Areas Related to Circles • Topic 3 of 3

Areas of segment of a circle

What is a segment of a circle? A segment is a region of a circle bounded by a straight chord and an arc. Unlike a sector, a segment does not connect back to the center of the circle. Think of cutting a small rounded piece off the side of a circular log with a single straight saw cut, or looking at the water line when a circular glass cup is tipped sideways.

Segments come in pairs:

  • Minor Segment: The smaller region chopped off by the chord line.
  • Major Segment: The massive remaining region of the circle left on the other side of the chord line.

How to Calculate the Area of a Segment We cannot find the area of a segment directly using a single basic formula. Instead, we use subtraction:

1. First, find the area of the entire sector connecting the chord ends to the circle's center point. This looks like a complete pizza slice. 2. Next, calculate the area of the triangle formed inside that slice by the two radii lines and the straight chord line. 3. Finally, subtract the area of the triangle from the area of the sector. The leftover curved piece on the edge is your segment!

Mathematical Subtraction Step Rule:

$$\text{Area of Minor Segment} = \text{Area of Sector } OAPB - \text{Area of Triangle } OAB$$

To find the area of the interior triangle with radius $r$ and central angle $\theta$, you can use the formula: $\frac{1}{2} \cdot r^2 \cdot \sin(\theta)$.

The triangle area, three ways

The interior triangle $OAB$ is isosceles with both sides equal to the radius $r$ and included angle $\theta$. Its area can be written as:

$$\text{Area of }\triangle OAB=\frac{1}{2}r^2\sin\theta.$$

Two special cases save a lot of time:

  • $\theta=90^{\circ}$: the triangle is right-angled with both legs $=r$, so its area is $\frac{1}{2}r^2$ (since $\sin 90^{\circ}=1$).
  • $\theta=60^{\circ}$: the triangle is equilateral with side $r$, so its area is $\frac{\sqrt{3}}{4}r^2$.

Putting it together

$$\text{Area of minor segment}=\frac{\theta}{360^{\circ}}\pi r^2-\frac{1}{2}r^2\sin\theta.$$

$$\text{Area of major segment}=\pi r^2-\text{area of minor segment}.$$

A chord splits a circle into exactly two segments, so the two areas always add back to the full circle $\pi r^2$.

Areas of combinations of plane figures

Many board problems show a design — a flower bed, a brooch, a tile, a hand-fan, a running track, or a shaded region — built from circles, sectors, semicircles, triangles, squares and rectangles. The reliable strategy is:

  1. Break the figure into standard shapes whose areas you know.
  2. Decide which areas to add (parts that build up the shaded region) and which to subtract (holes or gaps cut out).
  3. Watch for repeated equal pieces (four corner cut-outs, two semicircular ends) and use symmetry to combine them.

For example, four quarter-circles of radius $r$ at the corners of a square together make exactly one full circle of area $\pi r^2$. A running track is a rectangle with two semicircular ends; its area is the rectangle plus one full circle of the end-radius.

Common mistakes to avoid

  • Forgetting to subtract the triangle — the segment is not the same as the sector.
  • Using $\frac{\sqrt{3}}{4}r^2$ for the triangle when $\theta\neq 60^{\circ}$. That equilateral shortcut only applies at $60^{\circ}$.
  • In combination figures, double-counting an overlap or forgetting that a "hole" must be subtracted.
  • Mixing $\pi=\frac{22}{7}$ with $\sqrt{3}=1.73$ inconsistently — use the values the question gives.
A circle with a chord AB cutting off a shaded minor segment; the minor segment equals the sector OAB minus the triangle OABMinor segment = sector OAB − triangle OABOABchordθminorsegmentArea = (θ/360)×πr^2 − (1/2)r^2 sinθA combination of plane figures: a square with a quarter circle removed from each of its four corners, leaving a shaded regionCombination figure: square − four corner quadrantsshadedregion4 quadrants= one full circleShaded = side^2 − πr^2
Solved Examples
1
Worked Example
A chord of a circle of radius 10 cm subtends a right angle ($90^{\circ}$) at the centre. Find the area of the corresponding minor segment. (Use $\pi=3.14$).
Solution
  1. Step 1: Sector area $=\frac{90}{360}\times\pi r^2=\frac{1}{4}\times 3.14\times 100=78.5\ \text{cm}^2$.
  2. Step 2: For $\theta=90^{\circ}$ the triangle is right-angled with legs $=r$, so triangle area $=\frac{1}{2}\times 10\times 10=50\ \text{cm}^2$.
  3. Step 3: Minor segment $=$ sector $-$ triangle $=78.5-50=28.5\ \text{cm}^2$.

