Sectors of a Circle — Mathslive.ie

Section 1 of 10

What is a sector?

A sector is a slice of a circle — the region bounded by two radii and the arc between them.

Picture a pizza. The whole pizza is the circle. One slice is a sector. The crust of the slice is the arc; the two straight edges are the radii.

Greek letter $\theta$ ("theta") is used for the angle at the centre. The radius is $r$ as usual.

Section 2 of 10

The four formulas

A full circle has area $\pi r^{2}$ and circumference $2\pi r$. A sector is just a fraction of the full circle — the fraction $\dfrac{\theta}{360}$ when the angle is in degrees.

MUST LEARN — The four sector formulas

1.Area (degrees): $A = \pi r^{2}\left(\dfrac{\theta}{360}\right)$

2.Arc length (degrees): $\ell = 2\pi r\left(\dfrac{\theta}{360}\right)$

3.Area (radians): $A = \dfrac{1}{2}r^{2}\theta$

4.Arc length (radians): $\ell = r\theta$

If $\theta$ is in degrees, use formulas 1 and 2. If $\theta$ is in radians, use formulas 3 and 4. Mixing them is the easiest mistake to make.

Two facts worth tattooing on your hand: $\pi = 180°$ and $1°$ in radians is $\dfrac{\pi}{180}$.

Section 3 of 10

Working in degrees

Worked example 1

Sector $OPQ$ has radius $10$ m and the angle at $O$ is $55°$. Find:
(i) the area of the sector,
(ii) the length of arc $PQ$,
(iii) the perimeter of the sector.

(i) Area

$A = \pi r^{2}\left(\dfrac{\theta}{360}\right)$

$= \pi(10)^{2}\left(\dfrac{55}{360}\right)$

$A \approx 48.0 \text{ m}^{2}$

Round sensibly — to 1 decimal place is fine here. Always carry units.

(ii) Arc length $PQ$

The arc is just the curved bit — a fraction of the circumference.

$\ell = 2\pi r\left(\dfrac{\theta}{360}\right)$

$= 2\pi(10)\left(\dfrac{55}{360}\right)$

$\ell \approx 9.6 \text{ m}$

(iii) Perimeter

Perimeter is the total distance around the boundary — two radii and one arc.

$P = r + r + \ell$

$= 10 + 10 + 9.6$

$P = 29.6 \text{ m}$

Don't add $\pi r^{2}$ to anything when finding perimeter — area and perimeter are different beasts. Perimeter = two radii + one arc.

YOU TRY · §3

A sector has radius $12$ cm and central angle $75°$. Find its area and arc length.

Have a go on paper — write down which formulas you'll use first.

$A = \pi(12)^{2}\left(\dfrac{75}{360}\right)$

$= 144\pi \cdot \dfrac{5}{24}$

$= 30\pi \approx 94.2 \text{ cm}^{2}$

$\ell = 2\pi(12)\left(\dfrac{75}{360}\right)$

$= 24\pi \cdot \dfrac{5}{24} = 5\pi$

$A \approx 94.2 \text{ cm}^{2}$, $\ell \approx 15.7 \text{ cm}$

YOU TRY · §3

A pie slice has radius $9$ cm and angle $120°$. What is the perimeter of the slice?

Perimeter = two radii + arc. Find the arc first.

$\ell = 2\pi(9)\left(\dfrac{120}{360}\right) = 18\pi \cdot \dfrac{1}{3} = 6\pi$

$P = 9 + 9 + 6\pi$

$= 18 + 6\pi$

$P \approx 36.8 \text{ cm}$

$P = 18 + 6\pi \approx 36.8 \text{ cm}$

Section 4 of 10

Working in radians

When the angle is given in radians, the formulas get cleaner — no $\dfrac{\theta}{360}$ stuck on the end.

Reminder

A$A = \dfrac{1}{2}r^{2}\theta$

$\ell$$\ell = r\theta$

!$\theta$ must be in radians for these.

Worked example 2

A sector has radius $8$ m and the angle at the centre is $2.5$ radians. Find:
(i) the area,
(ii) the perimeter.

(i) Area

$A = \dfrac{1}{2}r^{2}\theta$

$= \dfrac{1}{2}(8)^{2}(2.5)$

$= \dfrac{1}{2}(64)(2.5)$

$A = 80 \text{ m}^{2}$

(ii) Perimeter

$\ell = r\theta = 8(2.5) = 20 \text{ m}$

$P = r + r + \ell$

$= 8 + 8 + 20$

$P = 36 \text{ m}$

Notice how much faster radians are. If the question gives you radians, use the radian formulas — don't convert to degrees first.

