Chapter 13, Problem 56A

To determine

To find the metal area of given washer.

Answer to Problem 56A

Area of given washer is $1225.58$${\text{mm}}^2$.

Given information:

A washer is given as below indicating dimensions in millimeter.

Calculation:

As we know that area of a circle is given by -

$A=\pi \times R^2$

Here, radius length of the smaller circle is $9.38$ millimeter. i.e. $R=9.38$ millimeter.

So, area of the smaller circle will be -

$\begin{array}{l}A=\pi \times R^2\\ \Rightarrow A=3.14\times {\left(9.38\right)}^2{\text{mm}}^2\text{ {Put }R=9.38\text{ mm}}\end{array}$

$\begin{array}{l}\Rightarrow A=3.14\times 87.9844{\text{mm}}^2\\ \Rightarrow A=276.2710{\text{mm}}^2\\ \Rightarrow A=276.27{\text{mm}}^2\text{ {Rounding off to two decimal point}}\end{array}$

Similarly, radius of the larger circle is $21.87$ millimeter. i.e. $R=21.87$ millimeter.

So, area of the larger circle will be -

$\begin{array}{l}A=\pi \times R^2\\ \Rightarrow A=3.14\times {\left(21.87\right)}^2{\text{mm}}^2\text{ {Put }R=21.87\text{ mm}}\end{array}$

$\begin{array}{l}\Rightarrow A=3.14\times 478.2969{\text{mm}}^2\\ \Rightarrow A=1501.8522{\text{mm}}^2\\ \Rightarrow A=1501.85{\text{mm}}^2\text{ {Rounding off to two decimal point}}\end{array}$

Now, area of the washer is area of smaller circle subtracted from area of larger circle. i.e.

Area of washer $=1501.82-276.27=1225.58{\text{ mm}}^2$.

Explanation of Solution

Given information:

A washer is given as below indicating dimensions in millimeter.

Calculation:

As we know that area of a circle is given by -

$A=\pi \times R^2$

Here, radius length of the smaller circle is $9.38$ millimeter. i.e. $R=9.38$ millimeter.

So, area of the smaller circle will be -

$\begin{array}{l}A=\pi \times R^2\\ \Rightarrow A=3.14\times {\left(9.38\right)}^2{\text{mm}}^2\text{ {Put }R=9.38\text{ mm}}\end{array}$

$\begin{array}{l}\Rightarrow A=3.14\times 87.9844{\text{mm}}^2\\ \Rightarrow A=276.2710{\text{mm}}^2\\ \Rightarrow A=276.27{\text{mm}}^2\text{ {Rounding off to two decimal point}}\end{array}$

Similarly, radius of the larger circle is $21.87$ millimeter. i.e. $R=21.87$ millimeter.

So, area of the larger circle will be -

$\begin{array}{l}A=\pi \times R^2\\ \Rightarrow A=3.14\times {\left(21.87\right)}^2{\text{mm}}^2\text{ {Put }R=21.87\text{ mm}}\end{array}$

$\begin{array}{l}\Rightarrow A=3.14\times 478.2969{\text{mm}}^2\\ \Rightarrow A=1501.8522{\text{mm}}^2\\ \Rightarrow A=1501.85{\text{mm}}^2\text{ {Rounding off to two decimal point}}\end{array}$

Now, area of the washer is area of smaller circle subtracted from area of larger circle. i.e.

Area of washer $=1501.82-276.27=1225.58{\text{ mm}}^2$.