Home › Learning Hub › A Level Maths › P1 4: Circular measure
P1 4

Circular measure

Pure Mathematics 1 · Paper 1 · AS and A Level

🎯What you need to be able to do

  • Understand what a radian is, and convert between radians and degrees.
  • Use \( s = r\theta \) for arc length and \( A = \tfrac{1}{2}r^{2}\theta \) for sector area.
  • Find the area of a segment, and the perimeter of a sector or segment.
  • Combine circular measure with the sine and cosine rules and with triangle areas in composite figures.
  • Give exact answers in terms of \( \pi \) and surds when asked.

📚The mathematics

What a radian is

One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. Since the whole circumference is \( 2\pi r \), a full turn is \( 2\pi \) radians, so

\[ \pi \text{ radians} = 180^{\circ} \]

To convert, multiply by \( \dfrac{\pi}{180} \) going to radians and by \( \dfrac{180}{\pi} \) coming back. The values worth knowing on sight are \( \dfrac{\pi}{6} = 30^{\circ} \), \( \dfrac{\pi}{4} = 45^{\circ} \), \( \dfrac{\pi}{3} = 60^{\circ} \), \( \dfrac{\pi}{2} = 90^{\circ} \).

Radians are not simply an alternative unit — they are the reason the two formulae below are as clean as they are, and later the reason \( \dfrac{\mathrm{d}}{\mathrm{d}x}\sin x = \cos x \) holds without a stray constant.

Arc length and sector area

\( s = r\theta \)
\( A = \tfrac{1}{2}r^{2}\theta \)

Both require \( \theta \) in radians. If a question gives you degrees, convert first. And set your calculator to radian mode at the start of the paper — a calculator left in degrees produces a confident wrong answer with no warning, which is worse than an error message.

These also let you work backwards: given an arc length and a radius, \( \theta = \dfrac{s}{r} \); given a sector area and an angle, \( r = \sqrt{\dfrac{2A}{\theta}} \).

Segments

A segment is the region between a chord and the arc it cuts off. Its area is the sector minus the triangle formed by the two radii and the chord:

\[ A_{\text{segment}} = \tfrac{1}{2}r^{2}\theta - \tfrac{1}{2}r^{2}\sin\theta = \tfrac{1}{2}r^{2}\left(\theta - \sin\theta\right) \]

That factorised form is worth remembering, but understand where it comes from: the triangle has two sides \(r\) with the angle \( \theta \) between them, so its area is \( \tfrac{1}{2}r^{2}\sin\theta \) by the standard formula. Since \( \theta > \sin\theta \) for \( \theta > 0 \), a segment area is always positive — a negative answer means you subtracted the wrong way round.

Perimeters

Perimeter questions are where marks quietly disappear, because you have to decide which pieces form the boundary:

  • Perimeter of a sector = arc + two radii = \( r\theta + 2r \).
  • Perimeter of a segment = arc + chord = \( r\theta + 2r\sin\!\left(\tfrac{\theta}{2}\right) \).

The chord length comes from splitting the isosceles triangle down the middle into two right-angled triangles with angle \( \tfrac{\theta}{2} \) — or equivalently from the cosine rule. The half is the part people forget.

“Perimeter” never means “the whole circle”. Read the diagram and trace the boundary of the shaded region with your finger, listing each piece as you go. Composite-figure questions are deliberately built so that a plausible-looking answer can be reached by including one piece too many or one too few, and the arithmetic gives you no clue.

Composite figures

Most 9709 questions on this topic are a circle sector combined with a triangle, and they are solved by decomposition: name each region, compute each area or length separately, then add and subtract. Two supporting tools appear constantly — the area of a triangle as \( \tfrac{1}{2}ab\sin C \), and the cosine rule for a chord or a third side.

Where the geometry involves a tangent, remember from P1 3 that a tangent meets the radius at right angles; that right angle is usually what lets you find the angle the question actually needs.

✏️Worked example

A sector \(OAB\) of a circle has centre \(O\), radius 12 cm, and angle \( AOB = 0.9 \) radians. (a) Find the arc length \(AB\) and the area of the sector. (b) Find the area of the segment cut off by the chord \(AB\). (c) Find the perimeter of that segment.

(a) Directly from the formulae, with \( \theta \) already in radians:

\( s = r\theta = 12 \times 0.9 = 10.8 \) cm
\( A = \tfrac{1}{2}r^{2}\theta = \tfrac{1}{2}(144)(0.9) = 64.8 \) cm\(^2\)

(b) The triangle \(OAB\) has two sides of 12 with 0.9 rad between them, so its area is \( \tfrac{1}{2}(12)(12)\sin 0.9 = 72 \times 0.78333 = 56.4 \) cm\(^2\). Hence

\[ A_{\text{segment}} = 64.8 - 56.4 = 8.40\ \text{cm}^{2} \]

Or in one step, \( \tfrac{1}{2}(144)(0.9 - \sin 0.9) = 72(0.9 - 0.78333) = 8.40 \) cm\(^2\).