YOU TRY · §4

A sector has radius $6$ cm and angle $\dfrac{\pi}{3}$ rads. Find its area.

$\theta$ is already in radians — go straight to $A = \dfrac{1}{2}r^{2}\theta$.

$A = \dfrac{1}{2}(6)^{2}\left(\dfrac{\pi}{3}\right)$

$= \dfrac{1}{2}(36)\left(\dfrac{\pi}{3}\right)$

$= 18 \cdot \dfrac{\pi}{3} = 6\pi$

$A = 6\pi \approx 18.85 \text{ cm}^{2}$

YOU TRY · §4

A sector has radius $5$ m and arc length $7.5$ m. What is the central angle in radians? Then find the area.

Use $\ell = r\theta$ to get $\theta$ first.

$\ell = r\theta \;\Rightarrow\; 7.5 = 5\theta$

$\theta = \dfrac{7.5}{5} = 1.5 \text{ rads}$

$A = \dfrac{1}{2}(5)^{2}(1.5)$

$= \dfrac{1}{2}(25)(1.5) = 18.75$

$\theta = 1.5$ rads, $A = 18.75 \text{ m}^{2}$

Section 5 of 10

Sector minus triangle

A common HL question: find a shaded area that is a sector with a triangle removed (or the bit of the sector left after slicing off a triangular chunk).

Strategy

1.Compute the full sector area.

2.Compute the triangle area using $\dfrac{1}{2}ab \sin C$.

3.Subtract.

Worked example 3

Triangle $OAB$ has $|OA| = |OB| = 8$ m and $\angle AOB = 60°$. A circular arc is drawn from $A$ to $B$ centred at $O$. Find the area of the region between the chord $AB$ and the arc, correct to $1$ decimal place.

The shaded region = (sector $OAB$) − (triangle $OAB$).

Sector area (degrees)

$A_{\text{sec}} = \pi r^{2}\left(\dfrac{\theta}{360}\right)$

$= \pi (8)^{2}\left(\dfrac{60}{360}\right)$

$= 64\pi \cdot \dfrac{1}{6}$

$A_{\text{sec}} \approx 33.51 \text{ m}^{2}$

Triangle area

For a triangle with two sides and the angle between them, use $\dfrac{1}{2}ab\sin C$.

$A_{\triangle} = \dfrac{1}{2}ab\sin C$

$= \dfrac{1}{2}(8)(8)\sin 60°$

$= 32 \cdot \dfrac{\sqrt{3}}{2}$

$A_{\triangle} \approx 27.71 \text{ m}^{2}$

Shaded area

$A_{\text{shaded}} = A_{\text{sec}} - A_{\triangle}$

$= 33.51 - 27.71$

$A_{\text{shaded}} \approx 5.8 \text{ m}^{2}$

Always do sector minus triangle when the shaded region sits between a chord and an arc. This is called a circular segment.

YOU TRY · §5

A sector has radius $10$ cm and angle $90°$. Find the area of the circular segment cut off by the chord joining the two endpoints of the arc.

Sector area, then triangle area (right-angled with legs $10$), then subtract.

$A_{\text{sec}} = \pi(10)^{2}\left(\dfrac{90}{360}\right) = 25\pi$

$A_{\triangle} = \dfrac{1}{2}(10)(10)\sin 90° = 50$

$A_{\text{shaded}} = 25\pi - 50$

$\approx 28.5 \text{ cm}^{2}$

$25\pi - 50 \approx 28.5 \text{ cm}^{2}$

YOU TRY · §5

A sector has radius $4$ cm and angle $\dfrac{\pi}{3}$ rads. Find the exact area of the circular segment cut off by the chord across the arc.

Use radian formula for the sector. Keep exact values — no calculator.

$A_{\text{sec}} = \dfrac{1}{2}(4)^{2}\left(\dfrac{\pi}{3}\right) = \dfrac{8\pi}{3}$

$A_{\triangle} = \dfrac{1}{2}(4)(4)\sin 60° = 4\sqrt{3}$

$A_{\text{shaded}} = \dfrac{8\pi}{3} - 4\sqrt{3}$

$= \dfrac{8\pi - 12\sqrt{3}}{3} \text{ cm}^{2}$

$\dfrac{8\pi}{3} - 4\sqrt{3} \approx 1.46 \text{ cm}^{2}$

Section 6 of 10

Working backwards

Sometimes the arc length is given and you need the area, or the area is given and you need the arc. Same four formulas — rearrange to find what you need.

