Mathematics 9709 · Pure Mathematics 1

Area of a sector and a segment

A sector is the pizza-slice shape between two radii and an arc.
A segment is the thin shape a chord cuts off. It sits between the straight chord and the curved arc.
Once you have the sector area, the segment is just one subtraction.

Sector area is $\tfrac12 r^2\theta$ . A segment is the sector with the triangle taken away. This gives $\tfrac12 r^2(\theta-\sin\theta)$ , with $\theta$ in radians.

The two big formulas are the same idea twice. A sector is the fraction $\tfrac{\theta}{2\pi}$ of the whole circle. Take that fraction of the edge $2\pi r$ and you get the arc $s=r\theta$ . Take that fraction of the area $\pi r^2$ and you get the sector $\tfrac12 r^2\theta$ . Learn one and you can work out the other.

See all three move together

Drag $\theta$ and $r$ : the arc and sector grow in step with $\theta$ , while the segment (green) stays small for small angles and only gets bigger as the chord moves away from the arc.

radius rangle θ

θ = 1.40 radarc s = 8.4sector 25.1segment 7.4

arc s = rθ, sector ½r²θ, segment ½r²(θ − sinθ) — all need θ in radians.

Area of a sector: ½r²θ

A sector is a fraction of the circle: angle $\theta$ out of the full turn $2\pi$ .
Take that fraction of the area $\pi r^2$ and the $2\pi$ cancels. This leaves a simple formula (with $\theta$ in radians, from §4.1):

$A_{\text{sector}}=\tfrac12 r^2\theta\qquad(\theta\text{ in radians})$

Where ½r²θ comes from (the fraction of the circle)

The full circle is $2\pi$ radians and has area $\pi r^2$ . A sector of angle $\theta$ is the fraction $\dfrac{\theta}{2\pi}$ of that whole circle. So:

$A=\dfrac{\theta}{2\pi}\times\pi r^2=\dfrac{\theta\,\pi r^2}{2\pi}=\tfrac12 r^2\theta$

The $2\pi$ on the bottom cancels the $\pi$ on top. That is why radians make it so simple. It is the twin of the arc length $s=r\theta$ from §4.2: same fraction, used on area instead of the edge.

Worked example

Sector of radius $9$ cm and angle $\tfrac{\pi}{6}$ , exact answer in terms of $\pi$

The angle is already in radians, so put it straight into $\tfrac12 r^2\theta$ and keep the $\pi$ in:

$A=\tfrac12\times 9^2\times\dfrac{\pi}{6}=\tfrac12\times 81\times\dfrac{\pi}{6}$

$A=\dfrac{81\pi}{12}=\dfrac{27\pi}{4}\text{ cm}^2$

Answer

A=\tfrac{27\pi}{4}\text{ cm}^2

The segment = sector − triangle

The one move for every segment question: the segment is the sector with the triangle of the two radii taken away.
So find the sector, find that triangle, and subtract.

Segment (gold) = sector − triangle. The triangle of the two radii sits below the chord. The segment is the thin piece of the sector left above it.

The triangle's two sides are both $r$ , with the angle $\theta$ between them. So its area uses the $\tfrac12 ab\sin C$ rule from Trigonometry with $a=b=r$ and $C=\theta$ :

$A_\triangle=\tfrac12 r^2\sin\theta$

Subtract the triangle from the sector:

$A_{\text{seg}}=\tfrac12 r^2\theta-\tfrac12 r^2\sin\theta=\tfrac12 r^2(\theta-\sin\theta)$

Use the formula in brackets, or just work out the two areas and subtract. Work line by line so one small slip does not spoil the rest:

Worked example

Real exam segment: arc $9.6$ cm, sector area $76.8\text{ cm}^2$ , find the shaded segment

From §4.2 the arc and sector gave $r=16$ cm and $\theta=0.6$ rad. The shaded region is the segment, so it is sector minus triangle.

