Segments and composite regions

AS · P1 · 13 min

Most circular measure questions on Paper 1 are not a single sector. They show a diagram built from sectors, triangles, segments and tangents, and ask for the perimeter or area of a shaded region, often in exact form. The formulas are those from Radians, arc length and sector area plus a little triangle trigonometry; the skill is seeing how the region is made from simple pieces. This note builds the segment formulas and then works through the standard composite diagrams.

Triangles inside circles

Join the ends $A$ and $B$ of an arc to the centre $O$ and to each other. Triangle $OAB$ has two sides equal to the radius $r$ and the angle $\theta$ between them. Two facts from GCSE trigonometry do the work here.

Area of the triangle. The area of a triangle with sides $a$ and $b$ and included angle $C$ is $\tfrac{1}{2}ab\sin C$ . So

\text{area of triangle } OAB = \tfrac{1}{2}r^2\sin\theta

Length of the chord. The perpendicular from $O$ to $AB$ bisects both the chord and the angle, making two right-angled triangles with hypotenuse $r$ and angle $\tfrac{\theta}{2}$ at $O$ . The half-chord is $r\sin\tfrac{\theta}{2}$ , so

AB = 2r\sin\tfrac{\theta}{2}

The cosine rule gives the same thing: $AB^2 = r^2 + r^2 - 2r^2\cos\theta$ . Use whichever you find quicker.

Segments

A segment is the region between a chord and its arc. The minor segment is the smaller one (for $\theta < \pi$ ); the major segment is the rest of the circle.

x^2 + y^2 = 25 (0, 0) -- (5, 0) (0, 0) -- (5*cos(2pi/3), 5*sin(2pi/3)) (5, 0) -- (5*cos(2pi/3), 5*sin(2pi/3))

The minor segment is the sector minus the triangle:

\tfrac{1}{2}r^2\theta - \tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2}r^2(\theta - \sin\theta)

Key result

For a chord subtending angle $\theta$ radians at the centre of a circle of radius $r$ :

\text{chord } AB = 2r\sin\tfrac{\theta}{2}, \qquad \text{triangle } OAB = \tfrac{1}{2}r^2\sin\theta

\text{minor segment area} = \tfrac{1}{2}r^2(\theta - \sin\theta), \qquad \text{segment perimeter} = r\theta + 2r\sin\tfrac{\theta}{2}

Major segment area $= \pi r^2 -$ minor segment area.

These are not in the formula list, and examiners are just as happy with "sector minus triangle" written out. Make sure your calculator is in radian mode for $\sin\theta$ .

Exact values

Many diagrams are built so that the angles are $\tfrac{\pi}{6}$ , $\tfrac{\pi}{4}$ , $\tfrac{\pi}{3}$ , $\tfrac{\pi}{2}$ or $\tfrac{2\pi}{3}$ , and the question asks for an exact answer such as $12\pi - 9\sqrt{3}$ . You need the exact values of sine, cosine and tangent of these angles (see Exact values and angles of any size), and the facts that produce them:

If $OA = OB = AB$ , triangle $OAB$ is equilateral, so angle $AOB = \tfrac{\pi}{3}$ .
If $\cos\angle AOT = \tfrac{1}{2}$ (for example $OA = r$ and $OT = 2r$ in a right-angled triangle), the angle is $\tfrac{\pi}{3}$ .
A tangent is perpendicular to the radius at the point of contact.
The angle in a semicircle is a right angle.

Breaking down a composite region

Area or perimeter of a shaded region

Copy the key lengths and angles onto the diagram. Mark every radius: equal radii create isosceles and equilateral triangles.
Look for right angles: tangent and radius, angle in a semicircle, a perpendicular dropped from the centre.
Find every angle you need, in radians.
Write the shaded area as a sum or difference of sectors, triangles and segments. Write the sum in words first, such as "triangle $OAT$ minus sector $OAB$ ".
For a perimeter, list every boundary piece: each arc ( $r\theta$ , with the correct radius and angle) and each straight piece.
Calculate, keeping exact values until the end.

