# 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.