Sector (already given, but here it is from the formula):
$A_{\text{sector}}=\tfrac12(16^2)(0.6)=76.8\text{ cm}^2$
Triangle of the two radii (keep the calculator in radian mode):
$A_\triangle=\tfrac12(16^2)\sin 0.6=72.27\ldots\text{ cm}^2$
Subtract, rounding only at the end:
$A_{\text{seg}}=76.8-72.27\ldots=4.53\text{ cm}^2$

Answer

A_{\text{seg}}\approx 4.53\text{ cm}^2

(9709/13 May/Jun 2023 Q6) If your calculator gives a very big number for $\sin 0.6$ , it is in degree mode. Fix that first.

The length of the chord: cosine rule

To find the chord $AB$ , look at the triangle $AOB$ : two sides are radii of length $r$ with the angle $\theta$ between them, so use the cosine rule (from Trigonometry):

$AB^2=r^2+r^2-2r^2\cos\theta=2r^2(1-\cos\theta)$

Take the square root to get the chord itself.
This is the method the exam asks for, and it works for any angle, so use it first.

Worked example

Chord of a circle: radius $7$ cm, angle $AOB=2$ rad

Two radii with $\theta=2$ between them, so use the cosine rule:

$AB^2=7^2+7^2-2(7^2)\cos 2=98-98\cos 2$

Keep the calculator in radian mode for $\cos 2$ :

$AB^2=98-98(-0.4161\ldots)=138.78\ldots$

Answer

AB=\sqrt{138.78\ldots}\approx 11.8\text{ cm}

A faster shortcut for the chord: 2r sin(θ/2)

Draw a line from the centre straight down to the chord. It cuts both the chord and the angle in half. This splits $AOB$ into two right-angled triangles, each with angle $\tfrac{\theta}{2}$ . So half the chord is $r\sin\tfrac{\theta}{2}$ , and the whole chord is

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

This gives the same answer as the cosine rule and is faster. But the cosine rule is the one the textbook teaches, so use that first and use this as a check.

The perimeter of a segment: arc + chord

A segment is made of the arc and the chord, not the two radii, so its perimeter is $\text{arc}+\text{chord}$ . The arc is $r\theta$ , and the chord comes from the cosine rule above (or the $2r\sin\tfrac{\theta}{2}$ shortcut):

$P_{\text{seg}}=r\theta+AB,\quad AB=\sqrt{2r^2(1-\cos\theta)}$

The same exam question above also asks for this perimeter. With $r=16$ and $\theta=0.6$ , the arc is $9.6$ (given) and the chord is $2(16)\sin 0.3=9.46\ldots$ , so $P=9.6+9.46\ldots\approx19.1\text{ cm}$ . (9709/13 May/Jun 2023 Q6(b))

Compare with the sector perimeter from §4.2, $r\theta+2r$ (arc plus two radii).
Different region, different straight edge: read the picture before you pick a formula.

“Show that” with exact values

One common type gives an exact angle and asks you to show that a length or area equals a printed answer in $r$ , $\pi$ and surds.
No rounding: every step stays exact, and you must reach the printed answer to get the marks.
Use the exact values $\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$ , $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ , $\sin\tfrac{\pi}{6}=\tfrac12$ .

Worked example

Real “show that”: sector radius $2r$ , angle $\tfrac{\pi}{6}$ , $C$ the midpoint of $OA$ — show $BC=r\sqrt{5-2\sqrt3}$

The radius is $OB=OA=2r$ . Since $C$ is the midpoint of $OA$ , we get $OC=r$ . Triangle $OBC$ has those two sides with the angle $\tfrac{\pi}{6}$ between them, so use the cosine rule on $BC$ :

