- Trending Categories
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
Figure below shows a sector of a circle, centre $O$, containing an angle $\theta$. Prove that:Perimeter of the shaded region is $r(\tan \theta+\sec \theta+\frac{\pi \theta}{180}-1)$
"
Given:
A sector of a circle with centre $O$, containing an angle $\theta$.
To do:
We have to prove that the perimeter of the shaded region is $r(\tan \theta+\sec \theta+\frac{\pi \theta}{180}-1)$.
Solution:
From the figure,
Radius of the circle $= r$
Arc $AC$ subtends $\theta$ at the centre of the circle.
$\angle OAB$ is a right triangle.
In right angled triangle $OAB$,
$\tan \theta=\frac{\mathrm{AB}}{\mathrm{OA}}$
$\Rightarrow \mathrm{AB}=\mathrm{OA} \times \tan \theta$
$=r \tan \theta$
Area of $\Delta \mathrm{OAB}=\frac{1}{2} \mathrm{OA} \times \mathrm{AB}$
$=\frac{1}{2} \times r \times (r \tan \theta)$
$=\frac{1}{2} r^{2} \tan \theta$
Area of sector $\mathrm{OAC}=\pi r^{2} (\frac{\theta}{360^{\circ}})$
$=\frac{\pi r^{2} \theta}{360}$
The length of the arc $\mathrm{AC}=2 \pi r \times \frac{\theta}{360^{\circ}}$
$=\frac{2 \pi r \theta}{360^{\circ}}$
$=\frac{\pi r \theta}{180^{\circ}}$
Perimeter of the shaded region $=$ Arc $AC+A B+B C$
$=\frac{\pi r \theta}{180}+r \tan \theta+(\mathrm{OB}-\mathrm{OC})$
$=\frac{\pi r \theta}{180}+r \tan \theta+(r \sec \theta-r)$ (Since $\mathrm{OB}=r \sec \theta$)
$=r(\frac{\pi \theta}{180}+\tan \theta+\sec \theta-1)$
$=r(\tan \theta+\sec \theta+\frac{\pi \theta}{180}-1)$
Hence proved.
74 Views