The most common structures:

Shaded region	Area
between a chord and its arc	sector $-$ triangle
between a tangent, a line through the centre and an arc	right-angled triangle $-$ sector
between two tangents from a point and the minor arc	kite (two right-angled triangles) $-$ sector
common to two overlapping circles	the sum of two segments, one from each circle
inside a sector but outside a polygon	sector $-$ polygon

Worked examples

A minor segment in exact form

A chord $AB$ of a circle with centre $O$ and radius $6$ cm subtends an angle of $\tfrac{2\pi}{3}$ at $O$ . Find the exact area and the exact perimeter of the minor segment.

Solution

\text{Area} = \tfrac{1}{2}(6^2)\left(\frac{2\pi}{3} - \sin\frac{2\pi}{3}\right) = 18\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = (12\pi - 9\sqrt{3})\ \text{cm}^2

Arc $= 6 \times \tfrac{2\pi}{3} = 4\pi$ . Chord $= 2(6)\sin\tfrac{\pi}{3} = 12 \times \tfrac{\sqrt{3}}{2} = 6\sqrt{3}$ .

\text{Perimeter} = (4\pi + 6\sqrt{3})\ \text{cm}

The angle from a chord

A chord of length $8$ cm is drawn in a circle of radius $5$ cm. Find the angle subtended at the centre and the area of the minor segment, giving your answers to 3 significant figures.

Solution

The perpendicular from the centre bisects the chord, giving a right-angled triangle with hypotenuse $5$ and opposite side $4$ :

\sin\frac{\theta}{2} = \frac{4}{5} \quad\Rightarrow\quad \frac{\theta}{2} = 0.9273 \quad\Rightarrow\quad \theta = 1.855 \text{ radians}

\text{Segment} = \tfrac{1}{2}(25)(1.8546 - \sin 1.8546) = 12.5(1.8546 - 0.96) = 11.2\ \text{cm}^2

( $\sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2} = 2 \times 0.8 \times 0.6 = 0.96$ , which is a useful check.)

A perpendicular inside a sector

$OAB$ is a sector of a circle with centre $O$ , radius $8$ cm and angle $AOB = \tfrac{\pi}{3}$ . The point $C$ on $OB$ is such that $AC$ is perpendicular to $OB$ . Find the exact area and the exact perimeter of the region bounded by the arc $AB$ and the lines $BC$ and $CA$ .

Solution

In the right-angled triangle $OCA$ , with hypotenuse $OA = 8$ :

OC = 8\cos\frac{\pi}{3} = 4, \qquad AC = 8\sin\frac{\pi}{3} = 4\sqrt{3}, \qquad CB = 8 - 4 = 4

Region $=$ sector $OAB$ $-$ triangle $OCA$ :

\tfrac{1}{2}(64)\left(\frac{\pi}{3}\right) - \tfrac{1}{2}(4)(4\sqrt{3}) = \left(\frac{32\pi}{3} - 8\sqrt{3}\right)\ \text{cm}^2

Perimeter $=$ arc $AB$ $+$ $BC$ $+$ $CA$ :

\frac{8\pi}{3} + 4 + 4\sqrt{3}\ \text{cm}

A tangent and a sector

A circle has centre $O$ and radius $10$ cm. The points $A$ and $B$ lie on the circle with angle $AOB = 0.8$ radians. The tangent at $A$ meets $OB$ extended at $T$ . Find the area and the perimeter of the region bounded by $AT$ , $TB$ and the minor arc $AB$ .

Solution

The tangent is perpendicular to the radius, so triangle $OAT$ has a right angle at $A$ :

AT = 10\tan 0.8 = 10.296, \qquad OT = \frac{10}{\cos 0.8} = 14.353, \qquad BT = OT - 10 = 4.353

\text{Area} = \text{triangle } OAT - \text{sector } OAB = \tfrac{1}{2}(10)(10.296) - \tfrac{1}{2}(100)(0.8) = 51.48 - 40 = 11.5\ \text{cm}^2

\text{Perimeter} = AT + TB + \text{arc } AB = 10.296 + 4.353 + 8 = 22.6\ \text{cm}

x^2 + y^2 = 100 (0, 0) -- (10, 0) (10, 0) -- (10, 10*tan(0.8)) (0, 0) -- (10, 10*tan(0.8))

Two tangents from a point

The tangents from a point $T$ touch a circle, centre $O$ and radius $6$ cm, at $A$ and $B$ . Angle $AOB = \tfrac{2\pi}{3}$ . Find the exact area of the region bounded by $TA$ , $TB$ and the minor arc $AB$ , and its exact perimeter.

