Chapter 12.6, Problem 16PSC

To determine

To compare: The volumes of a hemisphere and a cone with congruent bases and equal heights.

Answer to Problem 16PSC

The volume of a hemisphere is 2 times the volume of a cone.

Explanation of Solution

Given Information:

Bases of the hemisphere and the cone are congruent i.e. $d_1=d_2$

  $r_1=h$

Where, $d_1$ is the diameter of the hemisphere

  $d_2$ is the diameter of the cone

  $"r_1"$ is the radius of the hemisphere

  $"h"$ is the height of the cone

Formula used:

Volume of a hemisphere $\left(V\right)=\frac{2}{3}\pi {\left(r_1\right)}^3$

  $"r_1"$ is the radius of the hemisphere

Volume of a cone $\left(V\text{'}\right)=\frac{1}{3}\pi {\left(r_2\right)}^{2}h$

  $r_2\text{and}h$ are the radius and height of the cone

Calculation:

As $d_1=d_2$

  $\begin{array}{l}\therefore \frac{d_1}{2}=\frac{d_2}{2}\\ \therefore r_1=r_2\mathrm{...}(\text{i})\end{array}$

And, $r_1=h$ … (ii)

We know that, Volume of a hemisphere $\left(V\right)=\frac{2}{3}\pi {\left(r_1\right)}^3$

  $V=\frac{2}{3}\pi h^3\mathrm{...}\text{Using}\text{(ii)}$

Also, Volume of a cone $\left(V\text{'}\right)=\frac{1}{3}\pi {\left(r_2\right)}^{2}h$

  $\begin{array}{l}\therefore V\text{'}=\frac{1}{3}\pi {\left(r_1\right)}^{2}h\mathrm{...}\text{Using}\text{(i)}\\ \therefore V\text{'}=\frac{1}{3}\pi {\left(h\right)}^{2}h\mathrm{...}\text{Using}\text{(ii)}\\ \therefore V\text{'}=\frac{1}{3}\pi h^3\end{array}$

Hence, after comparing $V\text{and}V\text{'}$ , we get that volume of a hemisphere is 2 times the volume of a cone.