Worked example 4

Arc $PQ$ has length $6$ m on a circle of radius $5$ m, centred at $O$. Find the area of sector $OPQ$.

We will solve this two ways to compare. In a real exam, pick whichever is faster.

Method 1 — in degrees

Use $\ell = 2\pi r\left(\dfrac{\theta}{360}\right)$ to find $\theta$ first.

$2\pi(5)\left(\dfrac{\theta}{360}\right) = 6$

$\dfrac{10\pi \theta}{360} = 6$

$\theta = \dfrac{360(6)}{10\pi}$

$\theta \approx 68.75°$

Now use the area formula:

$A = \pi r^{2}\left(\dfrac{\theta}{360}\right)$

$= \pi(5)^{2}\left(\dfrac{68.75}{360}\right)$

$A = 15 \text{ m}^{2}$

Method 2 — in radians

Use $\ell = r\theta$ first — much cleaner.

$\ell = r\theta$

$6 = 5\theta$

$\theta = \dfrac{6}{5}$ rads

$A = \dfrac{1}{2}r^{2}\theta$

$= \dfrac{1}{2}(5)^{2}\left(\dfrac{6}{5}\right)$

$= \dfrac{1}{2} \cdot 25 \cdot \dfrac{6}{5}$

$A = 15 \text{ m}^{2}$

Same answer, half the work. Lesson: when you're given the arc length, switch to radians — $\theta = \dfrac{\ell}{r}$ falls out instantly.

Worked example 5

A sector has radius $5$ cm and area $6.25 \text{ cm}^{2}$. Find the arc length.

Find $\theta$ in radians using the area formula:

$\dfrac{1}{2}r^{2}\theta = A$

$\dfrac{1}{2}(5)^{2}\theta = 6.25$

$12.5\theta = 6.25$

$\theta = \dfrac{1}{2}$ rads

$\ell = r\theta$

$= 5 \cdot \dfrac{1}{2}$

$\ell = 2.5 \text{ cm}$

Worked example 6

An arc of length $12.5$ cm subtends an angle of $123°$ at the centre of a circle. Find the radius.

$\ell = 2\pi r\left(\dfrac{\theta}{360}\right)$

$2\pi r\left(\dfrac{123}{360}\right) = 12.5$

$2.147 \cdot r = 12.5$

$r \approx 5.8 \text{ cm}$

YOU TRY · §6

A sector has radius $8$ cm and arc length $10$ cm. Find its area.

Radians method. Get $\theta$ from $\ell = r\theta$ first.

$\theta = \dfrac{\ell}{r} = \dfrac{10}{8} = \dfrac{5}{4}$ rads

$A = \dfrac{1}{2}(8)^{2}\left(\dfrac{5}{4}\right)$

$= \dfrac{1}{2}(64)\left(\dfrac{5}{4}\right) = 40$

$A = 40 \text{ cm}^{2}$

YOU TRY · §6

A sector of area $50 \text{ cm}^{2}$ has central angle $\dfrac{\pi}{4}$ rads. Find the radius.

Rearrange $A = \dfrac{1}{2}r^{2}\theta$ to make $r^{2}$ the subject.

$\dfrac{1}{2}r^{2}\left(\dfrac{\pi}{4}\right) = 50$

$\dfrac{\pi r^{2}}{8} = 50$

$r^{2} = \dfrac{400}{\pi}$

$r \approx 11.28 \text{ cm}$

Section 7 of 10

Chords in sectors

When the question gives you an arc and asks for a chord, the chain is:

The chain

1.arc length $\to$ central angle (use $\ell = r\theta$)

2.central angle + two radii $\to$ chord (use the cosine rule)

The cosine rule: in a triangle with sides $a$, $b$, $c$ and angle $A$ opposite to side $a$,

$a^{2} = b^{2} + c^{2} - 2bc \cos A$

Worked example 7

$P$, $Q$, $R$ are points on a circle with centre $K$ and radius $2$ cm. The length of minor arc $PQ$ is $\dfrac{5\pi}{3}$ cm.
(i) Find the length of chord $[PQ]$, correct to two decimal places.
(ii) If $|PQ| = |PR|$, find $|RQ|$.

