Area of Minor and Major Segments of a Circle for CBSE Class 10

When you draw a straight chord across a circular disc, you slice the circle into two unequal regions. Unlike a sector (which connects to the center like a pie slice), the region trapped between the straight chord and the curved arc is called a segment of a circle. Calculating the area of a segment is one of the most intellectually rewarding topics in Class 10 geometry, synthesizing sector mensuration with triangle trigonometry.

In CBSE Class 10 Mathematics, Chapter 11 (Areas Related to Circles), segment area calculations for central angles of $60^\circ$, $90^\circ$, and $120^\circ$ represent standard 3-mark and 4-mark questions in board examinations.

What You Will Learn

• Formal definition of a segment: Minor segment vs. Major segment
• The Master Formula: $\text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle}$
• How to calculate the area of the corresponding triangle $\Delta OAB$:
  • When $\theta = 90^\circ$ (Right-angled isosceles triangle)
  • When $\theta = 60^\circ$ (Equilateral triangle: $\frac{\sqrt{3}}{4} r^2$)
  • When $\theta = 120^\circ$ (Trigonometric perpendicular partition)
• Calculating the area of a Major Segment
• Step-by-step solved CBSE board examination problems
• Presentation guidelines and radical handling

1. What is a Segment of a Circle?

Formal Definition

A segment of a circle is the region of the circular plane bounded by a chord and the corresponding arc subtended by the chord.

                                  O (Center)
                                 /                         Radius r/   \ Radius r
                               /  θ                                A-------B (Chord AB)
                             [  TRIANGLE  ]
                             ~~~~~~~~~~~~~~  <-- MINOR SEGMENT (Shaded)

1. Minor Segment: The region enclosed between the chord $AB$ and the minor arc $APB$.
2. Major Segment: The remaining larger portion of the circular region enclosed between chord $AB$ and the major arc $AQB$.

2. The Master Formula for the Area of a Minor Segment

To find the area of the shaded minor segment, look at the overall sector $OAPB$: The sector consists of two distinct non-overlapping parts: the interior triangle $\Delta OAB$ and the minor segment.

The Master Formula: $\mathbf{\text{Area of Minor Segment} = \text{Area of Sector } OAPB - \text{Area of } \Delta OAB}$ $\mathbf{\text{Area of Minor Segment} = \left(\frac{\theta}{360^\circ} \times \pi r^2\right) - \text{Area of } \Delta OAB}$

3. Calculating Area of $\Delta OAB$ for the Three Standard Angles

The sector area is straightforward, but how do we calculate $\text{Area}(\Delta OAB)$? The method depends strictly on the central angle $\theta$:

Case 1: When Central Angle $heta = 90^\circ$ (Right-Angled Triangle)

When $\theta = 90^\circ$, $OA \perp OB$. Since both sides are radii ($OA = OB = r$): $\mathbf{\text{Area}(\Delta OAB) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times r \times r = \frac{1}{2} r^2}$

Case 2: When Central Angle $heta = 60^\circ$ (Equilateral Triangle)

When $\theta = 60^\circ$:

• In $\Delta OAB$, $OA = OB = r \implies \angle OAB = \angle OBA$.
• Since $\angle AOB = 60^\circ$, the remaining two angles must sum to $120^\circ$, meaning $\angle OAB = \angle OBA = 60^\circ$.
• All three angles are $60^\circ$, which means $\Delta OAB$ is an equilateral triangle with side length $r$!
• The area of an equilateral triangle of side $r$ is: $\mathbf{\text{Area}(\Delta OAB) = \frac{\sqrt{3}}{4} r^2}$

Case 3: When Central Angle $heta = 120^\circ$ (Isosceles Triangle)

When $\theta = 120^\circ$, draw an altitude $OM \perp AB$:

