A sector is a fraction of the full circle area: A sector with angle θ covers a fraction θ/360 of the full circle area (πr²).
Interactive: switch to ''Sector Area'' tab to see the shaded area change with angle.
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Worked example — sector area
Find the area of a sector with radius 10 cm and angle 108°.
Step by step
- Substitute.
- Evaluate.
Final answer
Area = 30π ≈ 94.2 cm².
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Worked example — find the angle
A sector of radius 8 cm has area 32π cm².
Find its angle.
Step by step
- Substitute into the area formula.
- Cancel π and rearrange.
- Solve for θ.
Final answer
Angle = 180° (a semicircle).
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Worked example — find both arc and area
A sector has radius 6 cm and angle 150°.
Find:
(a) the arc length,
(b) the area.
Step by step
- (a) Arc length.
- (b) Sector area.
Final answer
Arc length ≈ 15.7 cm; Area ≈ 47.1 cm².
IB-style question — an annular sector (a ring)
A metal washer is shaped like an annular sector: a sector of a large circle with a smaller sector removed. The outer radius is 8 cm, the inner radius is 5 cm, and the central angle is 120°.
Find the area of the annular sector.
Step by step
- Take the big sector and subtract the small sector — both share the 120° angle, so subtract the radii squared.
- Substitute R = 8, r = 5, θ = 120°.
- Evaluate.
Final answer
A = 13π ≈ 40.8 cm².
The annular sector from the question: outer radius R = 8, inner radius r = 5, angle 120°. Its area is the big sector minus the small one.
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Worked example — pizza slice
A circular pizza has diameter 30 cm.
It is cut into 8 equal slices.
Find the area of one slice.
Step by step
- Each slice has angle θ = 360/8 = 45°. Radius = 15 cm (half of diameter).
- Area of one slice.
Final answer
Area of one slice ≈ 88.4 cm².
IB-style question — a curved track (area → volume)
A curved section of running track is an annular sector with outer radius 30 m, inner radius 24 m and central angle 100°.
(a) Find the area of the track surface.
(b) The surface is covered with rubber 0.05 m deep. Find the volume of rubber needed.
Step by step
- (a) Annular-sector area = (θ/360) × π(R² − r²) with R = 30, r = 24, θ = 100°.
- Evaluate the area.
- (b) The rubber is a thin layer: volume = surface area × depth.
Final answer
(a) ≈ 283 m². (b) ≈ 14.1 m³.
The track is an annular sector (R = 30 m, r = 24 m, angle 100°). Find its area, then multiply by the 0.05 m depth for the volume of rubber.
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