(i) Find the central angle

Switch the arc into a central angle using $\ell = r\theta$ in radians.

$\ell = r\theta$

$\dfrac{5\pi}{3} = 2\theta$

$\theta = \dfrac{5\pi}{6} = 150°$

Cosine rule on triangle $PKQ$

Sides $|KP| = |KQ| = 2$ are two radii. Angle at $K$ is $150°$. We want side $|PQ|$ opposite to that angle.

$|PQ|^{2} = |KP|^{2} + |KQ|^{2} - 2|KP||KQ|\cos K$

$= 2^{2} + 2^{2} - 2(2)(2)\cos 150°$

$= 4 + 4 - 8\left(-\dfrac{\sqrt{3}}{2}\right)$

$= 8 + 4\sqrt{3}$

$|PQ| = \sqrt{8 + 4\sqrt{3}}$

$|PQ| \approx 3.86 \text{ cm}$

(ii) Find $|RQ|$ given $|PQ| = |PR|$

Since $|PQ| = |PR|$ and both share the same chord-from-$P$ structure with radii $2$, the central angles must match: $\angle PKR = \angle PKQ = 150°$.

Total angle around $K$ is $360°$. So:

$\angle QKR = 360° - 150° - 150°$

$\angle QKR = 60°$

Cosine rule on triangle $QKR$:

$|QR|^{2} = 2^{2} + 2^{2} - 2(2)(2)\cos 60°$

$= 4 + 4 - 8\left(\dfrac{1}{2}\right)$

$= 8 - 4 = 4$

$|RQ| = 2 \text{ cm}$

Tip: if a triangle has two sides equal to the radius and an angle of $60°$ between them, it's equilateral — the third side also equals the radius. Quick mental check.

YOU TRY · §7

A circle has centre $O$ and radius $6$ cm. Points $A$ and $B$ lie on the circle so that arc $AB = 2\pi$ cm. Find the length of chord $[AB]$.

Get $\theta$ from $\ell = r\theta$. Then cosine rule on triangle $OAB$.

$2\pi = 6\theta \;\Rightarrow\; \theta = \dfrac{\pi}{3} = 60°$

$|AB|^{2} = 6^{2} + 6^{2} - 2(6)(6)\cos 60°$

$= 36 + 36 - 72 \cdot \dfrac{1}{2}$

$= 36$

$|AB| = 6 \text{ cm}$

$|AB| = 6 \text{ cm}$ (equilateral triangle)

YOU TRY · §7

A chord $[AB]$ in a circle of radius $5$ cm subtends an angle of $\dfrac{2\pi}{3}$ rads at the centre. Find the length of chord $[AB]$, in exact form.

$\cos\left(\dfrac{2\pi}{3}\right) = -\dfrac{1}{2}$. Keep surds — don't decimalise.

$|AB|^{2} = 5^{2} + 5^{2} - 2(5)(5)\cos\left(\dfrac{2\pi}{3}\right)$

$= 25 + 25 - 50\left(-\dfrac{1}{2}\right)$

$= 50 + 25 = 75$

$|AB| = \sqrt{75} = 5\sqrt{3}$

$|AB| = 5\sqrt{3} \text{ cm}$

Section 8 of 10

Circumscribed sector

A sector (solid line) is circumscribed by a circle (dashed line) — that is, the vertex of the sector and the two ends of its arc all lie on the circle.

Worked example 8

The diagram shows a sector with two straight sides of length $k$ and vertex angle $60°$ at $V$, circumscribed by a circle.
(i) Find the radius of the circle in terms of $k$.
(ii) Show that the circle encloses an area which is double that of the sector.

(i) Find the chord $|AB|$

First find the straight-line chord across the arc. Triangle $VAB$ has $|VA| = |VB| = k$ and $\angle V = 60°$ — that's isoceles with $60°$ at the apex, so it's equilateral.

$|AB| = k$

Quick check via cosine rule: $|AB|^{2} = k^{2} + k^{2} - 2k^{2}\cos 60° = 2k^{2} - k^{2} = k^{2}$ $\Rightarrow$ $|AB| = k$. ✓

Use the inscribed-angle & sine rule combo

Let $R$ be the radius of the circumscribing circle, and let $C$ be its centre. The inscribed angle at $V$ is $60°$, so the central angle $\angle ACB$ subtending the same arc $AB$ is $2 \times 60° = 120°$.