Cosine rule with sides $2r$ and $r$ and the included angle $\tfrac{\pi}{6}$ :
$BC^2=(2r)^2+r^2-2(2r)(r)\cos\dfrac{\pi}{6}$
Put in the exact cosine $\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$ ; the 4 and the $\tfrac12$ cancel cleanly:
$BC^2=4r^2+r^2-4r^2\cdot\dfrac{\sqrt3}{2}=5r^2-2\sqrt3\,r^2$
Factor and root, keeping it exact:
$BC^2=r^2(5-2\sqrt3)\ \Rightarrow\ BC=r\sqrt{5-2\sqrt3}$

Answer

BC=r\sqrt{5-2\sqrt3}\quad\text{as required}

(9709/12 May/Jun 2020 Q7(a)) The whole skill is exact values, no calculator. A decimal anywhere loses the “show that” marks.

The segment version: show that the area is r²(2π − 3√3)⁄12, with θ = π/3

Segment $=$ sector $-$ triangle, both kept exact. Sector $=\tfrac12 r^2\cdot\tfrac{\pi}{3}=\tfrac{\pi r^2}{6}$ ; triangle $=\tfrac12 r^2\sin\tfrac{\pi}{3}=\tfrac{\sqrt3\,r^2}{4}$ . Put both over a bottom of 12 and subtract:

$A_{\text{seg}}=\dfrac{2\pi r^2}{12}-\dfrac{3\sqrt3\,r^2}{12}=\dfrac{r^2(2\pi-3\sqrt3)}{12}$

The whole skill is common denominators with no calculator. It is the same move as the worked example, just one step longer. (9709-style)

Where the marks go

Common mistake

For the triangle you need

\tfrac12 r^2\sin\theta

, not

\tfrac12 r^2\theta

. The

\sin

is what makes it a triangle and not a second sector. And the calculator must be in radian mode for

\sin\theta

. This is the biggest slip in the topic. (9709/13 May/Jun 2023 Q6)

Common mistake

A segment's perimeter is the arc plus the chord, not arc plus two radii. The two radii make the sector, not the segment. Get the chord from the cosine rule

AB^2=2r^2(1-\cos\theta)

Common mistake

Do not round the sector or triangle too early. Keep the full triangle value (

72.27\ldots

) and only round the last subtraction. If you round early, a small segment can come out wrong in the second decimal place.

Now you try

A circle has centre $O$ and radius $6$ cm, $AB$ is a chord and angle $AOB=1.2$ radians. Find the area of the segment cut off by $AB$ . (9709-style)

Your turn— tap to reveal the worked answer (sector − triangle)

Sector: $\tfrac12(6^2)(1.2)=21.6$ .

Triangle (radian mode): $\tfrac12(6^2)\sin 1.2=16.78\ldots$ .

Segment: $21.6-16.78\ldots$ :

Answer

A_{\text{seg}}\approx 4.82\text{ cm}^2

PreviousLength of an arc NextApplications: shaded-region problems

Area of a sector and a segment

A sector is the pizza-slice shape between two radii and an arc.
A segment is the thin shape a chord cuts off. It sits between the straight chord and the curved arc.
Once you have the sector area, the segment is just one subtraction.

Sector area is $\tfrac12 r^2\theta$ . A segment is the sector with the triangle taken away. This gives $\tfrac12 r^2(\theta-\sin\theta)$ , with $\theta$ in radians.

See all three move together

Drag $\theta$ and $r$ : the arc and sector grow in step with $\theta$ , while the segment (green) stays small for small angles and only gets bigger as the chord moves away from the arc.

radius rangle θ

θ = 1.40 radarc s = 8.4sector 25.1segment 7.4

arc s = rθ, sector ½r²θ, segment ½r²(θ − sinθ) — all need θ in radians.

