Geometry Problem Solver Sector of a circle
A sector is the part of a circle enclosed by two radii and the arc between them: the pizza-slice shape. Its size is set by the angle at the centre, and its area is that fraction of the whole circle, θ/360 of πr².
Area and arc length
A sector is a fraction of the circle, and the fraction is θ/360 when θ is in degrees. So the area is (θ/360) x πr², and the arc length is the same fraction of the circumference, (θ/360) x 2πr.
A 90° sector of a circle of radius 6 has area (90/360) x π x 36 = 9π, a quarter of the whole, which is the sanity check: a quarter turn should give a quarter of the area.
In radians both formulas get shorter, because a radian is defined to make them so: area is ½r²θ and arc length is rθ. That tidiness is the main reason radians exist.
Sector against segment
A sector is bounded by two radii and an arc. A segment is bounded by a chord and an arc: it is what you get by slicing straight across, ignoring the centre.
The two are related. Take a sector and cut off the triangle formed by the two radii and the chord, and what remains is the segment. So the segment's area is the sector's area minus that triangle.
Common sectors of a circle of radius r
| Angle | Fraction | Area | Arc length |
|---|---|---|---|
| 90° | 1/4 | πr²/4 | πr/2 |
| 120° | 1/3 | πr²/3 | 2πr/3 |
| 180° | 1/2 (a semicircle) | πr²/2 | πr |
| 360° | the whole circle | πr² | 2πr |
Questions
What is the difference between a sector and a segment?
A sector is cut by two radii and includes the centre, like a pizza slice. A segment is cut by a straight chord and does not reach the centre, like the piece left after slicing the top off.
Is a semicircle a sector?
Yes, the sector whose angle is 180°. Its two radii lie in a straight line, forming the diameter.
How do you find the angle if you know the area?
Rearrange the formula: θ = 360 x area / (πr²). A sector of area 9π on a radius of 6 gives 360 x 9π / 36π = 90°.
More geometry reference
Stuck on the actual problem
Photograph it and get the worked solution, step by step, with the theorem named at each one.