Solution

By symmetry $OT$ bisects angle $AOB$ , so angle $AOT = \tfrac{\pi}{3}$ . Triangle $OAT$ is right-angled at $A$ :

AT = 6\tan\frac{\pi}{3} = 6\sqrt{3}

The kite $OATB$ is two such triangles: area $= 2 \times \tfrac{1}{2}(6)(6\sqrt{3}) = 36\sqrt{3}$ .

Sector $OAB = \tfrac{1}{2}(36)\left(\tfrac{2\pi}{3}\right) = 12\pi$ .

\text{Area} = (36\sqrt{3} - 12\pi)\ \text{cm}^2

\text{Perimeter} = 2 \times 6\sqrt{3} + 6 \times \frac{2\pi}{3} = (12\sqrt{3} + 4\pi)\ \text{cm}

Arcs centred on the vertices of a triangle

$ABC$ is an equilateral triangle of side $6$ cm. Three arcs are drawn: arc $BC$ with centre $A$ , arc $CA$ with centre $B$ , and arc $AB$ with centre $C$ , each of radius $6$ cm. Find the exact perimeter and area of the region enclosed by the three arcs.

Solution

Each arc has radius $6$ and subtends the angle of the equilateral triangle, $\tfrac{\pi}{3}$ , at its centre. Each arc length is $6 \times \tfrac{\pi}{3} = 2\pi$ , so

\text{Perimeter} = 3 \times 2\pi = 6\pi\ \text{cm}

The region is the triangle plus three equal segments, one on each side.

\text{Triangle} = \tfrac{1}{2}(6)(6)\sin\frac{\pi}{3} = 9\sqrt{3}, \qquad \text{each segment} = \tfrac{1}{2}(36)\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) = 6\pi - 9\sqrt{3}

\text{Area} = 9\sqrt{3} + 3\left(6\pi - 9\sqrt{3}\right) = (18\pi - 18\sqrt{3})\ \text{cm}^2

The region common to two circles

Two circles, each of radius $r$ , have centres $P$ and $Q$ with $PQ = r$ . Find, in terms of $r$ , the exact area of the region common to both circles.

Solution

Let the circles meet at $A$ and $B$ . Then $PA = PQ = QA = r$ , so triangle $PQA$ is equilateral and angle $APQ = \tfrac{\pi}{3}$ . By symmetry angle $BPQ = \tfrac{\pi}{3}$ too, so angle $APB = \tfrac{2\pi}{3}$ , and likewise angle $AQB = \tfrac{2\pi}{3}$ .

The chord $AB$ splits the common region into two equal segments, one from each circle, each with angle $\tfrac{2\pi}{3}$ :

2 \times \tfrac{1}{2}r^2\left(\frac{2\pi}{3} - \sin\frac{2\pi}{3}\right) = r^2\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)

x^2 + y^2 = 1 (x - 1)^2 + y^2 = 1 (0.5, -sqrt(3)/2) -- (0.5, sqrt(3)/2)

Watch out

Degree mode. $\sin\tfrac{2\pi}{3}$ in degree mode is $0.0366$ , not $0.866$ . For segments, check the mode before every calculation.

Using the wrong angle. In a tangent diagram, the angle in the right-angled triangle is often half the angle at the centre. In the overlapping circles, the angle at each centre is $\tfrac{2\pi}{3}$ , not $\tfrac{\pi}{3}$ .

Missing pieces of a perimeter. List the boundary in order round the region. Straight pieces such as $BT = OT - r$ are easy to forget.

Wrong radius for an arc. In a diagram with arcs centred at different points, each arc uses its own radius and its own angle.

Rounding early. Keep $\tan 0.8$ to at least 4 significant figures until the final line, or the third figure of the answer may be wrong.