Area of a sector: ½r²θ

A sector is a fraction of the circle: angle $\theta$ out of the full turn $2\pi$ .
Take that fraction of the area $\pi r^2$ and the $2\pi$ cancels. This leaves a simple formula (with $\theta$ in radians, from §4.1):

$A_{\text{sector}}=\tfrac12 r^2\theta\qquad(\theta\text{ in radians})$

Where ½r²θ comes from (the fraction of the circle)

The full circle is $2\pi$ radians and has area $\pi r^2$ . A sector of angle $\theta$ is the fraction $\dfrac{\theta}{2\pi}$ of that whole circle. So:

$A=\dfrac{\theta}{2\pi}\times\pi r^2=\dfrac{\theta\,\pi r^2}{2\pi}=\tfrac12 r^2\theta$

The $2\pi$ on the bottom cancels the $\pi$ on top. That is why radians make it so simple. It is the twin of the arc length $s=r\theta$ from §4.2: same fraction, used on area instead of the edge.

Worked example

Sector of radius $9$ cm and angle $\tfrac{\pi}{6}$ , exact answer in terms of $\pi$

The angle is already in radians, so put it straight into $\tfrac12 r^2\theta$ and keep the $\pi$ in:

$A=\tfrac12\times 9^2\times\dfrac{\pi}{6}=\tfrac12\times 81\times\dfrac{\pi}{6}$

$A=\dfrac{81\pi}{12}=\dfrac{27\pi}{4}\text{ cm}^2$

Answer

A=\tfrac{27\pi}{4}\text{ cm}^2

The segment = sector − triangle

The one move for every segment question: the segment is the sector with the triangle of the two radii taken away.
So find the sector, find that triangle, and subtract.

Segment (gold) = sector − triangle. The triangle of the two radii sits below the chord. The segment is the thin piece of the sector left above it.

The triangle's two sides are both $r$ , with the angle $\theta$ between them. So its area uses the $\tfrac12 ab\sin C$ rule from Trigonometry with $a=b=r$ and $C=\theta$ :

$A_\triangle=\tfrac12 r^2\sin\theta$

Subtract the triangle from the sector:

$A_{\text{seg}}=\tfrac12 r^2\theta-\tfrac12 r^2\sin\theta=\tfrac12 r^2(\theta-\sin\theta)$

Use the formula in brackets, or just work out the two areas and subtract. Work line by line so one small slip does not spoil the rest:

Worked example

Real exam segment: arc $9.6$ cm, sector area $76.8\text{ cm}^2$ , find the shaded segment

From §4.2 the arc and sector gave $r=16$ cm and $\theta=0.6$ rad. The shaded region is the segment, so it is sector minus triangle.

Sector (already given, but here it is from the formula):
$A_{\text{sector}}=\tfrac12(16^2)(0.6)=76.8\text{ cm}^2$
Triangle of the two radii (keep the calculator in radian mode):
$A_\triangle=\tfrac12(16^2)\sin 0.6=72.27\ldots\text{ cm}^2$
Subtract, rounding only at the end:
$A_{\text{seg}}=76.8-72.27\ldots=4.53\text{ cm}^2$

Answer

A_{\text{seg}}\approx 4.53\text{ cm}^2

(9709/13 May/Jun 2023 Q6) If your calculator gives a very big number for $\sin 0.6$ , it is in degree mode. Fix that first.

The length of the chord: cosine rule

To find the chord $AB$ , look at the triangle $AOB$ : two sides are radii of length $r$ with the angle $\theta$ between them, so use the cosine rule (from Trigonometry):

$AB^2=r^2+r^2-2r^2\cos\theta=2r^2(1-\cos\theta)$

Take the square root to get the chord itself.
This is the method the exam asks for, and it works for any angle, so use it first.

Worked example

Chord of a circle: radius $7$ cm, angle $AOB=2$ rad

Two radii with $\theta=2$ between them, so use the cosine rule:

$AB^2=7^2+7^2-2(7^2)\cos 2=98-98\cos 2$

Keep the calculator in radian mode for $\cos 2$ :

$AB^2=98-98(-0.4161\ldots)=138.78\ldots$

Answer

AB=\sqrt{138.78\ldots}\approx 11.8\text{ cm}

A faster shortcut for the chord: 2r sin(θ/2)

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

This gives the same answer as the cosine rule and is faster. But the cosine rule is the one the textbook teaches, so use that first and use this as a check.

