Home / IGCSE Add Maths 0606 / Circular measure / Compound shapes: segments and composite regions
Cambridge IGCSE · Topic 9.1
IGCSE Add Maths 0606 / Topic 9 / IGCSE

Sector minus triangle: the segment and its relatives.

Exam questions on circular measure rarely stop at one sector. They shade a segment, the gap between a tangent and an arc, or the overlap of two circles, and ask for its area and perimeter. Every one of these is a sector and a triangle added or taken away, with the angle in radians and the triangle handled by trigonometry.

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The key idea

Split every shaded region into sectors and triangles. A segment has area \(\tfrac{1}{2}r^{2}(\theta - \sin\theta)\) (sector minus triangle) and perimeter \(r\theta + 2r\sin\tfrac{\theta}{2}\) (arc plus chord). A tangent meets the radius at a right angle, so the right-angled triangle gives lengths by \(\tan\) and \(\cos\), and an unknown angle by an inverse trig function, which must be in radians. Keep \(\pi\) and surds for exact answers; otherwise give 3 significant figures.

O A B C r θ shaded tangent at A AC = r tan θ OC = r ÷ cos θ BC = OC − r arc AB = rθ area = ½r² tan θ − ½r²θ
Fig. 1: sector OAB with the tangent at A meeting OB extended at C. Angle OAC is a right angle, so triangle OAC gives AC and OC by trigonometry. The shaded region is triangle OAC minus sector OAB; its perimeter is AC + BC + arc AB
Section 01

The segment: sector minus triangle.

A chord \(AB\) cuts a circle into two segments. The minor segment is the sector \(OAB\) with the isosceles triangle \(OAB\) removed. The triangle has two sides \(r\) with angle \(\theta\) between them, so its area is \(\tfrac{1}{2}r^{2}\sin\theta\) from the formula \(\tfrac{1}{2}ab\sin C\). The formulas are not given in the exam, so learn how to build them.

Worked example Segment to 3 significant figures
A circle has radius 8 cm. The chord \(AB\) subtends an angle of 1.2 radians at the centre \(O\). Find the area and the perimeter of the minor segment.
  1. Set the calculator to radians.
  2. Area: \(\tfrac{1}{2}(8^{2})(1.2 - \sin 1.2) = 32(1.2 - 0.93204) = 8.5747\).
  3. Arc: \(8 \times 1.2 = 9.6\). Chord: \(2(8)\sin 0.6 = 16 \times 0.56464 = 9.0343\).
  4. Perimeter: \(9.6 + 9.0343 = 18.634\).
Answer: area \(8.57\) cm2, perimeter \(18.6\) cm
Worked example Exact segment with an angle of pi/3
In a circle of radius 6 cm the chord \(PQ\) subtends an angle of \(\tfrac{\pi}{3}\) at the centre. Find the exact area and the exact perimeter of the minor segment.
  1. Sector: \(\tfrac{1}{2}(36)\tfrac{\pi}{3} = 6\pi\). Triangle: \(\tfrac{1}{2}(36)\sin\tfrac{\pi}{3} = 18 \times \tfrac{\sqrt{3}}{2} = 9\sqrt{3}\).
  2. Area: \(6\pi - 9\sqrt{3}\).
  3. With an angle of \(\tfrac{\pi}{3}\) and two equal sides, the triangle is equilateral, so the chord is 6 cm. (Check: \(2(6)\sin\tfrac{\pi}{6} = 6\).)
  4. Arc: \(6 \times \tfrac{\pi}{3} = 2\pi\).
Answer: area \((6\pi - 9\sqrt{3})\) cm2, perimeter \((2\pi + 6)\) cm
Your turn
Find the area and the perimeter of the minor segment cut off by a chord that subtends 2 radians at the centre of a circle of radius 4 cm.
Examiner trap

A calculator in degree mode gives \(\sin 1.2 = 0.0209\), and the segment area comes out almost equal to the sector area. If your segment is nearly as big as the sector, check the mode. Also, the chord uses HALF the angle: \(2r\sin\tfrac{\theta}{2}\), not \(2r\sin\theta\).

Section 02

Tangents and right angles.

A tangent at \(A\) is perpendicular to the radius \(OA\). In Fig. 1 that makes triangle \(OAC\) right-angled at \(A\), with the angle \(\theta\) at \(O\). So \(AC = r\tan\theta\) and \(OC = \dfrac{r}{\cos\theta}\). When the lengths are given instead, find the angle from \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\) (or \(\tan\) or \(\sin\)) with the calculator in radians.

