Arcs and sectors

Arc length, perimeter and area

In 10 of 20 recent exams
2% of the marks
Paper 1 or 2
1 to 3 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator) about equally (13 parts on Paper 1 (no calculator), 12 on Paper 2 (calculator)), 1–3 marks a part: usually question 1 or 2 of Section A, 5–7 marks in two or three parts (an arc or a perimeter, a radius or an angle, then an area). Very common (10 of 20 exam versions).

Spot it in the exam

Exam phrases to spot (5)

“find the length of the arc $AB$ ”
“find the perimeter of the sector”
“find the area of the sector $OAB$ ”
“find the area of the shaded major sector”
“write down an expression for the arc length in terms of $r$ ”

In disguise (4)

a context with one curved edge and two straight edges from one point: a slice of cake, an opened fan, the region a windscreen wiper cleans
an angle with a ° sign: convert it to radians first, or use the degree versions
"the circumference is divided into $n$ equal arcs": each angle at the centre is $\frac{2\pi}{n}$
a hand or an arm turning at a steady rate: the angle is the angular speed times the time (Radians)

Variants (4)

an area (or an arc) and the angle → $r$ , then the perimeter or the area. The method mark needs an equation with $r^2$ (or $r$ ) and $\theta$ , rearranged to give $r^2$ or $r$ .
an expression in $r$ first, then a value of the perimeter gives $r$
the major sector, exact in $\pi$ : the reflex angle $2\pi - \theta$ , or the circle minus the minor sector
the region between two arcs with the same centre: a "show that" for its area, the inner radius, then a perimeter of two arcs and two straight edges

Looks like it, but isn't (3)

a chord $[AB]$ is given, asked for, or part of the region, or a triangle with a vertex at the centre is part of the region: Chords, segments and regions
the perimeter and the area are both given, and $r$ and $\theta$ are both unknown: Perimeter and area both given
the sector is bent round to make the curved surface of a solid: Volume and surface area (the arc becomes the base's circumference)

Worked example

Area and angle: the radius, then the perimeter

Paper 1 (no calculator)

Question

The sector $OAB$ of a circle with centre $O$ has angle $AOB = \frac{2}{3}$ radians and area $27$ cm².
(a) Find the radius of the sector. [3]
(b) Find the perimeter of the sector. [2]
(c) A second sector has the same perimeter as sector $OAB$ and radius $6$ cm. Find the angle of the second sector. [2]

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r

9 cm

$\theta = 1.20$ rad, about 69°, and $r = 9$ cm: arc $r\theta = 10.8$ cm, perimeter $2r + r\theta = 28.8$ cm, area $\frac{1}{2}r^{2}\theta = 48.6$ cm².
No ° sign means radians, and the booklet's formulas apply as they stand. One radian is about 57°: the angle whose arc is as long as the radius.
(a) Angle $\frac{2}{3}$ and area 27: $\frac{1}{2}r^{2} \times \frac{2}{3} = 27$ , so $r^{2} = 81$ and $r = 9$ cm. $r = -9$ is not a length.
(b) The perimeter is two radii plus the arc: $2 \times 9 + 9 \times \frac{2}{3} = 24$ cm. The arc alone is the classic lost mark. Check: $\frac{1}{2} \times 9 \times 6 = 27$ .
(c) The same perimeter with $r = 6$ : $2 \times 6 + 6\theta = 24$ , so $\theta = 2$ radians, less than $2\pi$ as an angle at the centre must be.

Steps

Get the angle in radians.
Convert a degree angle (or use the degree versions); for the major sector use $2\pi - \theta$ ; for $n$ equal arcs use $\frac{2\pi}{n}$ .
Sketch the sector
and mark $r$ , $r$ and the arc. Write the one formula that links what is given to what is asked: arc $r\theta$ , perimeter $2r + r\theta$ , area $\frac{1}{2}r^2\theta$ (or $\frac{1}{2} \times r \times \text{arc}$ ).
Solve
for the unknown. A radius is positive; an angle must be less than $2\pi$ .
Find the follow-up
quantity with the unrounded value (keep full-accuracy values). Exact ( $\pi$ , fractions) on ; 3 s.f. with units on .
Check:
$\frac{1}{2} \times r \times \text{arc}$ must equal the area, and the angle should look right on the sketch ( $1$ radian is about $57°$ ).

Common mistakes

The arc as the perimeter
the perimeter is $2r + r\theta$ . The two radii are a mark of their own.
A degree angle in $r\theta$ or $\frac{1}{2}r^2\theta$
convert first, or use $\frac{\theta}{360}$ of the circle.
The minor angle for the major sector
the major sector's angle is $2\pi - \theta$ (or $360° - \theta$ ).

More mistakes (5)

Leaving $r = \pm 9$
a radius is positive, so only $r = 9$ is the answer.
The straight edges of the region between two arcs
as $r$ or $R$ : each is $R - r$ .
Rounding the radius
before using it in the next part: keep the full value (keep full-accuracy values).
A "show that" with a rounded value
(e.g. "show that $\theta = 1.87$ , correct to 3 s.f."): write the exact or unrounded value first ( $\theta = \frac{28}{15} = 1.8666\ldots$ ), then the given value. The given value on its own shows nothing. In later parts keep the unrounded value (the given rounded one is also accepted).
Units
an arc and a perimeter are lengths (cm), an area is in cm².

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