Area of a Segment (Higher) — GCSE Maths Tutorial, Worked Examples & Practice

Revise Area of a Segment (Higher) (finding the area and perimeter of circle segments using sector area and 1/2 ab sin c (higher)) within Geometry and Measures — Circles, aligned to Edexcel, AQA & OCR GCSE (9–1) and International GCSE Maths. The interactive practice tool at the top of this page includes 30 self-marking practice questions across 3 structured skill steps, with instant diagnostic feedback and full worked solutions.

Step 1 of 310 interactive practice questions

Area of a Minor Segment (Higher)

Subtracting 1/2 r² sin θ from sector area to find the area of a segment

Key Method & Revision Notes

A chord $AB$ slices a circle into two segments. To find the area of the shaded minor segment, subtract the area of triangle $OAB$ from the area of sector $OAB$:
Area of a Minor Segment
$$\text{Area} = \frac{\theta}{360}\pi r^2 - \frac{1}{2}r^2\sin\theta$$
(Sector $OAB$ area $-$ Triangle $OAB$ area)
Notice both sides around angle $\theta$ are radii ($OA = OB = r$), so $\frac{1}{2}ab\sin C$ simplifies neatly to $\frac{1}{2}r^2\sin\theta$!

Worked Example

Question:

A sector $OAB$ has radius $9\text{ cm}$ and angle $\angle AOB = 100^\circ$. Calculate the area of the minor segment correct to 3 significant figures.

Step-by-Step Solution:

1. Sector area $= \dfrac{100}{360} \times \pi \times 9^2 = 70.686\text{ cm}^2$.

2. Triangle area $= \dfrac{1}{2} \times 9^2 \times \sin(100^\circ) = 39.885\text{ cm}^2$.

3. Segment area $= 70.686 - 39.885 = 30.801 \approx \mathbf{30.8}\text{ cm}^2$.

$$ 30.8\text{ cm}^2 $$

Ready to test this skill? Use the interactive tool at the top of the page to practise 10 self-marking exam-style questions (30 questions loaded in this topic module).

Step 2 of 310 interactive practice questions

Perimeter of a Segment (Higher)

Combining arc length and Cosine Rule chord length to find segment perimeter

Key Method & Revision Notes

The perimeter of a segment has two edges: the curved arc $AB$ plus the straight chord $AB$.
Common Exam Trap: Do not add the two radii $OA$ and $OB$ — they are inside the circle, not around the boundary of the segment!
1. Arc length: $\text{Arc } AB = \dfrac{\theta}{360} \times 2\pi r$
2. Chord length (Cosine Rule on $\triangle OAB$): $$AB = \sqrt{2r^2 - 2r^2\cos\theta}$$(Shortcuts: if $\theta = 60^\circ$, $\triangle OAB$ is equilateral so $AB = r$; if $\theta = 90^\circ$, $AB = r\sqrt{2}$.)
3. Perimeter: $\text{Arc } AB + \text{Chord } AB$.

Worked Example

Question:

A sector $OAB$ has radius $10\text{ cm}$ and angle $80^\circ$. Work out the perimeter of the minor segment correct to 3 significant figures.

Step-by-Step Solution:

1. Arc $AB = \dfrac{80}{360} \times 2\pi(10) = 13.963\text{ cm}$.

2. Chord $AB = \sqrt{10^2 + 10^2 - 2(10)(10)\cos(80^\circ)} = 12.856\text{ cm}$.

3. Perimeter $= 13.963 + 12.856 = 26.818 \approx \mathbf{26.8}\text{ cm}$.

$$ 26.8\text{ cm} $$

Ready to test this skill? Use the interactive tool at the top of the page to practise 10 self-marking exam-style questions (30 questions loaded in this topic module).

Step 3 of 310 interactive practice questions

Major Segments, Lenses & Reverse Problems (Higher)

Solving major segments, overlapping circles and working backwards to find the radius

Key Method & Revision Notes

Higher GCSE Paper 2/3 problem-solving questions extend segments in three ways:
1. Major Segment Area: Subtract the minor segment area from the full circle area $\pi r^2$.
2. Overlapping Identical Circles (Lens Shape): Two identical arcs back-to-back form two equal segments, so $\text{Lens Area} = 2 \times \text{Segment Area}$.
3. Reverse Segment Problems (Finding $r$): Factor out $r^2$ first: $$\text{Area} = r^2\left(\frac{\theta\pi}{360} - \frac{1}{2}\sin\theta\right)$$ Then divide by the bracket and take the square root!

Worked Example

Question:

A minor segment with central angle $90^\circ$ has area $18.2\text{ cm}^2$. Calculate the radius $r$ of the circle correct to 1 decimal place.

Step-by-Step Solution:

1. Factor out $r^2$: $18.2 = r^2\left(\dfrac{90\pi}{360} - \dfrac{1}{2}\sin 90^\circ\right) = r^2(0.28540)$.

2. $r^2 = 18.2 \div 0.28540 = 63.771$.

3. $r = \sqrt{63.771} = 7.986 \approx \mathbf{8.0}\text{ cm}$.

$$ r = 8.0\text{ cm} $$

Ready to test this skill? Use the interactive tool at the top of the page to practise 10 self-marking exam-style questions (30 questions loaded in this topic module).

Area of a Segment (Higher) — GCSE Maths Tutorial, Worked Examples & Practice

Area of a Minor Segment (Higher)

Key Method & Revision Notes

Worked Example

Perimeter of a Segment (Higher)

Key Method & Revision Notes

Worked Example

Major Segments, Lenses & Reverse Problems (Higher)

Key Method & Revision Notes

Worked Example

Related GCSE Maths Topics (Circles)