Arc length and sector area

A sector of a circle is the region enclosed by two radii and the arc between their endpoints. Think of a slice of pizza , the slice itself is a sector. The two radii are the straight cuts; the arc is the crust. The angle between the two radii at the centre is called the central angle of the sector.

A sector splits the disc into two parts: the minor sector (smaller piece, central angle less than $180^\circ$) and the major sector (larger piece, more than $180^\circ$). Together they make the whole disc.

Arc length

The total circumference of a full circle ($360^\circ$) is $2\pi r$. An arc that subtends a fraction of $360^\circ$ at the centre has that same fraction of the total length. If the central angle is $\theta$ (in degrees), then

$\ell_{\text{arc}} = \frac{\theta}{360^\circ} \cdot 2\pi r.$

Example. Arc length of a quarter circle ($\theta = 90^\circ$) of radius $14$: $\frac{90}{360} \cdot 2 \cdot 22/7 \cdot 14 = (1/4)(88) = 22$ cm.

Sector area

Similarly, the total area of a disc ($360^\circ$) is $\pi r^2$. A sector with central angle $\theta$ has

$A_{\text{sec}} = \frac{\theta}{360^\circ} \cdot \pi r^2.$

Alternative formula. Multiply numerator and denominator: $A_{\text{sec}} = (1/2) r \cdot \ell_{\text{arc}}$ , i.e., half of the radius times the arc length. This is the analogue of $A = (1/2) \cdot \text{base} \cdot \text{height}$ for a triangle. Sometimes it's the more convenient form.

Perimeter of a sector

The boundary of a sector consists of: two radii (each of length $r$) and the arc (length $\ell$). So the perimeter of a sector is

$P_{\text{sec}} = 2r + \ell = 2r + \frac{\theta}{360^\circ} \cdot 2\pi r.$

Classic computations

Quarter circle (90°):

• arc $= (1/4) \cdot 2\pi r = \pi r / 2$.
• area $= (1/4) \pi r^2$.
• perimeter $= 2r + \pi r/2$.

Semicircle (180°):

• arc $= \pi r$ (half circumference).
• area $= (1/2) \pi r^2$.
• perimeter $= 2r + \pi r$.

60° sector:

• arc $= (1/6) \cdot 2\pi r = \pi r/3$.
• area $= (1/6) \pi r^2$.
• perimeter $= 2r + \pi r/3$.

These are the most common sectors in board problems; memorise the multipliers.

Worked examples

Example 1. Find the arc length of a sector with central angle $60^\circ$ and radius $21$ cm.

$\ell = \frac{60}{360} \cdot 2 \cdot 22/7 \cdot 21 = (1/6)(132) = 22$ cm.

Example 2. A sector has central angle $30^\circ$ and radius $14$ cm. Find its area.

$A = (30/360) \cdot 22/7 \cdot 196 = (1/12) \cdot 616 = 51.33$ cm$^2$ (or $154/3$ cm$^2$).

Example 3. A sector of radius $10$ cm has arc length $15.7$ cm. Find the area.

Using $A = (1/2) r \ell = (1/2)(10)(15.7) = 78.5$ cm$^2$. (Notice we did not need the angle at all.)

Example 4. The hour and minute hands of a clock are $4$ cm and $6$ cm long respectively. Find the area swept by the minute hand in $10$ minutes.

The minute hand sweeps the whole face in $60$ min, so in $10$ min it covers $10/60 = 1/6$ of the disc.

Area $= (1/6) \pi \cdot 6^2 = 6\pi = 6 \cdot 22/7 \approx 18.86$ cm$^2$.

Example 5. A car wheel has radius $40$ cm. How many revolutions per minute does it make to travel at $66$ km/h?

Circumference $= 2 \cdot 22/7 \cdot 40 = 1760/7 \approx 251.4$ cm.

Speed $= 66$ km/h $= 6\,600\,000$ cm/h $= 110\,000$ cm/min.

Revolutions/min $= 110\,000 / 251.4 \approx 437.5$.

Example 6. A windscreen wiper has a $25$-cm blade attached to a $10$-cm arm. The blade sweeps through an angle of $115^\circ$. Find the swept area.

Outer radius $= 35$ cm (arm + blade), inner radius $= 10$ cm. Swept area (annular sector) $= (115/360) \pi (35^2 - 10^2) = (115/360)(22/7)(1125)$.

$= (115/360) \cdot 24750/7 = 2846250/2520 \approx 1129.5$ cm$^2$. (Approximately $1131$ cm$^2$ using $\pi = 22/7$ with care.)

Try it yourself

1. Find the arc length of a sector of radius $7$ cm and central angle $90^\circ$.
2. Find the area of a sector of radius $14$ cm and central angle $90^\circ$.
3. A sector has area $77$ cm$^2$ and central angle $45^\circ$. Find the radius.
4. A pendulum of length $80$ cm swings through an angle of $30^\circ$. Find the arc length traced by the tip.
5. The hour hand of a clock is $5$ cm long. Find the area swept by it from $3$ pm to $6$ pm.
6. A sector has perimeter $20$ cm and central angle $90^\circ$. Find the radius.
7. Find the central angle of a sector of area $51.33$ cm$^2$ and radius $14$ cm.
8. A pizza has radius $14$ cm and is cut into $8$ equal slices. Find the area of one slice and the length of its arc.
9. A sector has central angle $120^\circ$ and arc length $44$ cm. Find the radius.
10. A goat is tied to a corner of a square field by a $7$-m rope. Find the area it can graze.
11. The minute hand of a clock is $12$ cm long. Find the distance moved by its tip in $30$ minutes.
12. A circular flowerbed of radius $5$ m has a sector of $120^\circ$ planted with roses. Find the rose area.

Pitfalls / Insight

(1) Always use angle in degrees (with the $\theta/360$ formula) for board exam. Radians are introduced in Class XI.

(2) The formula $A = (1/2) r \ell$ is sometimes overlooked but is extremely handy when arc length is given but angle is not.

(3) For "swept area" problems (clock hands, pendulums, wipers), find the central angle from the time elapsed and then apply sector formulas. For wipers with a non-trivial arm, the swept region is an annular sector (subtract the inner sector from the outer).

(4) Goat-grazing problems: be careful about the angle available to the goat (often less than $360^\circ$ due to a wall or corner).