Circular measure for IGCSE Mathematics - Additional: Methods and Worked Examples

What you need to know about radians

Circular measure is one of the most mark-friendly topics in IGCSE Additional Mathematics. The formulas are short. The method is consistent. And the questions follow predictable patterns. Master three core skills and you'll pick up every available mark on this topic.

Those three skills are: converting between degrees and radians, calculating arc length and sector area, and handling segments and compound shapes. That's the whole topic. No hidden tricks.

Radians: the conversion toolkit

A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. One full turn equals 2pi radians. One half turn equals pi radians. That gives us the conversion relationship:

• pi radians = 180 degrees
• 1 radian = 180/pi degrees (approximately 57.3 degrees)
• 1 degree = pi/180 radians

Two conversion rules cover every situation:

Worked example 1: Converting between units

(a) Convert 135 degrees to radians. (b) Convert 2.4 radians to degrees.

1. (a) 135 x pi/180 = 135pi/180 = 3pi/4 radians
2. (b) 2.4 x 180/pi = 432/pi = 137.5 degrees (1 d.p.)

Leave answers in terms of pi when the question gives angles in terms of pi. Use decimals when the question gives decimal radians. Match the format the examiner uses.

Arc length

The formula is direct:

s = r x theta

where s is the arc length, r is the radius, and theta is the angle in radians.

That last part is critical. The formula only works with radians. If you're given degrees, convert first. This is the single most common error students make on circular measure questions.

Worked example 2: Finding arc length

A sector of a circle has radius 8 cm and angle 1.2 radians. Find the arc length.

1. s = r x theta
2. s = 8 x 1.2
3. s = 9.6 cm

Worked example 3: Finding the angle

An arc of length 15 cm belongs to a circle of radius 6 cm. Find the angle at the centre in radians.

1. s = r x theta, so theta = s/r
2. theta = 15/6
3. theta = 2.5 radians

Worked example 4: Finding the radius

A sector has arc length 10 cm and angle pi/5 radians. Find the radius.

1. s = r x theta, so r = s/theta
2. r = 10/(pi/5) = 50/pi
3. r = 15.9 cm (3 s.f.)

Sector area

The formula:

A = (1/2) x r² x theta

Again, theta must be in radians. Same warning as before: degrees will give a wrong answer with no error message from your calculator to alert you.

Worked example 5: Sector area

Find the area of a sector with radius 10 cm and angle 0.8 radians.

1. A = (1/2) x r² x theta
2. A = (1/2) x 100 x 0.8
3. A = 40 cm²

Worked example 6: Reverse calculation

A sector has area 27 cm² and radius 6 cm. Find the angle in radians.

1. A = (1/2) x r² x theta
2. 27 = (1/2) x 36 x theta
3. 27 = 18 x theta
4. theta = 27/18 = 1.5 radians

The perimeter of a sector

Students regularly forget this: the perimeter of a sector is not just the arc length. It includes two radii as well.

Perimeter of sector = 2r + s = 2r + r x theta

Worked example 7: Sector perimeter

A sector has radius 5 cm and angle 2.1 radians. Find its perimeter.

1. Arc length: s = 5 x 2.1 = 10.5 cm
2. Perimeter = 2(5) + 10.5 = 10 + 10.5 = 20.5 cm

Segments: the tricky part

A segment is the region between a chord and the arc it cuts off. This is where most IGCSE Additional Mathematics candidates lose marks, because it requires combining the sector formula with triangle area.

The key relationship:

Area of segment = Area of sector - Area of triangle

The triangle is formed by the two radii and the chord. Its area uses the standard formula:

Area of triangle = (1/2) x r² x sin(theta)

So the segment area becomes:

Area of segment = (1/2) x r² x theta - (1/2) x r² x sin(theta) = (1/2) x r² x (theta - sin(theta))

Worked example 8: Area of a segment

A chord subtends an angle of 1.4 radians at the centre of a circle with radius 12 cm. Find the area of the minor segment.

