Length of an arc and area of a sector
When we cut a slice of pizza, its curved outer edge is the arc and the slice is the sector. The bigger the angle of the slice, the longer the arc and the bigger the slice.
For radius r and central angle θ, the length of the arc is l = θ ÷ 360 × 2πr. The area of the sector is A = θ ÷ 360 × πr² = ½ × l × r. (Take π = 22/7.)
In the figure: r = 7, θ = 90°; arc l = 90 ÷ 360 × 2 × 22/7 × 7 = 11; area = 90 ÷ 360 × 154 = 38.5.
Rule 1: Length of the arc — θ ÷ 360 shows what fraction of the full circle it is; multiply it by the circumference 2πr.
Length of an arc and area of a sector
In the app you can move this figure with your finger.
Open in the app- Easy: r = 7, θ = 90° ⇒ l = 1/4 × 44 = 11.
- Medium: r = 21, θ = 60° ⇒ l = 1/6 × 132 = 22.
- Exam-type: r = 7, θ = 120° ⇒ l = 1/3 × 44 ≈ 14.67.
Rule 2: Area of the sector — Multiply the same fraction by the area of the circle πr².
Length of an arc and area of a sector
In the app you can move this figure with your finger.
Open in the app- Easy: r = 7, θ = 90° ⇒ A = 1/4 × 154 = 38.5.
- Medium: r = 14, θ = 60° ⇒ A = 1/6 × 616 ≈ 102.67.
- Exam-type: r = 21, θ = 120° ⇒ A = 1/3 × 1386 = 462.
Rule 3: When the arc is given — A = ½ × l × r. Also θ = l × 360 ÷ (2πr).
- Easy: l = 11, r = 7 ⇒ A = ½ × 11 × 7 = 38.5.
- Medium: l = 22, r = 21 ⇒ A = ½ × 22 × 21 = 231.
- Exam-type: l = 11, r = 7 ⇒ θ = 11 × 360 ÷ 44 = 90°.
θ is in degrees and is divided by 360. A semicircle is 180° and a quarter circle is 90°.
The length of the arc is in cm and the area is in cm²; do not mix the units.
Memory clue: "fraction × circumference = arc" and "fraction × area = sector", fraction = θ ÷ 360.
Spot the difference
| Point | Arc | Sector |
|---|---|---|
| What | Length | Area |
| Formula | θ ÷ 360 × 2πr | θ ÷ 360 × πr² |
| Other form | l | ½ × l × r |
| Unit | cm | cm² |
Practice questions
- 1If θ = 180°, what part of the circle is the arc?
- 2r = 14, θ = 90°; what is the length of the arc?
- 3r = 7, θ = 60°; what is the area of the sector?
- 4r = 21, θ = 90°; find the arc and the area.
- 5l = 15, r = 10; what is the area?
- 6In a circle of radius 14 an arc has length 22. Find θ.
- 7The minute hand of a clock is 14 cm long. How far does its tip move in 15 minutes?
- 8A wiper of length 21 sweeps a sector of 120°. What area does it clean?
Show answers (try them yourself first)
- 1Half: 180 ÷ 360 = 1/2.
- 21/4 × 88 = 22.
- 31/6 × 154 ≈ 25.67.
- 4l = 1/4 × 132 = 33; A = 1/4 × 1386 = 346.5.
- 5½ × 15 × 10 = 75.
- 6θ = 22 × 360 ÷ 88 = 90°.
- 715 minutes = 90°; l = 1/4 × 88 = 22 cm.
- 81/3 × 1386 = 462.
Area of a segment
When a round roti is cut by a straight knife, the piece between the chord and the arc is a segment.
A segment is the part between a chord and its arc. The area of the minor segment = area of the sector − area of the triangle formed by the centre and the ends of the chord. The major segment = area of the circle − the minor segment.
In the figure: r = 14, θ = 90°; sector 154, triangle ½ × 14 × 14 = 98; segment = 154 − 98 = 56.
Rule 1: Segment for θ = 90° — The triangle is right-angled and isosceles; its area is ½ × r × r. Segment = ¼πr² − ½r².
Area of a segment
In the app you can move this figure with your finger.
Open in the app- Easy: r = 7 ⇒ 38.5 − 24.5 = 14.
- Medium: r = 14 ⇒ 154 − 98 = 56.
- Exam-type: r = 21 ⇒ 346.5 − 220.5 = 126.
Rule 2: Segment for θ = 60° — The triangle is equilateral; area (√3 ÷ 4) × r². Take √3 = 1.73.
- Easy: r = 7 ⇒ sector 25.67, triangle 1.73 ÷ 4 × 49 = 21.19 ⇒ segment ≈ 4.48.
- Medium: r = 14 ⇒ sector 102.67, triangle 1.73 ÷ 4 × 196 = 84.77 ⇒ segment ≈ 17.9.
- Exam-type: r = 21 ⇒ sector 231, triangle 1.73 ÷ 4 × 441 = 190.73 ⇒ segment ≈ 40.27.
Rule 3: Major segment and shaded regions — Major segment = πr² − minor segment. For shaded regions subtract the areas of two figures.
- Easy: r = 7, minor segment 14 ⇒ major = 154 − 14 = 140.
- Medium: r = 14, minor segment 56 ⇒ major = 616 − 56 = 560.
- Exam-type: A circle (r = 7) inside a square of side 14 ⇒ area of the four corners = 196 − 154 = 42.
Segment = sector − triangle; the triangle is always formed by the centre and the two ends of the chord.
Do not add the sector and the triangle; the triangle must be subtracted.
Memory clue: "segment = sector − triangle" — first the sector, then the triangle, then subtract. θ = 90° ⇒ triangle ½r², θ = 60° ⇒ triangle (√3/4)r².
