11.2 Segments of a Circle
A straight cut across a circle — a chord — slices off a cap called a segment. Its area is a sector with a triangle taken away, and for central angles of 60°, 90° and 120° that triangle is easy to find exactly.
Slice a round cake with one straight cut that does not pass through the centre. The cut makes a small cap and a large remaining piece. Neither is a sector, because the straight cut is not a radius. How do we find the area of the cap? Join the ends of the cut to the centre: the cap together with a triangle makes a sector, whose area we know from Section 11.1. So the cap is the sector minus the triangle.
In this course the central angle of the cap is $60^\circ$, $90^\circ$ or $120^\circ$. These three angles are chosen because the triangle is then special: an equilateral triangle, a right-angled isosceles triangle, and an isosceles triangle with a $120^\circ$ angle at the centre. We use $\sqrt3\approx1.732$ and $\sqrt2\approx1.414$ and state the value of $\pi$ in every problem.
Segment = sector − triangle
Do not confuse a segment with a sector. A sector is bounded by two radii and an arc; a segment is bounded by one chord and an arc. If the chord is a diameter, both segments are semicircles.
Reason. The sector $OAB$ is made of two parts: the triangle $OAB$ (between the two radii and the chord) and the minor segment (between the chord and the arc). Take the triangle away and the segment is what is left. The major segment is everything else in the circle. $\blacksquare$
The sector area comes from Theorem 11.2: $\dfrac{\theta}{360}\pi r^2$. For the triangle we use the three central angles that the syllabus asks for.
| $\theta$ | chord $AB$ | area of $\triangle OAB$ | area of the minor segment |
|---|---|---|---|
| $60^\circ$ | $r$ | $\dfrac{\sqrt3}{4}r^2$ | $r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)$ |
| $90^\circ$ | $r\sqrt2$ | $\dfrac12r^2$ | $r^2\left(\dfrac{\pi}{4}-\dfrac12\right)$ |
| $120^\circ$ | $r\sqrt3$ | $\dfrac{\sqrt3}{4}r^2$ | $r^2\left(\dfrac{\pi}{3}-\dfrac{\sqrt3}{4}\right)$ |
Proof. Since $OA=OB=r$, the triangle $OAB$ is isosceles.
- $60^\circ$. The base angles are equal and the three angles add up to $180^\circ$, so each is $\dfrac{180^\circ-60^\circ}{2}=60^\circ$. The triangle is equilateral with side $r$, so $AB=r$ and its area is $\dfrac{\sqrt3}{4}r^2$. The sector is $\dfrac16\pi r^2$.
- $90^\circ$. The triangle is right-angled at $O$ with legs $r$ and $r$. Pythagoras gives $AB=\sqrt{r^2+r^2}=r\sqrt2$, and the area is $\dfrac12\times r\times r=\dfrac12r^2$. The sector is $\dfrac14\pi r^2$.
- $120^\circ$. Draw $OM\perp AB$ with $M$ on $AB$. The right triangles $OMA$ and $OMB$ are congruent (common side $OM$, equal hypotenuses $OA=OB$), so $\angle AOM=\angle BOM=60^\circ$ and $M$ is the midpoint of $AB$. In the right triangle $OMA$ the angle at $A$ is $30^\circ$, so $OM=r\cos60^\circ=\dfrac r2$ and $AM=r\sin60^\circ=\dfrac{\sqrt3}{2}r$. Hence $AB=\sqrt3\,r$ and the area of $\triangle OAB$ is $\dfrac12\times\sqrt3r\times\dfrac r2=\dfrac{\sqrt3}{4}r^2$. The sector is $\dfrac13\pi r^2$.
Subtracting the triangle from the sector gives the last column. $\blacksquare$
- Writing "segment = sector" and forgetting to subtract the triangle, or subtracting the wrong triangle: it must be the triangle with the centre as a vertex.
- Using $\dfrac{\sqrt3}{4}r^2$ for the $90^\circ$ triangle. Only the $60^\circ$ triangle is equilateral; for $120^\circ$ the same area $\dfrac{\sqrt3}{4}r^2$ happens to appear for a different reason (the height is $\dfrac r2$, not $\dfrac{\sqrt3}{2}r$).
- Taking the whole circle minus the sector for the major segment. The major segment is the circle minus the minor segment; equivalently, the major sector plus the triangle.
- Draw the circle, the chord $AB$, the centre $O$ and the radii $OA$, $OB$. Mark $\theta$.
- State $\pi$ and the root values you will use ($\sqrt3\approx1.732$, $\sqrt2\approx1.414$).
- Sector: $\dfrac{\theta}{360}\pi r^2$. Triangle: use the table of Theorem 11.6 (or half base × height).