1. Area of sector = (1/2) x 144 x 1.4 = 100.8 cm²
2. Area of triangle = (1/2) x 144 x sin(1.4) = 72 x 0.98545 = 70.95 cm² (4 s.f.)
3. Area of segment = 100.8 - 70.95 = 29.85 cm² (4 s.f.)

Alternatively, in one step: A = (1/2) x 144 x (1.4 - sin 1.4) = 72 x (1.4 - 0.98545) = 72 x 0.41455 = 29.85 cm²

Perimeter of a segment

The perimeter of a segment consists of the arc and the chord. You already know how to find the arc length. The chord length comes from the cosine rule or from basic trigonometry:

Chord length = 2r x sin(theta/2)

So: Perimeter of segment = r x theta + 2r x sin(theta/2)

Worked example 9: Perimeter of a segment

Using the same circle from worked example 8 (radius 12 cm, angle 1.4 radians), find the perimeter of the minor segment.

1. Arc length = 12 x 1.4 = 16.8 cm
2. Chord length = 2 x 12 x sin(0.7) = 24 x 0.6442 = 15.46 cm (4 s.f.)
3. Perimeter = 16.8 + 15.46 = 32.26 cm (4 s.f.)

Compound shapes

IGCSE Additional Mathematics papers regularly test compound shapes that combine sectors with triangles, rectangles, or other sectors. The approach is always the same: break the shape into parts, calculate each part, then add or subtract as needed.

Worked example 10: Compound shape

Two circles of radius 5 cm have their centres 8 cm apart. The circles overlap, forming a lens-shaped region. Each centre lies within the other circle. Find the area of the overlapping region.

1. Consider one circle. The chord of intersection and the two radii to its endpoints form an isosceles triangle with sides 5, 5, and a chord to be determined.
2. The distance from a centre to the midpoint of the chord is 4 cm (half of 8). By Pythagoras, the half-chord = sqrt(25 - 16) = 3 cm. So the full chord = 6 cm.
3. The angle at each centre: cos(theta/2) = 4/5, so theta/2 = 0.6435 rad, giving theta = 1.287 radians.
4. Area of one segment = (1/2) x 25 x (1.287 - sin 1.287) = 12.5 x (1.287 - 0.9602) = 12.5 x 0.3268 = 4.085 cm²
5. The overlap consists of two identical segments (one from each circle).
6. Total overlap area = 2 x 4.085 = 8.17 cm² (3 s.f.)

Do/Don't checklist

Common mistakes and mark losses

Self-check questions

Work through each one fully before checking the answers.

1. Convert 225 degrees to radians, giving your answer in terms of pi.
2. A sector has radius 7 cm and angle 2pi/3 radians. Find (a) the arc length and (b) the area.
3. A sector has arc length 12 cm and area 48 cm². Find the radius and the angle.
4. Find the area of the segment cut off by a chord that subtends an angle of 1.8 radians at the centre of a circle with radius 10 cm.
5. A sector OAB has radius 9 cm and angle 0.7 radians. Find the perimeter of the segment bounded by chord AB and arc AB.

Exam strategy for circular measure

Circular measure questions on IGCSE Additional Mathematics papers typically carry 5 to 8 marks. They're predictable in structure and generous in marks for method. Here's how to maximise your score:

• Write formulas first. Before any calculation, write s = r x theta or A = (1/2)r² x theta. This earns a method mark even if your arithmetic goes wrong.
• Sketch the shape. Even when one isn't required. A quick sketch prevents segment-versus-sector confusion and helps you spot whether to add or subtract areas.
• Check units consistency. If the radius is in centimetres, your arc length will be in centimetres and your area in square centimetres. Don't mix units mid-calculation.
• Use the relationship between the two formulas. Notice that A = (1/2) x r x s. This is useful when you're given arc length and need area without knowing the angle separately.
• Give answers to 3 significant figures unless told otherwise. Cambridge markers penalise premature rounding but accept 3 s.f. as the standard precision.

Circular measure rewards students who are methodical. The formulas are simple, the applications are repetitive, and the marks are there for the taking. Get comfortable with the three core skills, practise a handful of past paper questions, and this topic becomes one of the most reliable mark earners on the IGCSE Additional Mathematics exam.