The perimeter of a segment: arc + chord

$P_{\text{seg}}=r\theta+AB,\quad AB=\sqrt{2r^2(1-\cos\theta)}$

Compare with the sector perimeter from §4.2, $r\theta+2r$ (arc plus two radii).
Different region, different straight edge: read the picture before you pick a formula.

“Show that” with exact values

One common type gives an exact angle and asks you to show that a length or area equals a printed answer in $r$ , $\pi$ and surds.
No rounding: every step stays exact, and you must reach the printed answer to get the marks.
Use the exact values $\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$ , $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$ , $\sin\tfrac{\pi}{6}=\tfrac12$ .

Worked example

Real “show that”: sector radius $2r$ , angle $\tfrac{\pi}{6}$ , $C$ the midpoint of $OA$ — show $BC=r\sqrt{5-2\sqrt3}$

The radius is $OB=OA=2r$ . Since $C$ is the midpoint of $OA$ , we get $OC=r$ . Triangle $OBC$ has those two sides with the angle $\tfrac{\pi}{6}$ between them, so use the cosine rule on $BC$ :

Cosine rule with sides $2r$ and $r$ and the included angle $\tfrac{\pi}{6}$ :
$BC^2=(2r)^2+r^2-2(2r)(r)\cos\dfrac{\pi}{6}$
Put in the exact cosine $\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$ ; the 4 and the $\tfrac12$ cancel cleanly:
$BC^2=4r^2+r^2-4r^2\cdot\dfrac{\sqrt3}{2}=5r^2-2\sqrt3\,r^2$
Factor and root, keeping it exact:
$BC^2=r^2(5-2\sqrt3)\ \Rightarrow\ BC=r\sqrt{5-2\sqrt3}$

Answer

BC=r\sqrt{5-2\sqrt3}\quad\text{as required}

(9709/12 May/Jun 2020 Q7(a)) The whole skill is exact values, no calculator. A decimal anywhere loses the “show that” marks.

The segment version: show that the area is r²(2π − 3√3)⁄12, with θ = π/3

$A_{\text{seg}}=\dfrac{2\pi r^2}{12}-\dfrac{3\sqrt3\,r^2}{12}=\dfrac{r^2(2\pi-3\sqrt3)}{12}$

The whole skill is common denominators with no calculator. It is the same move as the worked example, just one step longer. (9709-style)

Where the marks go

Common mistake

For the triangle you need

\tfrac12 r^2\sin\theta

, not

\tfrac12 r^2\theta

. The

\sin

is what makes it a triangle and not a second sector. And the calculator must be in radian mode for

\sin\theta

. This is the biggest slip in the topic. (9709/13 May/Jun 2023 Q6)

Common mistake

A segment's perimeter is the arc plus the chord, not arc plus two radii. The two radii make the sector, not the segment. Get the chord from the cosine rule

AB^2=2r^2(1-\cos\theta)

Common mistake

Do not round the sector or triangle too early. Keep the full triangle value (

72.27\ldots

) and only round the last subtraction. If you round early, a small segment can come out wrong in the second decimal place.

Now you try

A circle has centre $O$ and radius $6$ cm, $AB$ is a chord and angle $AOB=1.2$ radians. Find the area of the segment cut off by $AB$ . (9709-style)

Your turn— tap to reveal the worked answer (sector − triangle)

Sector: $\tfrac12(6^2)(1.2)=21.6$ .

Triangle (radian mode): $\tfrac12(6^2)\sin 1.2=16.78\ldots$ .

Segment: $21.6-16.78\ldots$ :

Answer

A_{\text{seg}}\approx 4.82\text{ cm}^2