# Area of a sector

• Sep 20th 2009, 06:08 AM
Chinnie15
Area of a sector
The area of a circle is 72cm^2. Find the area of a sector of this circle that subtends a central angle of pi/6 rad.

Thanks!
• Sep 20th 2009, 06:53 AM
e^(i*pi)
Quote:

Originally Posted by Chinnie15
The area of a circle is 72cm^2. Find the area of a sector of this circle that subtends a central angle of pi/6 rad.

Thanks!

$\displaystyle A_{sector} = \frac{1}{2}r^2 \theta$

A whole circle may be considered to be a sector:
• $\displaystyle A_{sector} = 72$
• $\displaystyle \theta = 2\pi$

From this you can find r and then find the area of the sector.

Spoiler:
Rearranging the equation to find r and then plugging that expression in provides a way to find A in one step

$\displaystyle r = \sqrt{\frac{2A_{c}}{\theta_{c}}}$

$\displaystyle A_s = A_c \, \frac{\theta_{s}}{\theta_c}$

Where:

• $\displaystyle A_s$= Area of Sector
• $\displaystyle A_c$ = Area of Circle ($\displaystyle 72\, cm^2$)
• $\displaystyle \theta_{c}$ = Angle of Circle ($\displaystyle 2\pi$)
• $\displaystyle \theta_s$ = Angle subtended by sector ($\displaystyle \frac{\pi}{6}$

• Sep 20th 2009, 07:02 AM
Hello Chinnie15
Quote:

Originally Posted by Chinnie15
The area of a circle is 72cm^2. Find the area of a sector of this circle that subtends a central angle of pi/6 rad.

Thanks!

The whole circle ($\displaystyle 72\,cm^2$) subtends an angle of $\displaystyle 2\pi$ at the centre. The sector you want subtends an angle of $\displaystyle \frac{\pi}{6}$ which is $\displaystyle \frac{1}{12}$ of $\displaystyle 2\pi$. So its area is ...?