- Minor segment = sector − triangle. For a major segment subtract the minor segment from $\pi r^2$.
- If the chord length is given, find $r$ from the table ($AB=r$, $r\sqrt2$ or $r\sqrt3$) first.
Regular figures drawn inside a circle
Segments of $60^\circ$, $90^\circ$ and $120^\circ$ appear together when a regular polygon is drawn inside a circle. Take a circle of radius $r$ and mark points on it that are $60^\circ$ apart as seen from the centre: the chords join up to make a regular hexagon ($6$ segments of $60^\circ$). Points $90^\circ$ apart make a square ($4$ segments of $90^\circ$), and points $120^\circ$ apart make an equilateral triangle ($3$ segments of $120^\circ$). A polygon with $n$ sides has $n$ equal segments, so $$\text{area of circle}-\text{area of polygon}=n\times\text{area of one segment}.$$
Exercises
- 1 Find the area of the minor segment of a circle of radius $6$ cm cut off by a chord subtending $60^\circ$ at the centre. (Use $\pi=3.14$.)
- 2 A chord subtends $90^\circ$ at the centre of a circle of radius $21$ cm. Find the area of the minor segment. (Use $\pi=\tfrac{22}{7}$.)
- 3 A chord subtends $120^\circ$ at the centre of a circle of radius $7$ cm. Find the area of the minor segment. (Use $\pi=\tfrac{22}{7}$.)
- 4 Find the area of the minor segment of a circle of radius $20$ cm for a central angle of $60^\circ$. (Use $\pi=3.14$.)
- 5 Find the area of the minor segment of a $90^\circ$ chord in a circle of radius $28$ cm. (Use $\pi=\tfrac{22}{7}$.)
- 6 A chord of length $14$ cm subtends $60^\circ$ at the centre. Find the radius and the area of the minor segment. (Use $\pi=\tfrac{22}{7}$.)
- 7 A chord of length $6\sqrt3$ cm subtends $120^\circ$ at the centre. Find the radius and the area of the minor segment. (Use $\pi=3.14$.)
- 8 A chord of length $8\sqrt2$ cm subtends a right angle at the centre. Find the radius and the area of the minor segment. (Use $\pi=3.14$.)
- 1 Find the area of the major segment of a circle of radius $7$ cm when the chord subtends $120^\circ$. (Use $\pi=\tfrac{22}{7}$, $\sqrt3\approx1.732$.)
- 2 Find the areas of the minor and major segments when a chord subtends $90^\circ$ in a circle of radius $10$ cm. (Use $\pi=3.14$.)
- 3 Find the perimeter of the minor segment for a $90^\circ$ chord in a circle of radius $14$ cm. (Use $\pi=\tfrac{22}{7}$, $\sqrt2\approx1.414$.)
- 4 Find the perimeter of the minor segment for a $60^\circ$ chord in a circle of radius $21$ cm. (Use $\pi=\tfrac{22}{7}$.)
- 5 Find the perimeter of the major segment for a $120^\circ$ chord in a circle of radius $6$ cm. (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
- 6 Show that the minor segment for a $90^\circ$ chord has area $\dfrac{r^2(\pi-2)}{4}$, and with $\pi=3.14$ find what percentage of the circle's area it is (to $1$ decimal place).
- 7 A circle of radius $10$ cm has a chord that subtends $120^\circ$ at the centre. Find the ratio of the areas of the minor and the major segment, to $2$ decimal places. (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
- 1 A circular pond of radius $35$ m has a straight wooden bridge across it which subtends $90^\circ$ at the centre. Find the area of water on the smaller side of the bridge. (Use $\pi=\tfrac{22}{7}$.)
- 2 A pizza of radius $12$ cm is cut by a straight cut that subtends $60^\circ$ at the centre. Find the area of the small piece. (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
- 3 A circular rangoli of radius $14$ cm has an equilateral triangle with its corners on the circle. Find the total area of the three segments outside the triangle, and the area of each. (Use $\pi=\tfrac{22}{7}$, $\sqrt3\approx1.732$.)
- 4 A regular hexagon is inscribed in a circle of radius $10$ cm. Find the area of the circle outside the hexagon. (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
- 5 A circular signboard of radius $21$ cm has a red segment cut off by a chord subtending $120^\circ$. Find the red area. (Use $\pi=\tfrac{22}{7}$, $\sqrt3\approx1.732$.)
- 6 A circular garden of radius $14$ m has a straight path (a chord) cutting off a segment that subtends $60^\circ$ at the centre. The smaller piece is to be turfed at ₹ $40$ per m². Find the cost, to the nearest rupee. (Use $\pi=\tfrac{22}{7}$, $\sqrt3\approx1.732$.)