Worked example Region between a tangent and an arc
In Fig. 1, \(OA = 5\) cm and angle \(AOB = 1.1\) radians. Find the area and the perimeter of the shaded region.
  1. \(AC = 5\tan 1.1 = 9.8238\) and \(OC = \dfrac{5}{\cos 1.1} = 11.0230\), so \(BC = 11.0230 - 5 = 6.0230\).
  2. Triangle \(OAC\): \(\tfrac{1}{2}(5)(9.8238) = 24.5595\). Sector \(OAB\): \(\tfrac{1}{2}(25)(1.1) = 13.75\).
  3. Area: \(24.5595 - 13.75 = 10.8095\).
  4. Perimeter: \(AC + BC + \text{arc } AB = 9.8238 + 6.0230 + 5.5 = 21.347\).
Answer: area \(10.8\) cm2, perimeter \(21.3\) cm
Worked example Find the angle first
A tangent from the point \(C\) touches a circle, centre \(O\), radius 6 cm, at \(A\). The distance \(OC\) is 10 cm, and \(OC\) meets the circle at \(B\). Find angle \(AOC\) in radians, then the area and perimeter of the region bounded by \(AC\), \(CB\) and the arc \(AB\).
  1. Angle \(OAC = \tfrac{\pi}{2}\), so \(\cos\theta = \dfrac{6}{10}\), giving \(\theta = 0.92730\) radians.
  2. Pythagoras: \(AC = \sqrt{100 - 36} = 8\). Also \(CB = 10 - 6 = 4\).
  3. Area: \(\tfrac{1}{2}(6)(8) - \tfrac{1}{2}(36)(0.92730) = 24 - 16.691 = 7.3087\).
  4. Perimeter: \(8 + 4 + 6(0.92730) = 17.564\).
Answer: \(\theta = 0.927\) rad, area \(7.31\) cm2, perimeter \(17.6\) cm
Your turn
A tangent at \(A\) to a circle, centre \(O\), radius 4 cm, has length \(AC = 7\) cm. Find angle \(AOC\) in radians.
Examiner trap

Keep at least 5 significant figures in \(\theta\) and in every length until the last line. Rounding \(\theta\) to 0.93 in the second example gives an area of 7.26, which loses the accuracy mark. And \(BC\) is \(OC - r\), not \(OC\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Find the area of the minor segment of a circle of radius 10 cm cut off by a chord subtending 1 radian at the centre.

Quick check+10 XP

A chord subtends 2 radians at the centre of a circle of radius 5 cm. How long is the chord?

Quick check+10 XP

Find the exact area of the minor segment of a circle of radius 4 cm when the angle at the centre is \(\tfrac{\pi}{2}\).

Quick check+10 XP

The tangent at \(A\) to a circle, centre \(O\), radius 3 cm, meets the line \(OB\) extended at \(C\). Angle \(AOC = 1.2\) radians. Find \(AC\).

Section 03

Overlaps and cut-outs.

Bigger shapes follow the same plan: sketch, label every length and angle, then write the shaded region as a sum or difference of sectors, triangles and rectangles BEFORE you calculate anything. The overlap of two circles is two segments back to back. A sector removed from a triangle leaves the triangle's area \(\tfrac{1}{2}ab\sin C\) minus the sector.

Worked example Two overlapping circles, exact
Two circles, each of radius 6 cm, have centres \(P\) and \(Q\) with \(PQ = 6\) cm. Find the exact area of the region common to both circles.
  1. The circles meet at \(X\) and \(Y\). Triangles \(PQX\) and \(PQY\) have all sides 6, so they are equilateral and angle \(XPY = \tfrac{2\pi}{3}\).
  2. The common region is two equal segments, each with \(r = 6\) and \(\theta = \tfrac{2\pi}{3}\).
  3. One segment: \(\tfrac{1}{2}(36)\left(\tfrac{2\pi}{3} - \sin\tfrac{2\pi}{3}\right) = 18\left(\tfrac{2\pi}{3} - \tfrac{\sqrt{3}}{2}\right) = 12\pi - 9\sqrt{3}\).
  4. Double it.
Answer: \((24\pi - 18\sqrt{3})\) cm2 (about 44.2 cm2)
Worked example A sector cut from a triangle
Triangle \(ABC\) has \(AB = 9\) cm, \(AC = 8\) cm and angle \(BAC = 1.1\) radians. A sector of radius 5 cm, centre \(A\), is removed. Find the area and the perimeter of the region that remains.
  1. Triangle: \(\tfrac{1}{2}(9)(8)\sin 1.1 = 36(0.89121) = 32.083\). Sector: \(\tfrac{1}{2}(25)(1.1) = 13.75\).
  2. Area: \(32.083 - 13.75 = 18.333\).
  3. Cosine rule: \(BC^{2} = 81 + 64 - 2(9)(8)\cos 1.1 = 145 - 144(0.45360) = 79.682\), so \(BC = 8.9265\).
  4. Perimeter: \(BC + (9 - 5) + (8 - 5) + 5(1.1) = 8.9265 + 4 + 3 + 5.5 = 21.427\).
Answer: area \(18.3\) cm2, perimeter \(21.4\) cm
Examiner trap

The perimeter of a region is only its OUTSIDE edge. In the cut-out, the two radii of the removed sector are inside the original triangle and are not part of the boundary; only the parts \(AB - 5\) and \(AC - 5\) and the arc count.

Stage 2 · Exam

Exam-style questions

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A chord subtends an angle of 1.4 radians at the centre of a circle of radius 9 cm. Find the perimeter of the minor segment.

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A tangent from \(C\) touches a circle, centre \(O\), radius 5 cm, at \(A\), and \(OC = 13\) cm. \(OC\) meets the circle at \(B\). Find the area of the region bounded by \(AC\), \(CB\) and the arc \(AB\).

Exam style+20 XP

Sector \(OPQ\) has radius 6 cm and angle \(\tfrac{\pi}{3}\). Find the exact perimeter of the segment cut off by the chord \(PQ\).

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Two circles of radius 6 cm each pass through the other's centre. Find the exact area common to both circles.

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Compound shapes split and solved.

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Go deeper · practice
Six original Cambridge-style questions
Segments to 3 s.f. and exactly, the region between a tangent and an arc, an angle found from a right-angled triangle, an arc across a rectangle and two overlapping circles.