Triangle $ACB$ is isoceles with $|CA| = |CB| = R$ and apex $120°$ at $C$, so the base angles each measure $30°$.

Apply the sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$. Side $R$ is opposite to one of the $30°$ angles; side $k$ is opposite the $120°$ angle.

$\dfrac{R}{\sin 30°} = \dfrac{k}{\sin 120°}$

$R = \dfrac{k \sin 30°}{\sin 120°}$

$= \dfrac{k \cdot \tfrac{1}{2}}{\tfrac{\sqrt{3}}{2}}$

$= \dfrac{k}{\sqrt{3}}$

$R = \dfrac{\sqrt{3}\,k}{3}$

(ii) Show circle area = $2 \times$ sector area

Area of the circle:

$A_{\text{circle}} = \pi R^{2}$

$= \pi \left(\dfrac{\sqrt{3}\,k}{3}\right)^{2}$

$= \pi \cdot \dfrac{3k^{2}}{9}$

$A_{\text{circle}} = \dfrac{\pi k^{2}}{3}$

Area of the sector (radius $k$, angle $60°$):

$A_{\text{sector}} = \pi r^{2}\left(\dfrac{\theta}{360}\right)$

$= \pi k^{2}\left(\dfrac{60}{360}\right)$

$A_{\text{sector}} = \dfrac{\pi k^{2}}{6}$

Compare:

$\dfrac{A_{\text{circle}}}{A_{\text{sector}}} = \dfrac{\pi k^{2}/3}{\pi k^{2}/6} = 2$

$A_{\text{circle}} = 2 \cdot A_{\text{sector}}$ $\checkmark$

YOU TRY · §8

A sector has straight sides of length $6$ cm and vertex angle $90°$, and is inscribed in a circle (the vertex and the two arc-endpoints all lie on the circle). Find the radius of the circle.

Find the chord $|AB|$ across the arc (cosine rule or Pythagoras), then use the fact that the inscribed angle is $90°$, so $AB$ is a diameter.

Chord: $|AB|^{2} = 6^{2} + 6^{2} - 2(6)(6)\cos 90°$

$= 36 + 36 - 0 = 72$

$|AB| = 6\sqrt{2}$

Inscribed angle $90°$ means chord is a diameter:

$R = \dfrac{|AB|}{2} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2}$

$R = 3\sqrt{2} \approx 4.24$ cm

Section 9 of 10

Intersecting circles

Two circles of equal radius pass through each other's centres. This produces a beautiful lens-shaped region — and a classic HL question.

Worked example 9

$C_{1}$ is a circle with centre $a$ and radius $r$. $C_{2}$ is a circle with centre $b$ and radius $r$. They intersect at $k$ and $p$. Also $a \in C_{2}$ and $b \in C_{1}$.
(i) Find $\angle kap$ in radians.
(ii) Find the area of the shaded lens, in terms of $r$ and $\pi$.

(i) Find $\angle kap$

Every distance you can see is $r$. Pin that down first.

$|ak| = r$ ($k$ on $C_{1}$, centre $a$)

$|ab| = r$ ($b$ on $C_{1}$, centre $a$)

$|bk| = r$ ($k$ on $C_{2}$, centre $b$)

So triangle $akb$ has three sides of length $r$ — it's equilateral. Every angle is $60°$.

$\angle kab = 60° = \dfrac{\pi}{3}$

The same argument on triangle $abp$ gives:

$\angle pab = \dfrac{\pi}{3}$

Now $\angle kap$ is the angle at $a$ from $k$ across to $p$, which is $\angle kab + \angle bap$:

$\angle kap = \dfrac{\pi}{3} + \dfrac{\pi}{3}$

$\angle kap = \dfrac{2\pi}{3}$ rads ($= 120°$)

(ii) Area of the lens

By symmetry, the shaded lens is made of two equal circular segments: one from $C_{1}$ (bounded by chord $[kp]$ and the arc through $C_{1}$'s right side) and one from $C_{2}$ (mirror image).

Lens decomposition

!Lens area $= 2 \times$ (one segment)

!Segment $=$ sector $-$ triangle

Take the segment from $C_{1}$. It is the sector $kap$ (centred at $a$, angle $\dfrac{2\pi}{3}$, radius $r$) with the triangle $kap$ removed.

