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Area of a sector

Find sector and annular-sector areas in radians. Recover missing dimensions, combine area with perimeter, compare equal areas and handle measurement bounds with worked practice.

Before you startRadians, arc length, rearranging formulae and quadratics

01 / Take a fraction of the circle

The same angle selects the same fraction at every radius.

A central angle θ radians occupies θ/(2π) of a full turn. It selects that same fraction of the circle’s area. This gives a compact formula once the angle is in radians.

A = [θ/(2π)] ×πr² = ½r²θ

For a circle of radius r>0 and a selected angle 0≤θ≤2π. Area uses square length units.

A sector is bounded by two radii and an arc. A segment, bounded by a chord and an arc, has a different area; it needs an additional triangle calculation.

Compare the fraction and the sizeExplore

Radius 4. Sector area 4π ≈12.566371. Full circle area 16π ≈50.265482.

The sector occupies 1/4 of the circle.

Keep the angle fixed and double the radius. Every length doubles, but the shaded area becomes four times as large. Change the angle instead to change the fraction of the circle.

Radius 4Angle in steps of π/6 π/2

Watch radius doubling multiply area by four

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Use the radius squared

Check angle units before substituting.

Radius 8 cm, angle 1.25 radWorked example

A =½ ×8² ×1.25 =40 cm²

Square the radius, then multiply by half the angle.

Radius 6 cm, angle 150°Worked example

150° =5π/6 rad

Convert the degree input first.

A =½ ×36 ×5π/6 =15π cm²

Keep π for an exact result.

For fixed r, doubling θ doubles the area. For fixed θ, doubling r multiplies area by 4. These different scaling rules help diagnose a missing square.

03 / Recover a missing dimension

A radius is a positive square root.

r = √(2A/θ), for θ>0
θ =2A/r², for r>0

Area 45 cm² and angle 1.6 radWorked example

r² =2(45)/1.6 =225/4

Keep the rearrangement exact if possible.

r =15/2 =7.5 cm

The negative square root is not a physical radius.

Area 18π cm² and radius 9 cmWorked example

θ =2(18π)/81 =4π/9 rad

This is a minor sector; the angle lies between 0 and π.

If your computed angle exceeds 2π for an ordinary sector, recheck the data, units or interpretation. A single sector cannot have more area than its full circle.

04 / Use the arc when the angle is absent

Substitute l=rθ into the area formula.

A =½r²(l/r) =½rl

Radius 5 cm and arc length 8 cmWorked example

A =½ ×5 ×8 =20 cm²

No inverse trigonometry is needed.

The angle is 8/5 =1.6 rad if you need it

This is consistent with the same sector.

Both r and l are lengths. Their product has square units, as an area should. Do not substitute the whole perimeter for l; remove the two radii first.

05 / Combine area and perimeter

The same perimeter can produce two radii for a given area.

A proper sector has perimeter 24 cm and area 32 cm²Worked example

l =24 −2r

Only the remaining boundary length is the arc.

32 =½r(24 −2r)

Use A=½rl.

r² −12r +32 =0

Rearrange to a quadratic.

(r −4)(r −8) =0

Both positive roots need a geometry check.

r =4 gives l =16 and θ =4 rad

This is a valid major sector.

r =8 gives l =8 and θ =1 rad

This is a valid minor sector. Without further information, both answers work.

For perimeter 24, the area is A=12r−r²=36−(r−6)². Its largest possible value is 36 cm² at r=6, θ=2. Completing the square gives this maximum without assuming every quadratic root represents an allowed sector.

06 / Major sectors and area ratios

Use the angle belonging to the shaded region.

Minor angle π/3, radius 7 cmWorked example

Major angle =2π −π/3 =5π/3

A major sector uses the reflex angle.

Major area =½ ×49 ×5π/3 =245π/6 cm²

Alternatively subtract the minor area 49π/6 from the full area 49π.

Two complementary sectors have areas in ratio 2:5Worked example

The smaller sector is 2/7 of the circle

Its central angle is the same fraction of a full turn.

θ =(2/7) ×2π =4π/7

The common radius cancels; it is not needed.

Area ratios equal angle ratios only for sectors of the same radius. Sectors from different circles also depend on the squared radii.

07 / Subtract sectors with the same angle

The ring-shaped region is an outer sector minus an inner sector.

A =½(R² −r²)θ

For0≤r<R and0≤θ≤2π.

Outer radius 9 cm, inner radius 6 cm, angle 0.8 radWorked example

A =½(81 −36)(0.8) =18 cm²

Square each radius before subtracting.

(R −r)² =9 is not R² −r² =45

Using the radial thickness as if it were a circle radius gives the wrong shape.

The area subtracts the inner sector. By contrast, the perimeter adds both exposed arcs. Keep those two operations separate.

08 / Equal areas can have different perimeters

Compare the formulae for the actual shapes.

A right triangle has legs 3a and 4a, with a>0. A sector has radius 4a and the same area. Find its angle and compare perimeters.

Cancel the common square scaleWorked example

Triangle area =½(3a)(4a) =6a²

Its hypotenuse is 5a.

Sector area =½(4a)²θ =8a²θ

Set this equal to 6a².

θ =3/4 rad

Divide by 8a², which is nonzero.

Triangle perimeter =12a

Add 3a +4a +5a.

Sector perimeter =8a +4a(3/4) =11a

The equal-area sector has a smaller perimeter in this example.

09 / Measurement bounds for area and angle

Use the direction of change, and keep the endpoints honest.

A radius is recorded as 3.6 cm and an arc as 4.8 cm, both rounded to 1 decimal place. Using the usual rounding convention:

3.55 ≤ r <3.65
4.75 ≤ l <4.85

Bound the area A=½rlWorked example

A_min =½(3.55)(4.75) =8.43125 cm²

Both factors increase the product, and both lower endpoints are allowed.

Upper bound =½(3.65)(4.85) =8.85125 cm²

The upper endpoints are excluded, so this value is not attained.

8.43125 ≤ A <8.85125 cm²

The lower bound is a minimum; the upper bound is not a maximum.

Bound the angle θ=l/rWorked example

Lower bound =4.75/3.65 =95/73

Use a smaller numerator and larger denominator. The denominator cannot reach 3.65.

Upper bound =4.85/3.55 =97/71

Use a larger numerator and smaller denominator. The numerator cannot reach 4.85.

Suppose 2≤r<3 and 4≤l<5. Give exact bounds for A=½rl and θ=l/r.

Hint

Product bounds use matching extremes; quotient bounds use opposite extremes.

Worked solution

4≤A<15/2. For the angle,4/3<θ<5/2. Neither quotient endpoint can be attained because one required upper endpoint is excluded.

11 / Recap

Area depends on the angle and the square of the radius.

A=½r²θ requires radians and gives square units.
If an arc is known, use A=½rl.
Use the selected major or minor angle.
Annular area subtracts the inner sector.
Combined area/perimeter equations may have two admissible roots.
Check interval endpoints when using rounded measurements.

Section 1 of 11 · Take a fraction of the circle