Chapter 12.4, Problem 21PPS

To determine

Volume of given figure.

Answer to Problem 21PPS

Volume $\left(V\right)$ of given figure is $207.24i{n}^3$

Explanation of Solution

Given:

   Diameter $\left(d_1\right)$ of smaller cone $=6in$

Radius $\left(r_1\right)$ of larger cone $=\frac{\left(d_1\right)}{2}=3in$

Diameter $\left(d_2\right)$ of smaller cone $=6in$

Radius $\left(r_2\right)$ of larger cone $=\frac{\left(d_2\right)}{2}=3in$

Height $\left(h_1\right)$ of smaller cone $=7in$

Height $\left(h_2\right)$ of smaller cone $=15in$

Concept Used:

Volume of cone is calculated as,

  $V=\frac{1}{3}\pi r^{2}h$

Where,

  $V=$ volume of cone.

  $r=$ radius of cone.

  $h=$ height of cone.

Calculation:

Given figure is double cone so volume is calculated as,

  $\begin{array}{l}V=volumeofsmallercone+volumeofl\arg ercone\\ V=\frac{1}{3}\pi r_1{}^2{h}_1+\frac{1}{3}\pi r_2{}^2{h}_2\end{array}$

Put the values in the given equation,

  $\begin{array}{l}V=volumeofsmallercone+volumeofl\arg ercone\\ V=\frac{1}{3}\pi r_1{}^2{h}_1+\frac{1}{3}\pi r_2{}^2{h}_2\\ V=\frac{1}{3}\pi {\left(3\right)}^2\left(7\right)+\frac{1}{3}\pi {\left(3\right)}^2\left(15\right)\\ V=65.94+141.3\\ V=207.24i{n}^3\end{array}$

Conclusion:

Thus, volume $\left(V\right)$ of given figure is $207.24i{n}^3$