A curved boundary and a shaded region ask for different measurements. Arc length measures only the curved edge. Sector perimeter includes two radii as well as the arc. Sector area measures the wedge inside them. A segment is a different region, bounded by an arc and its chord, so its area needs another step.
Begin by naming the region and tracing its complete boundary. This lesson uses supplied central angles in degrees. Keep π in exact answers unless a value is specified; decimal answers below use a calculator’s π and are rounded to two decimal places. Set the calculator to degree mode for the sine calculations.
Use the central angle as a fraction of a turn
A full turn is 360°. A sector with central angle θ therefore occupies θ/360 of a circle. For radius r, the full circumference is 2πr and the full area is πr2. Multiply each by the same fraction, but keep the different units.
| Requested quantity | Calculation | Units |
|---|---|---|
| Arc length L | (θ/360) × 2πr | Length units |
| Sector perimeter P | L + 2r | Length units |
| Sector area S | (θ/360) × πr2 | Square units |
| Minor-segment area | S − (1/2)r2sin θ, for 0° < θ < 180° | Square units |
The last formula removes the triangle formed by the two radii and the chord. It applies here to the smaller segment and its central angle below 180°. A major segment requires identifying the complementary region; do not reuse the subtraction without checking which region is wanted.
Worked case 1: three measurements of one sector
A sector OAB has radius 8 cm and central angle AOB = 75°. Find the arc length AB, the perimeter of sector OAB and its area. The same given radius and angle serve all three calculations, but the answers describe different objects.
The curved edge only
L = (75/360) × 2π × 8 = 10π/3 cm ≈ 10.47 cm. This answer follows only the arc from A to B. No straight edge is included in arc length.
Arc diagram data: O is the centre, OA = OB = 8 cm and angle AOB = 75°. Only the minor arc AB is highlighted. The two radii are construction lines, not part of the requested arc.
The complete sector boundary
P = 8 + 8 + 10π/3 = 16 + 10π/3 cm ≈ 26.47 cm. Trace O to A, around the arc to B, and back to O. That walk contains two radii and one curved edge; it does not contain chord AB.
Perimeter diagram data: the same 8 cm, 75° sector has OA, the minor arc AB and BO highlighted. All three pieces belong to the boundary being measured.
The enclosed wedge
S = (75/360) × π × 82 = 40π/3 cm2 ≈ 41.89 cm2. The result is an area, so cm2 is essential. As a fraction check, the sector is 5/24 of the full 64π cm2 circle.
Area diagram data: the entire region between the two 8 cm radii and their 75° minor arc is shaded. No chord has been used to remove part of it.
Worked case 2: remove a triangle to find a minor segment
A circle has radius 6 cm. Its chord AB subtends a central angle of 120°. Find the area of the minor segment bounded by chord AB and the minor arc AB. The segment is the curved cap on the opposite side of the chord from the centre.
The 120° sector area is (120/360) × π × 62 = 12π cm2. Triangle OAB has two sides of length 6 cm and included angle 120°, so its area is (1/2) × 6 × 6 × sin 120° = 18 × (√3/2) = 9√3 cm2.
Subtract the triangle: segment area = 12π − 9√3 cm2 ≈ 22.11 cm2. Check the size: the triangle area is about 15.59 cm2, while the sector area is about 37.70 cm2. Their difference is positive and smaller than the sector.
Segment diagram data: OA = OB = 6 cm, angle AOB = 120°, and straight chord AB closes the shaded cap. The triangle OAB is unshaded. The radii form the triangle inside the sector, but neither radius borders the segment itself.
If asked for the segment perimeter, use its chord and arc. The chord has length 2r sin(θ/2) = 12 sin 60° = 6√3 cm. The arc is (120/360) × 12π = 4π cm. Their sum is 4π + 6√3 cm ≈ 22.96 cm. Adding two radii here would instead describe the sector boundary.
Worked case 3: recover an angle from arc length
A circle has radius 15 cm. Its minor arc AB has length 5π cm. Find the central angle and the area and perimeter of its sector. Write the known arc length into the arc formula before rearranging.
5π = (θ/360) × 2π × 15. Dividing by 30π gives θ/360 = 1/6, so θ = 60°. A quick interpretation is that the arc is one sixth of the full 30π cm circumference.
The sector area is (1/6) × π × 152 = 37.5π cm2 ≈ 117.81 cm2. Its perimeter is 5π + 30 cm ≈ 45.71 cm. Substituting θ = 60° back into the arc formula returns 5π cm, checking the rearrangement.
Reverse-question diagram data: OA = OB = 15 cm, arc AB = 5π cm, and the central angle is initially unknown. The calculated angle is 60°. The picture is a guide; the given arc and radius determine the answer.
Three practice tasks with full solutions
For each figure, name the requested boundary or region before calculating. The diagram labels below repeat all necessary information; the questions can be solved without measuring a drawing.
Practice 1: a quarter-sector comparison
A sector has radius 7 cm and central angle 90°. Using π = 22/7 for this task only, find its arc length, perimeter and area. The diagram shows two perpendicular radii and their quarter-circle arc.
Arc length = (1/4) × 2 × (22/7) × 7 = 11 cm. Perimeter = 11 + 7 + 7 = 25 cm. Area = (1/4) × (22/7) × 49 = 38.5 cm2. These are three answers with different meanings, even though they use the same fraction of a turn.
Practice 2: a smaller curved cap
A minor segment has radius 10 cm and central angle 60°. Find its area using calculator π. The region is bounded by the 60° arc and its chord. Sector area = (60/360) × 100π = 50π/3 cm2. Triangle area = (1/2) × 100 × sin 60° = 25√3 cm2. Segment area = 50π/3 − 25√3 ≈ 9.06 cm2.
Practice 3: repair an incomplete perimeter answer
A sector has radius 12 cm and angle 150°. A learner reports its perimeter as 10π cm. Identify the missing parts and correct the result. The figure shows a sector, not a segment.
The learner correctly calculated only the arc: (150/360) × 24π = 10π cm. Both radii also belong to the boundary, so the perimeter is 10π + 24 cm ≈ 55.42 cm. The repair is a boundary correction, not a change to the arc calculation.
Frequently asked questions
Are a sector and a segment the same shape?
No. A sector has two radii and an arc as boundary; a segment has a chord and an arc.
Why is the denominator 360?
The angle is in degrees and a full turn is 360°. Radian formulas use a different expression.
Should I always add two radii?
Add them for the perimeter of an ordinary sector. Arc length alone and segment perimeter have different boundaries.
Why subtract a triangle for segment area?
The minor sector consists of the triangle between the radii and chord plus the minor segment. Removing the triangle leaves the cap.
Which value of π should I use?
Follow the question. Otherwise retain exact π or use calculator π, stating the final rounding.
What if my segment area is negative?
Recheck the chosen region, angle mode and subtraction. The minor-segment areas in this lesson are positive and smaller than their sectors.
Can I measure the drawing to find a missing length?
Use the stated data and formulas unless a scale drawing is explicitly required. These illustrations are not measurement instruments.
Continue with clear boundaries
Before another calculation, trace the requested outline and mark each included edge. The Mathematics topical revision collection provides further navigation; some contents may require membership. This lesson does not establish access to a particular circular-measurement download.
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