Chapter 12.3, Problem 2CE

(a)

To determine

To find: The lateral areas of cylinders $\text{I,}\text{II}$ , and $\text{III}$ .

Answer to Problem 2CE

The lateral areas of cylinders $\text{I,}\text{II}$ , and $\text{III}$ is $24\pi ,48\pi ,$ and $48\pi$ respectively.

Explanation of Solution

Given information:

The given figure is shown below.

  

Figure (1)

Calculation:

The lateral area of a cylinder is give by formula.

  $2\pi rh$

Here, $r$ is the radius and $h$ is the height.

The values for the cylinder $\text{I}$ .

  $\begin{array}{l}r=3\\ h=4\end{array}$

Calculate the lateral area for the cylinder $\text{I}$ .

  $\begin{array}{c}2\pi rh=2\pi \times 3\times 4\\ =24\pi \end{array}$

The values for the cylinder $\text{II}$ .

  $\begin{array}{l}r=3\\ h=8\end{array}$

Calculate the lateral area for the cylinder $\text{II}$ .

  $\begin{array}{c}2\pi rh=2\pi \times 3\times 8\\ =48\pi \end{array}$

The values for the cylinder $\text{III}$ .

  $\begin{array}{l}r=6\\ h=4\end{array}$

Calculate the lateral area for the cylinder $\text{III}$ .

  $\begin{array}{c}2\pi rh=2\pi \times 6\times 4\\ =48\pi \end{array}$

Therefore, the lateral areas of cylinders $\text{I,}\text{II}$ , and $\text{III}$ is $24\pi ,48\pi ,$ and $48\pi$ respectively.

(b)

To determine

To find: Whether the lateral area of $\text{II}$ is twice the lateral areas of $\text{I}$ .

Answer to Problem 2CE

The yes, the lateral area of the $\text{II}$ is twice the lateral areas of $\text{I}$ .

Explanation of Solution

Given information:

The height of $\text{II}$ is twice the height of $\text{I}$ .

Calculation:

Calculate the lateral area for the cylinder $\text{II}$ .

  $\begin{array}{c}2\pi rh=2\pi \times 3\times 8\\ =48\pi \\ =2\left(24\pi \right)\\ =2\left(\text{L}\text{.A}\text{.}\text{of}\text{cylinder}\text{I}\right)\end{array}$

Therefore, yes, the lateral area of the $\text{II}$ is twice the lateral areas of $\text{I}$ .

(b)

To determine

To find: Whether the lateral area of $\text{III}$ is twice the lateral areas of $\text{I}$ .

Answer to Problem 2CE

The yes, the lateral area of the $\text{III}$ is twice the lateral areas of $\text{I}$ .

Explanation of Solution

Given information:

The height of $\text{II}$ is twice the height of $\text{I}$ .

Calculation:

Calculate the lateral area for the cylinder $\text{III}$ .

  $\begin{array}{c}2\pi rh=2\pi \times 6\times 4\\ =48\pi \\ =2\left(24\pi \right)\\ =2\left(\text{L}\text{.A}\text{.}\text{of}\text{cylinder}\text{I}\right)\end{array}$

Therefore, yes, the lateral area of the $\text{III}$ is twice the lateral areas of $\text{I}$ .