CDS Mathematics · Mensuration 2D

Arcs, Sectors & Segments

Arc length and sector area as a fraction of the whole circle, and the segment cut off by a chord as the sector minus a triangle.

All Mensuration 2D notes CDS Mathematics strategy

Why this matters

A steady source of easy marks: pendulums, clock hands and pie charts are all sectors in disguise. The one idea that costs marks is the segment, where the triangle has to be subtracted from the sector.

Concept 1 of 3: Arc length — radius times angle in radians

An arc is a fraction of the circumference: the fraction

\dfrac{\theta}{360^\circ}

. In radians that becomes simply

r\theta

, which is why a pendulum or a road curve asks you to convert the angle first.

Definition

Arc $= \dfrac{\theta}{360^\circ}\times 2\pi r = r\theta$ with $\theta$ in radians.
Degrees to radians: multiply by $\dfrac{\pi}{180}$ .
A pendulum swinging through $\theta$ : its bob traces an arc whose radius is the pendulum's length.
A road that turns through $\theta$ over a distance $s$ : $r = \dfrac{s}{\theta}$ .
Equal arcs in two circles: $r_1\theta_1 = r_2\theta_2$ , so the radii are in the inverse ratio of the angles.

Arc length

l = r\theta = \frac{\theta^\circ}{360^\circ}\times 2\pi r

Worked example

A pendulum

63

cm long swings through

40^\circ

. How long is the arc its bob traces?

(\pi = \tfrac{22}{7})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (II) 2023 — Elementary Mathematics · Q61Easy

Example 1 · Mensuration 2D · Arcs, Sectors and Segments

A pendulum swings through an angle of

9^\circ

and its end describes an arc of length 14.3 cm. What is the length of the pendulum ? (Take

\pi = \frac{22}{7}

)

Never put degrees into rθ

l = r\theta

needs radians. Multiplying the radius by

30

instead of

\dfrac{\pi}{6}

is the commonest slip, and the options rarely include anything that wild, so a strange result is the warning.

Drill 6 more on arc length — radius times angle in radians

Concept 2 of 3: Sector area — a slice of the disc

A sector is the same fraction of the disc's area as its angle is of

360^\circ

. If the arc length is known, there is a shortcut: the sector is half the radius times the arc, like a triangle with the arc as its base.

Definition

Sector $= \dfrac{\theta}{360^\circ}\times\pi r^2 = \dfrac12 r^2\theta$ (radians) $= \dfrac12 r\,l$ .
Perimeter of a sector $= 2r + l$ .
A minute hand sweeps $6^\circ$ a minute; an hour hand $\dfrac12^\circ$ a minute.
A pie chart with angles in the ratio $a : b : \ldots$ : each sector's angle is its share of $360^\circ$ .

Sector

A = \frac{\theta}{360^\circ}\,\pi r^2 = \tfrac12 r l

Worked example

The minute hand of a clock is

14

cm long. What area does it sweep in

15

minutes?

(\pi = \tfrac{22}{7})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (II) 2021 — Elementary Mathematics · Q65Moderate

Example 2 · Mensuration 2D · Arcs, Sectors and Segments

The minute hand of a clock is 21 cm long. What is the area on the face of the clock described by the minute hand between 10.10 a.m and 10.30 a.m ? (take

\pi = \frac{22}{7}

)

Drill 9 more on sector area — a slice of the disc

Concept 3 of 3: The segment — sector minus triangle

A chord cuts the disc into a minor and a major segment. The minor segment is the sector with the triangle formed by the two radii taken out; the major segment is the rest of the disc.

Definition

Minor segment $= \dfrac{\theta}{360^\circ}\pi r^2 - \dfrac12 r^2\sin\theta = \dfrac12 r^2(\theta - \sin\theta)$ (radians).
Major segment $= \pi r^2 - \text{minor segment}$ .
A chord at $90^\circ$ : minor segment $= r^2\left(\dfrac{\pi}{4} - \dfrac12\right)$ , and the chord is $r\sqrt2$ .
A chord at $120^\circ$ : minor segment $= r^2\left(\dfrac{\pi}{3} - \dfrac{\sqrt3}{4}\right)$ , and the chord is $r\sqrt3$ .
A chord at $60^\circ$ : the triangle is equilateral, area $\dfrac{\sqrt3}{4}r^2$ .

Minor segment

\text{Segment} = \tfrac12 r^2(\theta - \sin\theta)

Worked example

A chord subtends

60^\circ

at the centre of a circle of radius

6

cm. Find the area of the minor segment.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q74Moderate

Example 3 · Mensuration 2D · Arcs, Sectors and Segments

Consider the following for the items that follow : A chord of length

l

of a circle makes an angle

90^\circ

at the centre of the circle.

What is the area of the minor segment ?

Sector is not segment

The sector includes the triangle; the segment does not. The sector's value is almost always among the options for a segment question.

Drill 7 more on the segment — sector minus triangle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Arc length — radius times angle in radians
Arc length
$l = r\theta = \frac{\theta^\circ}{360^\circ}\times 2\pi r$
Sector area — a slice of the disc
Sector
$A = \frac{\theta}{360^\circ}\,\pi r^2 = \tfrac12 r l$
The segment — sector minus triangle
Minor segment
$\text{Segment} = \tfrac12 r^2(\theta - \sin\theta)$

Watch out for (2)

Never put degrees into rθ
→ Arc length — radius times angle in radians
Sector is not segment
→ The segment — sector minus triangle

Test yourself on Mensuration 2D

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Mensuration 2D chapter test Drill all 25 questions