- 1 A square of side $14$ cm is inscribed in a circle. Find the radius and the total area of the four segments outside the square. (Use $\pi=3.14$.)
- 2 A regular hexagon of side $8$ cm is inscribed in a circle. Find the area of the circle outside the hexagon. (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
- 3 Show that the minor segment for a $120^\circ$ chord has area $\dfrac{r^2(4\pi-3\sqrt3)}{12}$, and find its value for $r=6$ cm with $\pi=3.14$ and $\sqrt3\approx1.732$.
- 4 Two chords of a circle of radius $10$ cm subtend $90^\circ$ and $60^\circ$ at the centre. Which minor segment is larger, and by how much? (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
How this is examined: a segment question is typically worth $2$ to $5$ marks. Marks are given for the figure, the sector area, the triangle area and the subtraction. Questions on the major segment, or on a perimeter, add one more step. The central angle is $60^\circ$, $90^\circ$ or $120^\circ$, and the value of $\pi$ is given in the question.
E1. Board · 1 mark The area of the minor segment of a circle of radius $2$ cm cut off by a chord that subtends $90^\circ$ at the centre is (A) $(\pi-2)$ cm² (B) $(2\pi-2)$ cm² (C) $(\pi-1)$ cm² (D) $(4\pi-2)$ cm²
E2. Board · 1 mark Assertion (A): The area of the minor segment cut off by a $60^\circ$ chord is less than the area of the $60^\circ$ sector. Reason (R): Area of segment $=$ area of sector $-$ area of the triangle formed by the two radii and the chord.
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
E3. Board · 2 marks A chord of a circle of radius $20$ cm subtends a right angle at the centre. Find the area of the minor segment. (Use $\pi=3.14$.)
E4. Board · 3 marks A chord of a circle of radius $12$ cm subtends $120^\circ$ at the centre. Find the areas of the minor and the major segments. (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
E5. Board · 5 marks Find the perimeter and the area of the minor segment cut off by a chord that subtends $120^\circ$ at the centre of a circle of radius $9$ cm. (Use $\pi=3.14$, $\sqrt3\approx1.732$.)
E6. Board · Case study A circular fish pond has radius $14$ m. A straight wooden bridge $AB$ across the pond subtends a right angle at the centre $O$ of the pond. (Use $\pi=\tfrac{22}{7}$, $\sqrt2\approx1.414$.)
(i) Find the length of the bridge. (1 mark)
(ii) Find the area of the water on the smaller side of the bridge. (2 marks)
(iii) Find the area of the water on the larger side. (1 mark)
Research question. If you draw a regular hexagon inside a circle with only a compass and a ruler, how close is the area outside the hexagon to the value $6\times r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)$ predicted by the segment formula?
Why it matters. Designers of tiles, windows, wheels and floor patterns build shapes from circles and chords. Stepping a compass round a circle gives six equal chords because the chord of a $60^\circ$ arc equals the radius (Theorem 11.6). Knowing how much material is cut away or left over (the segments) helps in costing.
What to do.
- Draw a circle of radius $r=6$ cm on card. Without changing the compass, put the point on the circle and mark an arc across it; move to that mark and repeat, going round. You should return almost exactly to the start with six marks.
- Join consecutive marks to make the hexagon, and cut along the sides to separate the six segments from the hexagon.
- Weigh the whole circle (mass $M$) and the six segments together (mass $m$). Estimate the segment area as $\dfrac{m}{M}\times\pi r^2$ with $\pi=3.14$.
- Compute the same area from the formula: $6\times r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)$ and compare.
- Repeat with a square (mark $90^\circ$ with a protractor) and an equilateral triangle ($120^\circ$).
Tools: Compass, ruler, protractor, card, scissors and a kitchen scale (grams)
Hand in: A photo of your cut-out pieces, a table of measured and calculated areas for the three polygons, and one paragraph on the error you found.
Going further. Which shape leaves the least material outside it — the triangle, the square or the hexagon? What would a regular polygon with very many sides leave?
🧪 Self-check quiz
Q1. The area of the minor segment of a circle of radius $14$ cm, for a central angle of $90^\circ$ (take $\pi=\tfrac{22}{7}$), is
Q2. A chord subtends $60^\circ$ at the centre of a circle. The triangle formed by the chord and the two radii is
Q3. In a circle of radius $r$, the chord that subtends $120^\circ$ at the centre has length
Q4. A chord subtends $90^\circ$ at the centre of a circle of radius $7$ cm. The area of the major segment is (take $\pi=\tfrac{22}{7}$)
Q5. The area of the minor segment for a $60^\circ$ central angle in a circle of radius $r$ is