side $OM$ in an equilateral triangle.

  $\begin{array}{c}AO=2\left(OM\right)\\ =2\left(3\right)\\ =6\end{array}$

So, the length of the side $AO$ is equal to $6$ .

The side $AM$ is the altitude of the equilateral triangle $ABC$ with side $BC$ . The relation between both is given below:

  $\begin{array}{c}AM=\frac{\sqrt{3}}{2}\left(BC\right)\\ AO+OM=\frac{\sqrt{3}}{2}\left(BC\right)\\ 6+3=\frac{\sqrt{3}}{2}\left(BC\right)\\ 9=\frac{\sqrt{3}}{2}\left(BC\right)\end{array}$

Further simplify,

  $\begin{array}{c}9=\frac{\sqrt{3}}{2}\left(BC\right)\\ BC=9\times \frac{2}{\sqrt{3}}\\ BC=6\sqrt{3}\end{array}$

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

Therefore, the length of the sides $OM$ , $OA$ and $BC$ is equal to $3$ , $6$ and $6\sqrt{3}$ respectively.

(b)

To determine

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

Answer to Problem 23WE

The lateral area and volume of the pyramid is equal to $45\sqrt{3}$ and $36\sqrt{3}$ respectively.

Explanation of Solution

Given information:

The length of $h$ is equal to $4$ and the length of $l$ is equal to $5$ in the given pyramid:

  

Calculation:

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

  $\text{A}=\frac{1}{2}pl$

The perimeter of the base is the three times the sum of the length of side $BC$ of equilateral triangle $ABC$ .

  $\begin{array}{c}p=3\left(BC\right)\\ =3\left(6\sqrt{3}\right)\\ =18\sqrt{3}\end{array}$

As found in part(a), the length $l$ is equal to $5$ .

Substitute $5$ for $l$ and $18\sqrt{3}$ for $p$ in the formula for lateral area.

  $\begin{array}{c}\text{A}=\frac{1}{2}pl\\ =\frac{1}{2}\left(18\sqrt{3}\right)\left(5\right)\\ =45\sqrt{3}\end{array}$

So, the lateral area of the pyramid is equal to $45\sqrt{3}$ .

The formula for the volume of the pyramid is equal to the one third of the product of area of base and the length $h$ .

  $\text{V}=\frac{1}{3}\text{B}h$

The base of the pyramid is equilateral triangle with side equal to $6\sqrt{3}$ .

The area of the equilateral triangle is:

  $\begin{array}{c}\text{B}=\frac{\sqrt{3}}{4}{\left(BC\right)}^2\\ =\frac{\sqrt{3}}{4}{\left(6\sqrt{3}\right)}^2\\ =27\sqrt{3}\end{array}$

As found in part(a), the length $h$ is equal to $4$ .

Substitute $4$ for $h$ and $27\sqrt{3}$ for $\text{B}$ in the formula for volume.

  $\begin{array}{c}\text{V}=\frac{1}{3}\text{B}h\\ =\frac{1}{3}\left(27\sqrt{3}\right)\left(4\right)\\ =36\sqrt{3}\end{array}$

So, the volume of the pyramid is equal to $36\sqrt{3}$ .

Therefore, the lateral area and volume of the pyramid is equal to $45\sqrt{3}$ and $36\sqrt{3}$ , respectively.