Exam tip

Write the plan in words. "Area $=$ triangle $OAT$ $-$ sector $OAB$ " earns the method mark even if a later number slips.
Exact answers. "Exact" means leave $\pi$ and surds in, simplified: $12\pi - 9\sqrt{3}$ . Never convert to a decimal.
Show angles. If an angle comes from an equilateral triangle or from $\cos^{-1}\tfrac{1}{2}$ , say so. "Show that angle $AOB = \tfrac{2\pi}{3}$ " needs a reason, not a calculator value.
Typical marks. A composite question is usually 6 to 8 marks: one or two for an angle or length, two or three for a perimeter, three for an area.
Accuracy. Unless told otherwise, 3 significant figures. Carry 4 or more figures through intermediate steps.

Summary

Triangle $OAB$ : area $\tfrac{1}{2}r^2\sin\theta$ ; chord $AB = 2r\sin\tfrac{\theta}{2}$ .
Minor segment area $\tfrac{1}{2}r^2(\theta - \sin\theta)$ ; perimeter $r\theta + 2r\sin\tfrac{\theta}{2}$ .
Tangent $\perp$ radius gives right-angled triangles; two tangents from a point form a kite.
Equal radii create isosceles and equilateral triangles; an equilateral triangle gives $\tfrac{\pi}{3}$ .
Break every shaded region into sectors, triangles and segments; write the plan in words first.
Radian mode, exact values where asked, and full accuracy until the final answer.

Practice questions

Question

A chord of a circle of radius $6$ cm subtends an angle of $1$ radian at the centre. Find the area of the minor segment.
A chord of a circle of radius $10$ cm subtends a right angle at the centre. Find the exact area and exact perimeter of the minor segment.
A chord of length $12$ cm is drawn in a circle of radius $10$ cm. Find the area of the minor segment.
A chord of a circle of radius $4$ cm subtends an angle of $\tfrac{\pi}{3}$ at the centre. Find the exact area of the major segment.
A chord $AB$ of a circle with radius $7$ cm subtends an angle of $2.2$ radians at the centre. Find the perimeter of the minor segment.
$AB$ is a diameter of a circle of radius $5$ cm, and $C$ is a point on the circle with angle $ABC = \tfrac{\pi}{6}$ . Find the exact area of the region inside the semicircle on the side of $C$ but outside triangle $ABC$ .
Two circles each have radius $5$ cm, and their centres are $5\sqrt{2}$ cm apart. Show that each centre subtends a right angle at the common chord, and find the exact area common to the two circles.
A point $P$ is $8\sqrt{2}$ cm from the centre $O$ of a circle of radius $8$ cm. Tangents from $P$ touch the circle at $A$ and $B$ . Find angle $AOB$ , and the exact area and perimeter of the region bounded by $PA$ , $PB$ and the minor arc $AB$ .
$OAC$ is a sector of a circle with centre $O$ , radius $6$ cm and angle $AOC = \tfrac{2\pi}{3}$ . The point $B$ is such that $OABC$ is a rhombus. (a) Show that $B$ lies on the arc $AC$ . (b) Find the exact area of the region inside the sector but outside the rhombus.
$OAB$ is a sector with centre $O$ , radius $r$ and angle $\tfrac{\pi}{3}$ . A circle is drawn inside the sector touching $OA$ , $OB$ and the arc $AB$ . (a) Show that the radius of this circle is $\tfrac{r}{3}$ . (b) Given that $r = 9$ cm, find the exact area of the region of the sector outside the circle.

