Chapter 12, Problem 21RP

To determine

To calculate: The volume of a frustum of a cone.

Answer to Problem 21RP

The volume of a frustum of a cone is $1811.986{\text{ unit}}^{\text{3}}$ .

Explanation of Solution

Given information:

Given frustum of a cone has upper base’s radius $\left(r\right)=9\text{ units}$ , lower base’s radius $\left(R\right)=12\text{ units}$ and slant height $\left(l\right)=6\text{ units}$ . $$

Formula used:

Volume of frustum of a cone $(V)=\left(\frac{1}{3}\right)\pi h\left(r^2+rR+R^2\right)$ ………… (i)

where r = radius of upper base of the frustum of a cone

R = radius of lower base of the frustum of a cone

h = height of the frustum of a cone

Slant height of frustum of a cone $\left(l\right)=\sqrt{{\left(R-r\right)}^2+h^2}$ …………… (ii)

Calculation:

For calculating the volume of frustum of a cone we have the values for upper and lower base’s radius but the height of the frustum is not given.

So, let us calculate its height by using equation (ii) as slant height is given in the problem

  $l=\sqrt{{\left(R-r\right)}^2+h^2}$

Squaring both sides,

  $\begin{array}{l}{l}^2={\left(R-r\right)}^2+h^2\\ h^2=l^2-{\left(R-r\right)}^2\end{array}$

Taking square root of the terms

  $h=\sqrt{l^2-{\left(R-r\right)}^2}$ ………………………………………………… (iii)

Substituting the given values $r=9\text{ units}$ , $R=12\text{ units}$ and $l=6\text{ units}$ in equation (iii) to get,

  $\begin{array}{l}h=\sqrt{6^2-{(12-9)}^2}\\ h=\sqrt{6^2-3^2}\\ h=\sqrt{36-9}\\ h=\sqrt{27}\\ h=3\sqrt{3}\text{ units}\end{array}$

Now using the given values $r=9\text{ units}$ , $R=12\text{ units}$ and the calculated value $h=3\sqrt{3}\text{ units}$ in equation (i) to calculate the volume of frustum of a cone

  $\begin{array}{l}V=\left(\frac{1}{3}\right)\pi h\left(r^2+rR+R^2\right)\\ V=\left(\frac{1}{3}\right)\times \pi \times 3\sqrt{3}\times \left(9^2+9\times 12+{12}^2\right)\\ V=\left(\frac{1}{3}\right)\times \pi \times 3\sqrt{3}\times \left(81+108+144\right)\\ V=\left(\frac{1}{3}\right)\times \pi \times 3\sqrt{3}\times 333\\ V=1811.986{\text{ unit}}^{\text{3}}\end{array}$

Hence, the volume of the given frustum of a cone is $1811.986{\text{ unit}}^{\text{3}}$ .