# What is the area of a sector of a circle with r = 18”, given that its arc length is 6π?

**Solution**:

The area of the sector AOB in the diagram of the circle of radius r shown below and subtended by an angle θ is given by the equation:

Area of the sector of angle θ = (θ/360°) × πr^{2 }------>(1),

where r is the radius of the circle, θ in degrees

Similarly, the length of an arc of a sector of angle θ = (θ/360) × 2πr ---------->(2)

where r is the radius of the circle.

In problem given the length of the segment given is 6π and the radius is 18 inches(“).

6π = (θ/360°) × 2π(18)

3 = (θ/360°) × 18

⇒ θ = 60°

Substituting the above value of θ in equation (1) we have:

Area of the sector subtended by θ = (60/360) ×πr^{2}

= (60/360) x π(18)^{2}

= (1/6) x 3.14 x 324

= 169.56 inches^{2}

## What is the area of a sector of a circle with r = 18”, given that its arc length is 6π?

**Summary**:

The area of the sector enclosed by the two radii and the arc is 1/6^{th} (60°/360°) of the total area of the sector which is 169.56 inches^{2}.