Spot the difference
| Point | Minor segment | Major segment |
|---|---|---|
| Size | Small | Big |
| Formula | Sector − triangle | Circle − minor segment |
| r = 7, θ = 90° | 14 | 140 |
| r = 14, θ = 90° | 56 | 560 |
Practice questions
- 1Write the formula for the area of a segment.
- 2r = 10, θ = 90°; what is the area of the triangle?
- 3Sector 154, triangle 98; what is the segment?
- 4Find the minor segment for r = 7, θ = 90°.
- 5Find the minor segment for r = 14, θ = 60° (√3 = 1.73).
- 6Find the major segment of a circle of r = 7 whose minor segment is 14.
- 7A circle (r = 14) is inside a square of side 28. What is the area of the four corners?
- 8Find the minor and the major segment for r = 21, θ = 90°.
Show answers (try them yourself first)
- 1Area of sector − area of triangle.
- 2½ × 10 × 10 = 50.
- 3154 − 98 = 56.
- 438.5 − 24.5 = 14.
- 5102.67 − 84.77 ≈ 17.9.
- 6154 − 14 = 140.
- 7784 − 616 = 168.
- 8Minor = 346.5 − 220.5 = 126; major = 1386 − 126 = 1260.
Combined solids, frustum and recasting
An ice-cream cone is a cone joined to a hemisphere; a tea glass is a frustum. The volume of such joined shapes is found part by part.
The volume of a combined solid = the sum of the volumes of its parts. Hemisphere: ⅔πr³; cone: ⅓πr²h; cylinder: πr²h. For a frustum (bigger radius r₁, smaller r₂, height h, slant height l = √[h² + (r₁ − r₂)²]): curved area = πl(r₁ + r₂); volume = πh ÷ 3 × (r₁² + r₁r₂ + r₂²).
In the figure: A cone (h = 12) on a hemisphere (r = 7); volume = ⅔π × 343 + ⅓π × 49 × 12 ≈ 718.67 + 616 = 1334.67.
Rule 1: Joined solids — Find the volume / area of each part separately and add. Leave out the area of the joint surface from both.
Combined solids, frustum and recasting
In the app you can move this figure with your finger.
Open in the app- Easy: Cone (r = 7, h = 12) + hemisphere ⇒ 718.67 + 616 = 1334.67.
- Medium: Capsule: cylinder (r = 7, h = 10) + two hemispheres ⇒ volume = 1540 + 1437.33 = 2977.33; surface area = 440 + 616 = 1056.
- Exam-type: A cube of side 14 with a hemispherical hollow (r = 7) on top ⇒ surface area = 6 × 196 − 154 + 308 = 1330.
Rule 2: Frustum of a cone — First find l, then curved area πl(r₁ + r₂) and volume πh ÷ 3 × (r₁² + r₁r₂ + r₂²).
- Easy: r₁ = 10, r₂ = 4, h = 8 ⇒ l = √(64 + 36) = 10; curved area = 22/7 × 10 × 14 = 440.
- Medium: r₁ = 14, r₂ = 7, h = 24 ⇒ l = √(576 + 49) = 25; curved area = 22/7 × 25 × 21 = 1650; volume = 22/7 × 8 × 343 = 8624.
- Exam-type: r₁ = 15, r₂ = 6, h = 12 ⇒ l = √(144 + 81) = 15; curved area = 22/7 × 15 × 21 = 990.
Rule 3: Melting into a new shape (same volume) — The volume does not change when the shape changes. Volume of the big one ÷ volume of one small one = the number of small shapes.
- Easy: A sphere of r = 3 is melted into spheres of r = 1 ⇒ 27 ÷ 1 = 27 spheres.
- Medium: A cylinder r = 7, h = 18 melted into cones r = 7, h = 6 ⇒ 18 ÷ (6 ÷ 3) = 9 cones.
- Exam-type: A sphere r = 6 melted into cones r = 3, h = 8 ⇒ 288π ÷ 24π = 12 cones.
In joined solids the joint circle is hidden; do not count it in the surface area.
In a frustum the slant height l and the height h are different; use l in the curved area and h in the volume.
Memory clue: "split the shape first, then add". For melting, "volume = volume".
Spot the difference
| Point | Frustum | Cone |
|---|---|---|
| Radii | r₁ and r₂ | r |
| Curved area | πl(r₁ + r₂) | πrl |
| Volume | πh ÷ 3 × (r₁² + r₁r₂ + r₂²) | ⅓πr²h |
| If r₂ = 0 | a cone | — |
Practice questions
- 1What is the volume of a hemisphere of r = 7? (π = 22/7)
- 2What is the volume of a cone with r = 7, h = 6?
- 3Write the formula for the slant height l of a frustum.
- 4r₁ = 10, r₂ = 4, h = 8; what is l?
- 5A sphere of r = 3 is melted into spheres of r = 1; how many spheres?
- 6A capsule: cylinder r = 7, h = 14 and two hemispheres; find the surface area.
- 7Frustum r₁ = 14, r₂ = 7, h = 24; find the curved area and the volume.
- 8A sphere r = 6 is melted to make cones of r = 3, h = 8; how many cones?
Show answers (try them yourself first)
- 1⅔ × 22/7 × 343 ≈ 718.67.
- 2⅓ × 154 × 6 = 308.
- 3l = √[h² + (r₁ − r₂)²].
- 4√(64 + 36) = 10.
- 527.
- 62π r h + 4π r² = 616 + 616 = 1232.
- 7l = 25; curved area = 22/7 × 25 × 21 = 1650; volume = 22/7 × 8 × 343 = 8624.
- 8Volume of the sphere 288π, of a cone 24π ⇒ 12.