# Arc Length, Area of a Circular Sector

## Arc of a circle:

An arc is a part of the circumference of a circle. The red part AB in the figure below shows an arc.

Note: The length ''l'' of an arc that subtends an angle \theta at the center of a circle of radius r is given by

# l = r\theta

i.e. Arc length is angle times radius of the circle. (Here \theta is in radians, not in degrees)

##### Example 1:

The radius of a circle is r = 10 cm. Find the length of an arc that subtends an angle of 1.4 radians at the center of the circle.

Solution: We are given that r = 10 cm

\theta = 1.4 radians

=> Arc Length = l = r.\theta

= (10cm)(1.4)

= 14cm

##### Example 2:

A cyclist is moving along a circular path of radius 18m. How much distance does he travel if he makes a total of 4 revolutions?

Solution: We know that 1 revolution = 2\pi radians

=> 4 revolutions = 4 x 2\pi radians = 8\pi radians

Since the cyclist moves along the circular path, he moves along a trajectory that is part of the circumference of the circle (i.e. an arc) as shown in the figure.

=> Distance travelled = arc length = radius x angle in radians

=> Distance travelled = (18m) (8\pi) = 1448\pi m

## Segment of a Circle:

A segment is the part of a circle bounded by an arc and a chord of the circle. The shaded part in the following figure indicates the segment of the circle.

## Sector of a Circle:

A sector is the part of a circle bounded by two radii and an arc of the circle as shown in the figure. The part shown in maroon is a sector of the circle.

## Area of a circular sector

Let r be the radius of a circle and let there be an arc of length l on that circle that subtends an angle \theta at the center of the circle.

We know that,

The area of the circle = \pir^2

Angle of the circle = 2\pi radians

Angle of sector POQ = \theta

Then according to the elementary geometry

(text(Area of Sector) POQ)/(text(Area of Circle)) = (text(Angle of sector))/(text(Angle of Circle))

(text(Area of Sector) POQ)/(\pir^2 ) = \theta/( 2\pi )

=> Area of Sector POQ = \theta/( 2\pi )  x \pi r^2

=> Area of Sector POQ = 1/( 2 ) \theta r^2

##### Example 3:

Find area of the sector having a central angle of 20Â° in a circle of radius 7m.

Solution: We are given that r = 7m

\theta = 20^\circ x \pi/180 = \pi/( 9 ) radians

Area of sector = ( 1 )/2 r^2 \theta

= ( 1 )/2 (7)^2 (\pi/( 9 ))

= ( 1 )/2 (49) (\pi/( 9 ))

since, (\pi = 3.1416)

= ( 49\pi )/( 18) m^2

##### Example 4:

Imagine a circular track of radius 500m and let a train be running on this track at a rate of 30km per hour. Calculate the distance covered by the train in a time interval of 10 seconds. Also find the angle through which it turns in this interval.

Solution: We are given that r = 500m

v = 30km/hr

=> Distance travelled in 1 second = 30 x 1000 x 1/( 3600 ) m

= ( 25 )/3 m = 8.33m

=> Distance travelled in 10 seconds = l = ( 250 )/3 m = 83.3m

Therefore, The train travels 83.3m in 10 seconds.

Let \theta be the angle through which the train turns in 10 seconds.

=> \theta = ( l )/r = ( 250 )/3 Ã· 500

=> \theta = ( 1 )/6 radian

Therefore, The train turns through an angle of 1/6 radian in the interval of 10 seconds.

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