Composite shapes: shaded regions with arcs, sectors and triangles

What you need to be able to do

What you should already know

The idea: every region is built from three shapes

However complicated a shaded region looks, its curved edges are arcs of circles and its straight edges are lines. So its area is always some combination of sectors, triangles and segments (a segment is itself sector − triangle), added or subtracted, and its perimeter is a sum of arcs and straight lengths. The work is in spotting which pieces, and finding the angle at the centre of each arc.

Example: the overlap of two circles. Two circles of radius 6 have their centres 6 apart. Each centre lies on the other circle, so the two centres and either crossing point form an equilateral triangle: the angle at each centre is \(2\times\tfrac\pi3=\tfrac{2\pi}3\). The overlap is two equal segments: \(2\times\tfrac12(36)\left(\tfrac{2\pi}3-\sin\tfrac{2\pi}3\right)=24\pi-18\sqrt3\approx44.2\).

A strategy for any shaded region

1. Label the diagram: every centre, every radius ("which circle is this arc part of, and what is its radius?"), every right angle and every equal length.
2. Write the plan in words before any numbers: "shaded = sector \(ADC\) − triangle \(ABC\)". It stops you computing pieces you don't need.
3. Find the angles you need at the centres of the arcs, in radians (next section). Never guess an angle from the look of the diagram.
4. Perimeter = each arc (\(r\theta\) with its own \(r\) and \(\theta\)) + each straight edge (often a difference of lengths, e.g. \(BD=AD-AB\)).
5. Area = add and subtract sector areas \(\tfrac12r^2\theta\), triangles \(\tfrac12ab\sin C\), segments.
6. Keep accuracy: store angles in the calculator; round only the final answer.

Finding the angle at the centre

Case by case

Key facts at a glance

• Every region = sectors, triangles and segments added or subtracted; every perimeter = arcs + straight lengths.
• Label every centre and radius; write the plan in words; then find each angle at a centre.
• Angles: tangent ⟂ radius (right triangles), isosceles base angles \(\tfrac{\pi-\theta}2\), the cosine rule, equilateral \(\tfrac\pi3\).
• Store values in the calculator; round only the final answer.

How this is examined

The June 2021 report: "candidates would be helped by writing down an overall strategy first". Schemes give M0 for assuming a triangle is equilateral, and for using \(180-2\theta\) as a base angle when the base angle is \(\theta\). The November 2021 report on a two-sector plate: "a common error was to calculate half the area of the triangle rather than the whole area".

Mark-scheme focus

Find the area of the shaded region (exact)

Credited

• angle \(\tfrac\pi3\) seen B1
• sector \(\tfrac12\times36\times\tfrac\pi3\) M1
• a correct strategy with values: 2 sectors − triangle, or sector + segment, or triangle + 2 segments M1 A1
• \(12\pi-9\sqrt3\): "must be simplified exact value" A1

Not credited

• decimals (25.1)
• \(\tfrac12\times\tfrac12\times36\sin\tfrac\pi3\) (half the triangle)
• \(6\pi-9\sqrt3+6\pi\) left unsimplified for the final mark

From: 9709/11/O/N/21 Q6(b) and the November 2021 report.

Worked examples

Example 1 · An arc centred at a vertex of a square (Paper 1, 8 marks) ★★ Medium

1 Read it with me what is actually happening?

The arc has centre \(A\), so its radius is \(AE\), not 12. Angle \(EAF\) is what is left of the right angle at \(A\) after the two small angles \(BAE\) and \(DAF\). The shaded region is the square minus two triangles minus the sector.

• centre Aradius \(AE=\sqrt{12^2+4^2}\)
• Show that … 4 s.f.show more figures than 0.9273

2 What is it asking for? the target, part by part

• (a)angle EAF2
• (b)perimeter3
• (c)area3

3 What do we know? everything the question gives us

• side 12
• BE = DF = 4EC = FC = 8

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)angle
  • Use\(\tfrac\pi2-2\tan^{-1}\tfrac4{12}\)
• (b)perimeter
  • r\(\sqrt{160}\)
  • Usearc + 8 + 8
• (c)area
  • Use144 − 24 − 24 − ½(160)(0.9273)
  • Trapr = 12 (scores M0)

tan⁻¹(4/12) in radians?

0.32175…

Area of triangle ABE?

\(\tfrac12\times12\times4=24\).

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \(\tfrac\pi2-2\tan^{-1}\tfrac13=1.570796-0.643501=0.927295\ldots=0.9273\) M1 A1
2. (b) \(AE=\sqrt{160}=12.65\) B1; arc \(=12.65\times0.9273=11.73\) M1; perimeter \(=11.73+8+8=27.7\) cm A1
3. (c) sector \(=\tfrac12(160)(0.9273)=74.18\) M1; shaded \(=144-24-24-74.18=21.8\) cm² M1 A1

6 Check it does the answer make sense?

21.8 cm² is about 15% of the square: about right for the small corner region in the diagram.