Sector area

$A_{\text{sec}} = \dfrac{1}{2}r^{2}\theta$

$= \dfrac{1}{2}r^{2}\left(\dfrac{2\pi}{3}\right)$

$A_{\text{sec}} = \dfrac{\pi r^{2}}{3}$

Triangle area

Triangle $kap$ has two sides of length $r$ (the radii $|ak|$ and $|ap|$) with angle $\dfrac{2\pi}{3}$ between them.

$A_{\triangle} = \dfrac{1}{2}ab \sin C$

$= \dfrac{1}{2}r^{2}\sin\left(\dfrac{2\pi}{3}\right)$

$= \dfrac{1}{2}r^{2}\cdot \dfrac{\sqrt{3}}{2}$

$A_{\triangle} = \dfrac{\sqrt{3}\,r^{2}}{4}$

Put it together

$A_{\text{segment}} = A_{\text{sec}} - A_{\triangle}$

$= \dfrac{\pi r^{2}}{3} - \dfrac{\sqrt{3}\,r^{2}}{4}$

$A_{\text{lens}} = 2 \cdot A_{\text{segment}}$

$= 2\left(\dfrac{\pi r^{2}}{3} - \dfrac{\sqrt{3}\,r^{2}}{4}\right)$

$= \dfrac{2\pi r^{2}}{3} - \dfrac{\sqrt{3}\,r^{2}}{2}$

$A_{\text{lens}} = \dfrac{r^{2}}{6}\left(4\pi - 3\sqrt{3}\right)$

Factor of $r^{2}/6$ pulled out because $\dfrac{2\pi}{3} = \dfrac{4\pi}{6}$ and $\dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{6}$ — common denominator trick.

YOU TRY · §9

Two circles each of radius $4$ cm intersect so that each passes through the other's centre. Find the exact area of the lens-shaped intersection.

Same setup, just plug $r = 4$ into the formula derived above.

$A_{\text{lens}} = \dfrac{r^{2}}{6}(4\pi - 3\sqrt{3})$

$= \dfrac{16}{6}(4\pi - 3\sqrt{3})$

$= \dfrac{8}{3}(4\pi - 3\sqrt{3})$

$\approx 19.65 \text{ cm}^{2}$

$\dfrac{8}{3}(4\pi - 3\sqrt{3}) \approx 19.65 \text{ cm}^{2}$

YOU TRY · §9 (challenge)

Two circles of equal radius $r$ overlap so that the central angle subtended by the lens at each centre is $\dfrac{\pi}{2}$ rads (instead of $\dfrac{2\pi}{3}$). Find the area of the lens, in terms of $r$.

Same recipe: lens = 2(sector − triangle). Angle is now $\pi/2$ instead of $2\pi/3$.

$A_{\text{sec}} = \dfrac{1}{2}r^{2}\left(\dfrac{\pi}{2}\right) = \dfrac{\pi r^{2}}{4}$

$A_{\triangle} = \dfrac{1}{2}r^{2}\sin\left(\dfrac{\pi}{2}\right) = \dfrac{r^{2}}{2}$

$A_{\text{segment}} = \dfrac{\pi r^{2}}{4} - \dfrac{r^{2}}{2}$

$A_{\text{lens}} = 2 \cdot A_{\text{segment}}$

$= \dfrac{\pi r^{2}}{2} - r^{2}$

$A_{\text{lens}} = \dfrac{r^{2}}{2}(\pi - 2)$

Section 10 of 10

Pulling it together

FOUR-STEP CHECKLIST

1.Read carefully — is the angle in degrees or radians?

2.Pick the matching formula. Degrees: $\pi r^{2}\left(\tfrac{\theta}{360}\right)$ & $2\pi r\left(\tfrac{\theta}{360}\right)$. Radians: $\tfrac{1}{2}r^{2}\theta$ & $r\theta$.

3.Given an arc and asked for an area? Switch to radians via $\theta = \tfrac{\ell}{r}$ first — it's almost always faster.

4.Shaded region between chord and arc? Sector minus triangle. Lens between two arcs? Twice a segment.

For HL chord problems, remember the cosine rule is your best friend once you've got the central angle:

$|\text{chord}|^{2} = r^{2} + r^{2} - 2r^{2}\cos \theta = 2r^{2}(1 - \cos \theta)$

This compact form is handy: $|\text{chord}|^{2} = 2r^{2}(1 - \cos\theta)$ — derive it once, recognise it forever.

End of lesson

Sectors of a Circle — HL · Mathslive.ie