# Area of a Sector

A sector is a portion of a circle bounded by two radii and the arc joining their extremities. It is thus a form of a triangle with a covered base. In the figure, the portion $AOB$ is a sector. Arc $AB$ is called the arc of the sector and $\angle AOB$ is called the angle of the sector. The area of a sector of a circle is equal to a fraction of the area of the circle determined by dividing the size of the angle by ${360^ \circ }$. So, if the angle of the sector is ${90^ \circ }$, the area of the sector is $\frac{{90}}{{360}}$ or $\frac{1}{4}$ the area of the circle. Thus, the area of a sector of the circle divided by the area of the whole circle is proportional to the angle of the sector divided by ${360^ \circ }$.

If the angle $\angle AOB$ is given in degrees, say${N^ \circ }$, then

Area of the sector: $A = \frac{{\pi {r^2}}}{{360}} \times {N^ \circ }$

Length of the arc: $l = \frac{{2\pi r}}{{360}} \times {N^ \circ }$

If the angle $\angle AOB$ is given in radians, say $\theta$ radians, then

Area of the sector, $A = \frac{{\pi {r^2}}}{{360}} \times {\left( {\frac{{180}}{\pi } \times \theta } \right)^ \circ } = \frac{1}{2}{r^2}\theta$

If the length $l$ of the arc and the radius $r$ of the circle are given, then

Area of the sector, $A = \frac{1}{2}{r^2} \times \theta = \frac{1}{2}{r^2} \times \frac{1}{l}$   where $\theta = \frac{l}{r}$

$A = \frac{1}{2}r \times l$

Example:

Find the area of a sector of ${60^ \circ }$ in a circle of radius $10$cm.

Solution:

The sector $AOB$ has angle ${60^ \circ }$. Its area is equal to $\frac{{60}}{{360}} \times \pi {r^2}$
Since the radius $= 10$cm
Therefore $A = \frac{1}{6} \times 3.1416 \times {(10)^2} = 52.36$ square

Example:

The minute hand of a clock is $12$cw long. Find the area which is described on the clock face between $6$A.M to $6.20$A.M.

Solution:

Given that the length of the minute hand’s radius is $= 12$cm
$\therefore$ $60$minutes $= {360^ \circ }$

$\therefore$ $20$minutes $= {120^ \circ }$

Since the area of sector $= \frac{{\pi {r^2}}}{{360}} \times {N^ \circ }$

$= \frac{{3.1416 \times {{(12)}^2}}}{{360}} \times {120^ \circ }$

$= \frac{1}{3} \times 3.1416 \times 144 = 150.7$ square cm