Example 2 · Two circles: a crescent in terms of r (Paper 1, 9 marks) ★★★ Hard

Why it is hard: you have to see the equilateral triangle, find two different centre angles, and organise four areas into one plan.

1 Read it with me what is actually happening?

\(BC=CD=s\) (radii of the big circle) and \(BD=s\): triangle \(BCD\) is equilateral, so \(\angle BCD=\tfrac\pi3\). \(A\) is the centre of the circle through \(B,C,D\), so \(\angle BAD=2\times\tfrac\pi3=\tfrac{2\pi}3\). The shaded crescent is the small circle's segment beyond \(BD\) minus the big circle's segment beyond \(BD\).

• in the form (a + bπ)r²exact: everything in terms of r, special angles
• Show that s = √3ra convincing proof with exact trig values

2 What is it asking for? the target, part by part

• (a)s = √3 r2
• (b)shaded area7

3 What do we know? everything the question gives us

• small circle: centre A, r
• big circle: centre C, s
• BD = s

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)s
  • Usecosine rule in ABD with angle \(\tfrac{2\pi}3\)
• (b)crescent
  • Small seg.\(\tfrac12r^2(\tfrac{2\pi}3-\sin\tfrac{2\pi}3)\)
  • Big seg.\(\tfrac12s^2(\tfrac\pi3-\sin\tfrac\pi3)\)
  • Shadedsmall − big

Why is angle BAD = 2π/3?

By symmetry \(A\) is the centre of equilateral triangle \(BCD\); the three angles at \(A\) are equal and sum to \(2\pi\).

s² in terms of r?

\(3r^2\).

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \(s^2=r^2+r^2-2r^2\cos\tfrac{2\pi}3=3r^2\Rightarrow s=\sqrt3r\) M1 A1
2. (b) small segment \(=\tfrac12r^2\left(\tfrac{2\pi}3-\tfrac{\sqrt3}2\right)=\left(\tfrac\pi3-\tfrac{\sqrt3}4\right)r^2\) B1 B1 B1
3. big segment \(=\tfrac12(3r^2)\left(\tfrac\pi3-\tfrac{\sqrt3}2\right)=\left(\tfrac\pi2-\tfrac{3\sqrt3}4\right)r^2\) DB1 DB1
4. shaded \(=\left(\tfrac\pi3-\tfrac{\sqrt3}4-\tfrac\pi2+\tfrac{3\sqrt3}4\right)r^2=\left(\tfrac{\sqrt3}2-\tfrac\pi6\right)r^2\) B1

6 Check it does the answer make sense?

\(\tfrac{\sqrt3}2-\tfrac\pi6\approx0.342\), positive and smaller than the small segment (0.614): the crescent is thin ✓.

Example 3 · Arc from the other end of an isosceles triangle (Paper 1, 8 marks) ★★ Medium

1 Read it with me what is actually happening?

Arc \(CD\) is centred at \(A\), so its radius is \(AC\) and its angle is \(\angle CAB\). Find both from the isosceles triangle. The shaded region is sector \(ACD\) minus triangle \(ABC\).

• centre Aradius AC, angle CAB
• ABD straightBD = AD − AB = AC − 6

2 What is it asking for? the target, part by part

• (a)perimeter5
• (b)area3

3 What do we know? everything the question gives us

• AB = BC = 6
• angle ABC = 1.8

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)perimeter
  • AC\(2\times6\sin0.9\)
  • angle\(\tfrac{\pi-1.8}2\)
  • Sumarc CD + BD + BC
• (b)area
  • Usesector ACD − triangle ABC

Why 6 sin 0.9?

Drop the perpendicular from B: it halves AC and the 1.8 angle.

Base angle?

\(\tfrac{\pi-1.8}2=0.6708\).

5 Now solve it have a go on paper first, then open the worked solution

1. \(AC=12\sin0.9=9.40\) M1 A1; \(\angle CAB=0.6708\) M1
2. arc \(CD=9.40\times0.6708=6.306\) M1; perimeter \(=6+3.40+6.306=15.7\) cm A1
3. (b) \(\tfrac12(9.40)^2(0.6708)-\tfrac12(36)\sin1.8=29.64-17.53=12.1\) cm² M1 M1 A1

6 Check it does the answer make sense?

Using 6 for the arc's radius would give a much smaller sector than the triangle: impossible, since the sector contains the triangle.

Summary

• Label centres and radii; write the plan (add/subtract) in words first.
• Angles: tangent ⟂ radius, isosceles base angles (π − θ)/2, cosine rule, equilateral π/3. Never guess from the picture.
• Each arc uses its own radius and angle; straight edges are often differences.
• Store values; 3 s.f. or exact at the end.

Can I do this?

Quiz

Concept checks

Answer each one in your head first, then open it.

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.