# 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