How is the area of a Segment (minor segment) calculated in CBSE Class 10?

\text{Area of Minor Segment} = \text{Area of Sector } OAPB - \text{Area of } \Delta OAB. For sector angle \theta: \text{Area} = \frac{\theta}{360^\circ}\pi r^2 - \frac{1}{2}r^2\sin\theta. For \theta = 60^\circ, \text{Area}(\Delta) = \frac{\sqrt{3}}{4}r^2; for \theta = 90^\circ, \text{Area}(\Delta) = \frac{1}{2}r^2.

How do you find the area of the Major Segment and Major Sector?

\text{Area of Major Sector} = \pi r^2 - \text{Area of Minor Sector} = \frac{360^\circ - \theta}{360^\circ} \times \pi r^2. \text{Area of Major Segment} = \pi r^2 - \text{Area of Minor Segment}.

What is the angle swept by the minute hand of a clock in $1$ minute and in $5$ minutes?

In 60 minutes, the minute hand completes 360^\circ. In 1 minute, it sweeps \frac{360^\circ}{60} = 6^\circ. In 5 minutes, it sweeps 5 \times 6^\circ = 30^\circ. In 15 minutes, it sweeps 90^\circ.

What is the difference between $\pi = \frac{22}{7}$ and $\pi = 3.14$ in exam scoring?

Always use \pi = \frac{22}{7} unless the question explicitly instructs "Use \pi = 3.14". If \pi = 3.14 is specified, using \frac{22}{7} will lead to slight decimal discrepancies that may cost ½ mark in final answer accuracy.

Are combinations of complex plane figures (e.g. shaded regions with triangles and squares) included in CBSE 2026–27?

The rationalized syllabus has simplified the scope to areas of sectors and segments of a circle. Elaborate shaded composite designs from older Exercise 12.3 have been trimmed, focusing on clear sectors and segments with central angles of 60^\circ, 90^\circ, 120^\circ.

How do you find the distance travelled by a circular wheel in $N$ revolutions?

In 1 revolution, the wheel travels its circumference C = 2\pi r = \pi d. Total distance travelled in N revolutions is \text{Distance} = N \times 2\pi r. Conversely, \text{Number of Revolutions} = \frac{\text{Total Distance}}{2\pi r}.

All Revision Notes

MathematicsCh-11 15 min comprehensive revision

NCERT Class 10 Mathematics — Chapter 11

Areas Related to Circles

Perimeter and area of circles, length of an arc, area of sector (major and minor), area of segment of a circle, and combinations of plane figures.

Quick Key Takeaways:

Perimeter & Area: Circumference

= 2\pi r

, Area

= \pi r^2

Length of an Arc: For a sector of angle

\theta

l = \frac{\theta}{360^\circ} \times 2\pi r

Area of Sector: For sector of central angle

\theta

\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{1}{2} l r

Area of Segment (Minor):

\text{Area of Minor Segment} = \text{Area of Sector } OAPB - \text{Area of } \Delta OAB = \frac{\theta}{360^\circ}\pi r^2 - \frac{1}{2}r^2\sin\theta

Area of Major Sector / Segment: Major Sector

= \pi r^2 - \text{Minor Sector}

; Major Segment

= \pi r^2 - \text{Minor Segment}

Jump to Concept:1. 1. Circle Geometry, Arc Length & Sector Area Formulations 2. 2. Area of Segment Formulations & Combinations of Plane Figures 3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Circular Brooch Jewelry Design Modeling 5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. (Use $\pi = 22/7$ )

•

Step 1 (Angle Calculation): Angle swept in 60 minutes

= 360^\circ

\text{Angle swept in 5 minutes } \theta = \frac{360^\circ}{60} \times 5 = 30^\circ

•

Step 2 (Area of Sector Calculation):

\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2

\text{Area} = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{1}{12} \times 616 = \frac{154}{3}\text{ cm}^2

•

Final Boxed Answer:

\mathbf{\text{Area Swept} = \frac{154}{3}\text{ cm}^2 \approx 51.33\text{ cm}^2}

• 5-Mark Heavyweight Board Problem / Rider: A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$ )

•

Step 1 (Area of Sector $OAPB$ ):

\text{Area of Minor 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}

•

Step 2 (Area of Equilateral $\Delta OAB$ with $\theta = 60^\circ$ ):

\text{Area}(\Delta OAB) = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 225 = \frac{389.25}{4} = \mathbf{97.3125\text{ cm}^2}

•

Step 3 (Area of Minor Segment):

\text{Area of Minor Segment} = 117.75 - 97.3125 = \mathbf{20.4375\text{ cm}^2}

•

Step 4 (Area of Major Segment):

\text{Total Circle Area} = \pi r^2 = 3.14 \times 225 = 706.5\text{ cm}^2

\text{Area of Major Segment} = 706.5 - 20.4375 = \mathbf{686.0625\text{ cm}^2}

•

Final Boxed Answer:

\mathbf{\text{Minor Segment Area} = 20.44\text{ cm}^2, \quad \text{Major Segment Area} = 686.06\text{ cm}^2}

Authentic Board Question (3 Marks)Topic: Areas Related to Circles Mathematical Proof & Computations

State the governing mathematical theorem/formula, show complete geometric or computational steps, and find the exact result for Areas Related to Circles.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Given / To Prove / Formula Setup: State the governing theorem (BPT, Tangents, Pythagoras), mensuration formula, or trigonometric ratios with given values.

1 Mark

Step 2: Step-by-Step Derivation / Arithmetic Working: Show complete deductive steps or numerical calculations with

\pi = 22/7

or trigonometric standard values.

1 Mark

Step 3: Final Boxed Answer with Units / Q.E.D.: State the final numerical result with proper square/cubic units (

\text{cm}^2, \text{m}^3

) or formal proof conclusion.

1 Mark

Model Student Answer (Target: Full 3/3 Marks):

To score full 3 marks on Areas Related to Circles in CBSE Mathematics:

1. Theorem / Formula: Write the standard governing formula or state the geometric theorem clearly.
2. Calculation / Proof: Substitute dimensions systematically or provide deductive step justifications.
3. Final Result: Box the final answer with required units (

\text{cm}, \text{cm}^2, \text{m}^3

) or conclude with "Hence Proved".

Examiner Mark Deduction Traps:

•Always write intermediate calculation lines to earn partial credit under CBSE step-marking.

•State reasons in parentheses (e.g., "[Tangents from an external point are equal]") during geometric proofs.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

1. Area of Sector:

A = \frac{\theta}{360^\circ} \times \pi r^2

2. Length of Arc of Sector:

l = \frac{\theta}{360^\circ} \times 2\pi r

3. Relationship:

A = \frac{1}{2} \times l \times r

Areas Related to Circles

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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