Answer: $28.5\ \text{cm}^2$.

2
Worked Example
A chord of a circle of radius 14 cm subtends $60^{\circ}$ at the centre. Find the area of the minor segment. (Take $\pi=\frac{22}{7}$ and $\sqrt{3}=1.73$).
Solution
  1. Step 1: Sector area $=\frac{60}{360}\times\frac{22}{7}\times 196=\frac{1}{6}\times 616=\frac{308}{3}\approx 102.67\ \text{cm}^2$.
  2. Step 2: For $\theta=60^{\circ}$ the triangle is equilateral, area $=\frac{\sqrt{3}}{4}r^2=\frac{1.73}{4}\times 196=1.73\times 49=84.77\ \text{cm}^2$.
  3. Step 3: Minor segment $=102.67-84.77=17.9\ \text{cm}^2$.

Answer: $\approx 17.9\ \text{cm}^2$.

3
Worked Example
In a circle of radius 7 cm, the minor segment cut by a chord has area $14\ \text{cm}^2$. Find the area of the major segment. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Total circle area $=\pi r^2=\frac{22}{7}\times 49=154\ \text{cm}^2$.
  2. Step 2: Major segment $=$ circle $-$ minor segment.
  3. Step 3: $=154-14=140\ \text{cm}^2$.

Answer: $140\ \text{cm}^2$.

4
Worked Example
A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the minor segment. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Sector area $=\frac{90}{360}\times\frac{22}{7}\times 14\times 14=\frac{1}{4}\times 22\times 28=154\ \text{cm}^2$.
  2. Step 2: Triangle area $=\frac{1}{2}r^2=\frac{1}{2}\times 196=98\ \text{cm}^2$.
  3. Step 3: Minor segment $=154-98=56\ \text{cm}^2$.

Answer: $56\ \text{cm}^2$.

5
Worked Example
A chord of a circle of radius 12 cm subtends $120^{\circ}$ at the centre. Find the area of the corresponding minor segment. (Take $\pi=3.14$ and $\sqrt{3}=1.73$).
Solution
  1. Step 1: Sector area $=\frac{120}{360}\times 3.14\times 144=\frac{1}{3}\times 452.16=150.72\ \text{cm}^2$.
  2. Step 2: Triangle area $=\frac{1}{2}r^2\sin 120^{\circ}=\frac{1}{2}\times 144\times\frac{\sqrt{3}}{2}=36\times 1.73=62.28\ \text{cm}^2$.
  3. Step 3: Minor segment $=150.72-62.28=88.44\ \text{cm}^2$.

Answer: $88.44\ \text{cm}^2$.

6
Worked Example
A square ABCD has side 14 cm. A quarter circle of radius 14 cm is drawn with centre A, sweeping from B to D. Find the area of the shaded region inside the square but outside the quarter circle. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Area of square $=14\times 14=196\ \text{cm}^2$.
  2. Step 2: Area of quadrant $=\frac{1}{4}\pi r^2=\frac{1}{4}\times\frac{22}{7}\times 196=\frac{1}{4}\times 616=154\ \text{cm}^2$.
  3. Step 3: Shaded area $=$ square $-$ quadrant $=196-154=42\ \text{cm}^2$.

Answer: $42\ \text{cm}^2$.

7
Worked Example
From each corner of a square of side 28 cm, a quarter circle of radius 14 cm is cut off. A circle of diameter 14 cm is also removed from the centre. Find the area of the remaining part. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Area of square $=28\times 28=784\ \text{cm}^2$.
  2. Step 2: Four quarter circles of radius 14 cm together make one full circle: area $=\frac{22}{7}\times 14\times 14=616\ \text{cm}^2$.
  3. Step 3: Central circle has diameter 14 cm, so radius 7 cm; area $=\frac{22}{7}\times 49=154\ \text{cm}^2$.
  4. Step 4: Remaining area $=784-616-154=14\ \text{cm}^2$.

Answer: $14\ \text{cm}^2$.

