Chords, segments and regions

Made of sectors and triangles

In 11 of 20 recent exams
2% of the marks
Paper 1 or 2
1 to 7 marks

How it is set in the exam

Paper 2 (calculator) in 19 of its 20 parts, 1–7 marks a part: a whole Section A question of 5–7 marks, or several parts of a Section B question. The usual Paper 2 (calculator) shape is a "show that" for the area in terms of $\theta$ (or $r$ ), then "the area is …; find $\theta$ " on the GDC, in radians. Paper 1 (no calculator) once: an exact area from triangles with special angles. Very common (11 of 20 exam versions).

Spot it in the exam

Exam phrases to spot (6)

“find the length of the chord $[AB]$ ”
“find the area of the shaded segment”
“show that the area of the shaded region is $\frac{1}{2}r^2(\ldots)$ ”
“given that the area of the shaded region is …, find $\theta$ ”
“the tangents to the circle at $A$ and $B$ meet at $P$ ”
“a regular hexagon is inscribed in the circle”

In disguise (4)

a rotating sprinkler, searchlight or radar whose reach meets a straight edge or a line: the chord is never named (the perpendicular from the centre, then half the angle, then a time)
a cross-section with a curved top: a tunnel, an arched doorway
a sector joined to a triangle across a diameter: the triangle's angle at the centre is $\pi - \theta$
a logo or a design cut from a rectangle by arcs

Variants (10)

a chord on its own, by the cosine rule or by half the triangle
a segment's area as an expression in $r$ and $\theta$
composite regions: a rectangle minus segments, a segment on a rectangle, a kite of tangents minus a sector, a sector plus a triangle
the angle at the external point $P$ is given: the tangent comes from triangle $OAP$ with half that angle, $\beta$ , at $P$ ( $AP = \frac{r}{\tan\beta}$ ), and the angle at the centre, $\pi$ minus the angle at $P$ , is a step with its own marks (in degrees it scores )
"area = value" → $\theta$ on the GDC, with $r$ cancelled first when it is unknown
a segment's area and a ratio of two areas: $r$ cancels from the ratio, which gives $\theta$ , then $r$
a triangle of radii with a known area → $r$ → the major arc
a time from an angular speed and the angle swept across a chord
$n$ segments around an inscribed regular polygon
a area in surd form from triangles of radii with special angles

Looks like it, but isn't (4)

only two radii and an arc are involved, with no chord given, asked for or drawn: Arcs and sectors
a triangle whose own sides and angles are the data, with no vertex at the centre and no radius among the facts: Find a side or an angle: choose the rule
the area of the region is to be made as large or as small as possible: Optimisation (this type's area in terms of $\theta$ is its set-up)
an edge is the graph of a function, not an arc: Area between two curves

Worked example

A tunnel: the chord first, then a segment on a rectangle

Paper 2 (calculator)

Question

The cross-section of a tunnel is a rectangle $PQRS$ with a circular segment on top. $[PQ]$ is the floor and $PS = QR = 3$ m. The top of the rectangle, $[SR]$ , is a chord of a circle with centre $O$ and radius $5$ m, and the minor arc $SR$ above it is the roof. Angle $SOR = 2.4$ radians.
(a) Find the width of the tunnel, $SR$ . [3]
(b) Find the area of the cross-section. [4]
(c) Find the height of the tunnel at its highest point. [3]

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See it move: A segment is a sector minus a triangle

$\theta = 1.60$ : $SR = 2 \times 5\sin 0.80 = 7.17$ m, $OM = 5\cos 0.80 = 3.48$ m, and the top is $3 + 5 - 3.48 = 4.52$ m above the floor.
Roof: $\frac{1}{2}(5^{2})(1.60) - \frac{1}{2}(5^{2})\sin 1.60 = 20.00 - 12.49 = 7.51$ m². With the rectangle $3 \times 7.17$ the cross-section is 29.0 m².
Draw every radius: the roof is a segment, the sector $OSR$ minus the triangle $OSR$ , both with the angle $\theta$ at $O$ . Past $\pi$ the triangle counts the other way, and $\sin\theta < 0$ does that by itself.
(a) The chord by the cosine rule: $SR^{2} = 5^{2} + 5^{2} - 2(5)(5)\cos 2.4$ , so $SR = 9.3203\ldots \approx 9.32$ m. Keep it unrounded: it is the rectangle's width too.
(b) Segment $30 - 8.4432\ldots = 21.556\ldots$ m², plus the rectangle $3 \times 9.3203\ldots$ : $49.517\ldots \approx 49.5$ m². In degree mode $\sin 2.4$ comes out as 0.042, a sign the mode is wrong.
(c) $M$ is the midpoint of $[SR]$ , so angle $SOM = 1.2$ and $OM = 5\cos 1.2 = 1.8117\ldots$ . The top of the roof is 5 m above $O$ : height $3 + 5 - 1.8117\ldots \approx 6.19$ m. Check: between $3 \times 9.32 = 28.0$ and $9.32 \times 6.19 = 57.7$ m².

Which method?