(c) The segment is bounded by the arc and the chord. The chord is

\[ AB = 2r\sin\!\left(\frac{\theta}{2}\right) = 2(12)\sin 0.45 = 24(0.43497) = 10.44\ \text{cm} \]

so the perimeter is \( 10.8 + 10.44 = 21.2 \) cm.

Check it. Two sanity checks are available and both take seconds. The chord must be slightly shorter than the arc it subtends — 10.44 against 10.8. ✓ And the segment must be a small fraction of the sector, since 0.9 rad is a modest slice: 8.4 out of 64.8 is about 13%, which matches the picture. A segment area coming out close to the sector area means the triangle was computed wrongly.
The calculator mode, and the half-angle. If your calculator is in degrees, \( \sin 0.9 \) returns 0.0157 instead of 0.7833, the triangle area becomes 1.13, and the segment comes out as 63.7 cm\(^2\) — larger than it should be and almost equal to the whole sector, which is the tell. In part (c), using \( 2r\sin\theta \) rather than \( 2r\sin\!\left(\tfrac{\theta}{2}\right) \) gives a chord longer than the arc, which is geometrically impossible.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Convert \( 135^{\circ} \) to radians in terms of \( \pi \), and 2.4 radians to degrees to 1 decimal place.
\( 135 \times \dfrac{\pi}{180} = \dfrac{135\pi}{180} = \dfrac{3\pi}{4} \) radians. And \( 2.4 \times \dfrac{180}{\pi} = \dfrac{432}{\pi} = 137.5^{\circ} \). When a question says "in terms of \( \pi \)", cancel the fraction fully — \( \dfrac{135\pi}{180} \) left unsimplified may not earn the mark.
2. An arc of a circle of radius 9 cm has length 15 cm. Find the angle it subtends at the centre, and the area of the sector.
From \( s = r\theta \), \( \theta = \dfrac{15}{9} = \dfrac{5}{3} = 1.667 \) radians. Then \( A = \tfrac{1}{2}(81)\left(\tfrac{5}{3}\right) = 67.5 \) cm\(^2\). A neat shortcut: since \( A = \tfrac{1}{2}r^{2}\theta = \tfrac{1}{2}r(r\theta) = \tfrac{1}{2}rs \), you can go straight to \( \tfrac{1}{2}(9)(15) = 67.5 \) without finding \( \theta \) at all.
3. A sector has radius 7 cm and angle \( \dfrac{2\pi}{5} \). Find its perimeter, giving your answer in exact form.
Arc \( = r\theta = 7 \times \dfrac{2\pi}{5} = \dfrac{14\pi}{5} \) cm. The perimeter of a sector also includes the two radii: \( P = \dfrac{14\pi}{5} + 14 \) cm (which is 22.8 cm). Leaving out the \( +14 \) is the standard error — the boundary of a sector is not just its arc.
4. A circle has radius 10 cm. A chord subtends an angle of 1.2 radians at the centre. Find the area of the minor segment.
\( A = \tfrac{1}{2}r^{2}(\theta - \sin\theta) = \tfrac{1}{2}(100)(1.2 - \sin 1.2) = 50(1.2 - 0.93204) = 50(0.26796) = 13.4 \) cm\(^2\). Note \( \sin 1.2 \) is evaluated in radians. As a check, \( \theta > \sin\theta \) so the bracket is positive, as a segment area must be.
5. In a circle of radius \(r\), a sector has angle \( \theta \). The perimeter of the sector is 40 cm and its area is 84 cm\(^2\). Find \(r\) and \( \theta \).
Perimeter: \( r\theta + 2r = 40 \). Area: \( \tfrac{1}{2}r^{2}\theta = 84 \), so \( r\theta = \dfrac{168}{r} \). Substituting into the first: \( \dfrac{168}{r} + 2r = 40 \), so \( 168 + 2r^{2} = 40r \), giving \( 2r^{2} - 40r + 168 = 0 \) and \( r^{2} - 20r + 84 = 0 \). Factorising, \( (r-6)(r-14) = 0 \), so \( r = 6 \) or \( r = 14 \). Then \( \theta = \dfrac{168}{r^{2}} \): for \( r = 6 \), \( \theta = 4.67 \) rad; for \( r = 14 \), \( \theta = 0.857 \) rad. Both are valid, since each angle is less than \( 2\pi \), so give both pairs. Discarding one without a reason loses a mark — the only grounds for rejecting a root here would be \( \theta > 2\pi \), which does not occur.
6. Two circles of radius 5 cm have centres 6 cm apart. Find the angle subtended at the centre of one circle by the common chord.
By symmetry the common chord is the perpendicular bisector of the line joining the centres, meeting it 3 cm from each centre. That gives a right-angled triangle with hypotenuse 5 (a radius) and adjacent side 3, so the half-angle satisfies \( \cos\!\left(\tfrac{\theta}{2}\right) = \dfrac{3}{5} \), giving \( \tfrac{\theta}{2} = 0.9273 \) rad. Hence \( \theta = 1.85 \) radians. Doubling at the end is essential — the right-angled triangle only ever gives you half the angle.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • GeoGebra — an interactive sector where dragging the angle updates arc, sector and segment together
  • Khan Academy — radians, arc length and sector area
  • Better Explained — an intuitive account of why radians are the natural unit