8
Worked Example
A round table cover has six equal designs as shown, formed between a circle of radius 28 cm and a regular hexagon inscribed in it. Find the total area of the six designs given each design is a segment of central angle $60^{\circ}$. (Take $\pi=\frac{22}{7}$, $\sqrt{3}=1.73$).
Solution
  1. Step 1: Each design is the minor segment for $\theta=60^{\circ}$, $r=28$ cm.
  2. Step 2: Sector area $=\frac{60}{360}\times\frac{22}{7}\times 784=\frac{1}{6}\times 2464=410.67\ \text{cm}^2$.
  3. Step 3: Triangle (equilateral) $=\frac{\sqrt{3}}{4}r^2=\frac{1.73}{4}\times 784=1.73\times 196=339.08\ \text{cm}^2$.
  4. Step 4: One segment $=410.67-339.08=71.59\ \text{cm}^2$; six designs $=6\times 71.59=429.54\ \text{cm}^2$.

Answer: $\approx 429.54\ \text{cm}^2$.

9
Worked Example
A semicircular flower bed has diameter 14 m. A rectangular lawn 14 m by 7 m is attached to its straight edge. Find the total area of the flower bed and lawn together. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Semicircle radius $=7$ m; area $=\frac{1}{2}\pi r^2=\frac{1}{2}\times\frac{22}{7}\times 49=77\ \text{m}^2$.
  2. Step 2: Rectangle area $=14\times 7=98\ \text{m}^2$.
  3. Step 3: Total area $=77+98=175\ \text{m}^2$.

Answer: $175\ \text{m}^2$.

10
Worked Example
A brooch is made from a circular gold sheet of radius 35 mm. The boundary is decorated and the silver wire used along the diameter splits it into sectors of $30^{\circ}$ each. Find the total length of wire required for the circumference and for all five diameters. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Circumference $=2\pi r=2\times\frac{22}{7}\times 35=220$ mm.
  2. Step 2: A full circle of $360^{\circ}$ split into $30^{\circ}$ sectors needs $\frac{360}{30}=12$ radii, i.e. $\frac{12}{2}=6$ full diameters; but along 5 diameters here, wire $=5\times(2r)=5\times 70=350$ mm.
  3. Step 3: Total wire $=220+350=570$ mm.

Answer: $570$ mm.

11
Worked Example
A running track is shaped as a rectangle 70 m long with a semicircle at each end of radius 21 m. Find the total area enclosed by the track. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: The two semicircular ends together form one full circle of radius 21 m.
  2. Step 2: Circle area $=\pi r^2=\frac{22}{7}\times 21\times 21=22\times 3\times 21=1386\ \text{m}^2$.
  3. Step 3: Rectangle area $=$ length $\times$ width $=70\times(2\times 21)=70\times 42=2940\ \text{m}^2$.
  4. Step 4: Total enclosed area $=2940+1386=4326\ \text{m}^2$.

Answer: $4326\ \text{m}^2$.

12
Worked Example
On a square handkerchief of side 42 cm, nine equal circular designs each of radius 7 cm are embroidered (3 rows of 3). Find the area of the cloth left uncovered. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Area of square $=42\times 42=1764\ \text{cm}^2$.
  2. Step 2: Area of one circle $=\pi r^2=\frac{22}{7}\times 49=154\ \text{cm}^2$; nine circles $=9\times 154=1386\ \text{cm}^2$.
  3. Step 3: Uncovered area $=1764-1386=378\ \text{cm}^2$.

Answer: $378\ \text{cm}^2$.

Key Points

  • A segment of a circle is the space trapped between a straight chord line and an outer arc.
  • Segments do not naturally touch the center point of the circle on their own.
  • To find a minor segment's area, you must calculate Area of Sector minus Area of Triangle.
  • If the central angle is $90^\circ$, the interior triangle area simplifies to $\frac{1}{2}r^2$.
  • If the central angle is $60^\circ$, the interior triangle becomes equilateral with an area of $\frac{\sqrt{3}}{4}r^2$.
  • The major segment area is found by subtracting the minor segment area from the circle's total area.
  • The interior triangle area is $\frac{1}{2}r^2\sin\theta$; it equals $\frac{1}{2}r^2$ at $90^{\circ}$ and $\frac{\sqrt{3}}{4}r^2$ at $60^{\circ}$.
  • For combination figures: split into known shapes, then add the built-up parts and subtract the cut-out holes.
  • Four corner quarter-circles of equal radius combine into one full circle; a running track is a rectangle plus one full end-circle.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The area of a minor segment is:
Explanation: Segment $=$ sector $-$ triangle.
Q2.The area of a major segment is:
Explanation: Whole circle minus the minor segment.
Q3.A chord divides a circle into:
Explanation: Two segments.
Q4.Segment area uses a sector and a:
Explanation: The triangle formed by the two radii and the chord.
Practice more MCQs → 📝 Assignment · 35 marks