What bounds the region decides the pieces.

a chord and an arc
Pieces
sector $-$ triangle (a segment)
Area
$\frac{1}{2}r^2(\theta - \sin\theta)$
an arc and straight edges that are not all radii (for example a diameter and a chord)
Pieces
sector $+$ triangle(s) of radii; find the missing angle at the centre first ( $\pi - \theta$ on a diameter, $2\pi$ minus the others at a point)
Area
$\frac{1}{2}r^2\theta + \frac{1}{2}r^2\sin(\pi - \theta)$ , …
two tangents from a point $P$ and the arc between them
Pieces
kite $OAPB$ (two right triangles) $-$ sector
Area
$r^2\tan\alpha - r^2\alpha$ , with $2\alpha$ the angle at the centre. If the angle at $P$ is given instead, say $2\beta$ : triangle $OAP$ has angle $\beta$ at $P$ , so $AP = \frac{r}{\tan\beta}$ and the kite is $\frac{r^2}{\tan\beta}$ ; the angle at the centre is $\pi - 2\beta$ (the kite's angles add up to $2\pi$ , and two of them are right angles), so the sector is $\frac{1}{2}r^2(\pi - 2\beta)$
a circle and an inscribed regular $n$ -sided polygon
Pieces
circle $-$ $n$ triangles, or $n$ segments
Area
$\pi r^2 - n \times \frac{1}{2}r^2\sin\frac{2\pi}{n}$
a rectangle with pieces cut off (or added) by chords and arcs
Pieces
rectangle $\pm$ segments
Area
rectangle $\pm$ $\frac{1}{2}r^2(\theta - \sin\theta)$ each

The region is bounded by	Pieces	Area
a chord and an arc	sector $-$ triangle (a segment)	$\frac{1}{2}r^2(\theta - \sin\theta)$
an arc and straight edges that are not all radii (for example a diameter and a chord)	sector $+$ triangle(s) of radii; find the missing angle at the centre first ( $\pi - \theta$ on a diameter, $2\pi$ minus the others at a point)	$\frac{1}{2}r^2\theta + \frac{1}{2}r^2\sin(\pi - \theta)$ , …
two tangents from a point $P$ and the arc between them	kite $OAPB$ (two right triangles) $-$ sector	$r^2\tan\alpha - r^2\alpha$ , with $2\alpha$ the angle at the centre. If the angle at $P$ is given instead, say $2\beta$ : triangle $OAP$ has angle $\beta$ at $P$ , so $AP = \frac{r}{\tan\beta}$ and the kite is $\frac{r^2}{\tan\beta}$ ; the angle at the centre is $\pi - 2\beta$ (the kite's angles add up to $2\pi$ , and two of them are right angles), so the sector is $\frac{1}{2}r^2(\pi - 2\beta)$
a circle and an inscribed regular $n$ -sided polygon	circle $-$ $n$ triangles, or $n$ segments	$\pi r^2 - n \times \frac{1}{2}r^2\sin\frac{2\pi}{n}$
a rectangle with pieces cut off (or added) by chords and arcs	rectangle $\pm$ segments	rectangle $\pm$ $\frac{1}{2}r^2(\theta - \sin\theta)$ each

Steps

Draw every radius
from the centre to the corners of the region. Split the region into sectors and triangles with a vertex at the centre (the table above).
Angles at the centre:
use the angle facts ( $\pi$ on a line, $2\pi$ at a point or in a quadrilateral), or half the angle from a perpendicular: $\cos(\text{half}) = \frac{\text{distance from } O \text{ to the chord}}{r}$ (angle from a ratio, in radians).
Lengths:
a chord by the cosine rule or $2r\sin\frac{\theta}{2}$ ; a tangent from $P$ by $r\tan\alpha$ ; a half-chord by Pythagoras.
Areas:
each triangle of radii is $\frac{1}{2}r^2\sin(\text{its angle at the centre})$ , each sector $\frac{1}{2}r^2(\text{its angle})$ . Add or subtract as the table says. Write $\sin(\pi - \theta) = \sin\theta$ as a line of its own when you use it.
An area is given:
write the equation. If $r$ is unknown and every term has $r^2$ (a ratio of two areas, or an area given as a fraction of the circle), divide by $r^2$ first. Solve on the GDC in radians (solve an equation, or zeros and intersections; Solving equations graphically). Give only the root the diagram allows ( $\theta = 0$ and angles outside the stated interval go).
Follow-ups:
an arc, a perimeter (arcs plus straight edges, never an inside radius), a time (angle $\div$ angular speed).

Common mistakes

Degrees in the segment formula,
or the GDC in degree mode: $\frac{1}{2}r^2(\theta - \sin\theta)$ needs $\theta$ in radians, and degree working caps the marks. Check the settings at the start of every part.
The triangle across a diameter with $\sin\theta$ and no reason
its angle is $\pi - \theta$ . Write $\sin(\pi - \theta) = \sin\theta$ as a step; skipping it has cost a mark.
Segment $=$ sector $+$ triangle,
or the triangle as $\frac{1}{2}r^2\theta$ : the segment is the sector minus the triangle, and the triangle has $\sin\theta$ .

More mistakes (5)

The chord as $2r\sin\theta$
it is $2r\sin\frac{\theta}{2}$ (half the angle, in half the triangle), or use the cosine rule.
More than one $\theta$ in the answer
$\theta = 0$ , or a root outside the stated interval, left in scores for the final mark.
An inside radius in the perimeter
the perimeter goes round the outside of the region only.
The booklet formula copied, not applied
writing $A = \frac{1}{2}r^2\theta$ without this region's angle and radius earns nothing.
Rounded intermediate values
in Example 3.4, the half-angle rounded to $0.93$ gives a time of $\frac{1.86}{\pi / 6} = 3.55$ seconds, not $3.54$ . Keep the full values (keep full-accuracy values).

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