Answers

$\tfrac{1}{2}(36)(1 - \sin 1) = 18(1 - 0.8415) = 2.85\ \text{cm}^2$ .
Area $= \tfrac{1}{2}(100)\left(\tfrac{\pi}{2} - 1\right) = (25\pi - 50)\ \text{cm}^2$ . Arc $= 10 \times \tfrac{\pi}{2} = 5\pi$ ; chord $= 2(10)\sin\tfrac{\pi}{4} = 10\sqrt{2}$ . Perimeter $= (5\pi + 10\sqrt{2})$ cm.
$\sin\tfrac{\theta}{2} = \tfrac{6}{10}$ , so $\theta = 2\sin^{-1}0.6 = 1.287$ . Segment $= \tfrac{1}{2}(100)(1.287 - \sin 1.287) = 50(1.2870 - 0.96) = 16.4\ \text{cm}^2$ .
Minor segment $= \tfrac{1}{2}(16)\left(\tfrac{\pi}{3} - \tfrac{\sqrt{3}}{2}\right) = \tfrac{8\pi}{3} - 4\sqrt{3}$ . Major segment $= 16\pi - \tfrac{8\pi}{3} + 4\sqrt{3} = \left(\tfrac{40\pi}{3} + 4\sqrt{3}\right)\ \text{cm}^2$ .
Arc $= 7 \times 2.2 = 15.4$ ; chord $= 2(7)\sin 1.1 = 12.48$ . Perimeter $= 27.9$ cm.
Angle $ACB = \tfrac{\pi}{2}$ (angle in a semicircle). $AC = 10\sin\tfrac{\pi}{6} = 5$ and $BC = 10\cos\tfrac{\pi}{6} = 5\sqrt{3}$ , so the triangle has area $\tfrac{25\sqrt{3}}{2}$ . The semicircle has area $\tfrac{25\pi}{2}$ . Region $= \dfrac{25\pi - 25\sqrt{3}}{2}\ \text{cm}^2$ .
Each centre, the other centre and an intersection point form a triangle with sides $5$ , $5$ and $5\sqrt{2}$ . Since $5^2 + 5^2 = (5\sqrt{2})^2$ , the angle at an intersection point is $\tfrac{\pi}{2}$ and the base angles are $\tfrac{\pi}{4}$ , so each centre subtends $2 \times \tfrac{\pi}{4} = \tfrac{\pi}{2}$ at the chord. Common area $= 2 \times \tfrac{1}{2}(25)\left(\tfrac{\pi}{2} - 1\right) = \left(\tfrac{25\pi}{2} - 25\right)\ \text{cm}^2$ .
$\cos\angle AOP = \tfrac{8}{8\sqrt{2}} = \tfrac{1}{\sqrt{2}}$ , so $\angle AOP = \tfrac{\pi}{4}$ and $\angle AOB = \tfrac{\pi}{2}$ . $PA = 8\tan\tfrac{\pi}{4} = 8$ . Kite $= 2 \times \tfrac{1}{2}(8)(8) = 64$ ; sector $= \tfrac{1}{2}(64)\left(\tfrac{\pi}{2}\right) = 16\pi$ . Area $= (64 - 16\pi)\ \text{cm}^2$ . Perimeter $= 8 + 8 + 8 \times \tfrac{\pi}{2} = (16 + 4\pi)$ cm.
(a) In the rhombus, $OA = AB = 6$ and angle $OAB = \pi - \tfrac{2\pi}{3} = \tfrac{\pi}{3}$ (adjacent angles of a parallelogram add to $\pi$ ). So triangle $OAB$ is isosceles with apex angle $\tfrac{\pi}{3}$ , hence equilateral, and $OB = 6$ : $B$ is on the arc. (b) Rhombus area $= 6 \times 6 \times \sin\tfrac{2\pi}{3} = 18\sqrt{3}$ . Sector area $= \tfrac{1}{2}(36)\left(\tfrac{2\pi}{3}\right) = 12\pi$ . Region $= (12\pi - 18\sqrt{3})\ \text{cm}^2$ .
(a) Let the small circle have centre $P$ and radius $\rho$ . By symmetry $P$ lies on the bisector of angle $AOB$ , at angle $\tfrac{\pi}{6}$ to $OA$ . The radius to the point of contact with $OA$ is perpendicular to $OA$ , so $\rho = OP\sin\tfrac{\pi}{6} = \tfrac{1}{2}OP$ . The circle also touches the arc, so $OP + \rho = r$ . Then $OP = 2\rho$ gives $3\rho = r$ , so $\rho = \tfrac{r}{3}$ . (b) Sector $= \tfrac{1}{2}(81)\left(\tfrac{\pi}{3}\right) = \tfrac{27\pi}{2}$ . Circle $= \pi(3^2) = 9\pi$ . Region $= \tfrac{27\pi}{2} - 9\pi = \tfrac{9\pi}{2}\ \text{cm}^2$ .

Triangles inside circles

Segments

Exact values

Breaking down a composite region

Worked examples

Practice questions

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