• In $\Delta OMA$ and $\Delta OMB$, by RHS congruence, $OM$ bisects $\angle AOB$ and bisects base $AB$: $\angle AOM = \angle BOM = 60^\circ$
• Using trigonometry in right triangle $\Delta OMA$: $\cos 60^\circ = \frac{OM}{r} \implies OM = r \cos 60^\circ = \frac{r}{2} \quad (\text{Height})$ $\sin 60^\circ = \frac{AM}{r} \implies AM = r \sin 60^\circ = \frac{r\sqrt{3}}{2}$
• Total base $AB = 2 \times AM = 2 \times \frac{r\sqrt{3}}{2} = r\sqrt{3}$.
• Area of $\Delta OAB$: $\mathbf{\text{Area}(\Delta OAB) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (r\sqrt{3}) \times \left(\frac{r}{2}\right) = \frac{\sqrt{3}}{4} r^2}$

The Universal Shortcut: <u>For any central angle $heta$, the area of the isosceles triangle formed by two radii $r$ is given by: ext{Area}(\Delta OAB) = rac{1}{2} r^2 \sin heta.</u>

4. Calculating Area of the Major Segment

Once the area of the minor segment is known:

$\mathbf{\text{Area of Major Segment} = \text{Total Area of Circle } (\pi r^2) - \text{Area of Minor Segment}}$

5. Solved CBSE Board Examination Problems

Solved Example 1: Segment with $heta = 60^\circ$ (NCERT Classic)

Problem: A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the center. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$).

Solution:

1. Given: $r = 15\text{ cm}$ and $\theta = 60^\circ$.
2. Calculate Area of Sector $OAPB$: $\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times 3.14 \times 15 \times 15 = \frac{1}{6} \times 706.5 = \mathbf{117.75\text{ cm}^2}$
3. Calculate Area of $\Delta OAB$ (Equilateral Triangle): $\text{Area}(\Delta OAB) = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 15 \times 15 = \frac{1.73 \times 225}{4} = \frac{389.25}{4} = \mathbf{97.3125\text{ cm}^2}$
4. Calculate Area of Minor Segment: $\text{Area of Minor Segment} = 117.75 - 97.3125 = \mathbf{20.4375\text{ cm}^2}$
5. Calculate Area of Major Segment:
   • Total area of circle $= \pi r^2 = 3.14 \times 15 \times 15 = 706.5\text{ cm}^2$. $\text{Area of Major Segment} = 706.5 - 20.4375 = \mathbf{686.0625\text{ cm}^2}$
6. Therefore, <u>the area of the minor segment is $20.44\text{ cm}^2$ and the area of the major segment is $686.06\text{ cm}^2$</u>.

Solved Example 2: Segment with $heta = 120^\circ$ (CBSE Board Classic)

Problem: A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the center. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$).

Solution:

1. Given: $r = 12\text{ cm}$ and $\theta = 120^\circ$.
2. Calculate Area of Sector: $\text{Area of Sector} = \frac{120^\circ}{360^\circ} \times 3.14 \times 12 \times 12 = \frac{1}{3} \times 3.14 \times 144 = 3.14 \times 48 = \mathbf{150.72\text{ cm}^2}$
3. Calculate Area of $\Delta OAB$: $\text{Area}(\Delta OAB) = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 12 \times 12 = 1.73 \times 36 = \mathbf{62.28\text{ cm}^2}$
4. Calculate Area of Minor Segment: $\text{Area of Segment} = \text{Sector Area} - \text{Triangle Area} = 150.72 - 62.28 = \mathbf{88.44\text{ cm}^2}$
5. Therefore, <u>the area of the segment is $88.44\text{ cm}^2$</u>.

6. Summary and Examination Tips

Exam Tip: Remember that both $\theta = 60^\circ$ and $\theta = 120^\circ$ yield the exact same formula for triangle area: $\frac{\sqrt{3}}{4} r^2$! (Since $\sin 60^\circ = \sin 120^\circ = \frac{\sqrt{3}}{2}$).

Common Mistake: Subtracting the minor segment from the sector instead of from the whole circle to find the major segment. To find the major segment, subtract the minor segment from the ENTIRE circle ($\pi r^2$)!