- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
A chord of a circle of radius $ 14 \mathrm{~cm} $ makes a right angle at the centre. Find the areas of the minor and major segments of the circle.
Given:
A chord of a circle of radius \( 14 \mathrm{~cm} \) makes a right angle at the centre.
To do:
We have to find the areas of the minor and major segments of the circle.
Solution:
Radius of the circle $r = 14\ cm$
Angle at the centre $\theta = 90^o$
Area of the circle $=\pi r^{2}$
$=\frac{22}{7} \times(14)^{2}$
$=616 \mathrm{~cm}^{2}$
$\mathrm{AB}$ is the chord.
Area of the minor segment $\mathrm{ACB}=(\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}) r^{2}$
$=(\frac{\pi \times 90^{\circ}}{360^{\circ}}-\sin \frac{90^{\circ}}{2} \cos \frac{90^{\circ}}{2})(14)^{2}$
$=(\frac{\pi}{4}-\sin 45^{\circ} \cos 45^{\circ}) \times 196$
$=196(\frac{22}{7 \times 4}-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}})$
$=(\frac{22 \times 196}{28}-196 \times \frac{1}{2})$
$=154-98$
$=56 \mathrm{~cm}^{2}$
Therefore,
Area of the major segment $\mathrm{ADB}=$ Area of the circle $-$ Area of the minor segment
$=616-56$
$=560 \mathrm{~cm}^{2}$
The areas of the minor and major segments of the circle are $56\ cm^2$ and $560 \mathrm{~cm}^{2}$ respectively.
To Continue Learning Please Login
Login with Google