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By rohit.pandey1
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Updated on 27 Jul 2026, 11:34 IST
Class 10 Maths Chapter 11 Areas Related to Circles Notes PDF covers arc length, area and perimeter of sectors, minor and major segments, shaded regions, and practical applications of circular figures.
These detailed CBSE notes include important formulas, simple definitions, labelled-diagram placeholders, step-by-step solved examples, important exam questions, common mistakes, MCQs, and quick revision points. The topic is Chapter 11 in the current NCERT textbook, although some older books and videos may show it as Chapter 12.
Also Check: Areas Related to Circles NCERT solutions
Areas Related to Circles is based on finding the length, perimeter, or area of a complete circle or a selected part of it.
The six core ideas of the chapter are:
A major sector or major segment can usually be calculated by subtracting the corresponding minor region from the complete circle.
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Shaded-region questions are solved by dividing the figure into familiar shapes and then adding or subtracting their areas.
The Areas Related to Circles Class 10 Notes PDF provides complete chapter revision material in a printable and exam-friendly format.
Also Check: Complete Class 10 Maths notes PDF
Areas Related to Circles is Chapter 11 in the current NCERT Class 10 Mathematics textbook.
Older editions placed this topic in Chapter 12. This is why many older videos, notes, and question banks still use the heading “Chapter 12 Areas Related to Circles.”
Students may use an older video to understand a formula, but they should match the questions with the latest textbook and syllabus.
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The current CBSE Class 10 Maths syllabus covers the areas of sectors and segments and questions involving their areas, perimeters, arcs, and circumferences.
CBSE currently restricts segment-area questions to central angles of 60°, 90°, and 120°. These angles allow the corresponding triangle area to be calculated using familiar geometric or trigonometric methods.
Students should revise the following topics before beginning this chapter:
Also Check: Class 10 Trigonometry Notes | Circles Class 10 Notes
The main parts of a circle are the centre, radius, diameter, chord, arc, sector, segment, and central angle.
Centre: The fixed point inside a circle that is at the same distance from every point on the circumference.
Radius: A line segment joining the centre of a circle to any point on its circumference.
The radius is usually represented by the letter r.
Diameter: A chord that passes through the centre of a circle.
Diameter = 2 × radius
d = 2 × r
Radius = diameter ÷ 2
r = d ÷ 2
Chord: A line segment joining any two points on the circumference of a circle.
Every diameter is a chord, but every chord is not a diameter.
Circumference: The complete boundary or outer length of a circle.
The circumference of a circle is similar to the perimeter of a polygon.
Arc: A part of the circumference between two points.
Minor arc: The shorter arc between two points.
Major arc: The longer arc between the same two points.
Sector: The region enclosed by two radii and the arc between them.
A sector looks like a slice of pizza.
Minor sector: The smaller sector formed by two radii.
Major sector: The larger remaining sector.
Segment: The region enclosed by a chord and its corresponding arc.
Minor segment: The smaller segment.
Major segment: The larger segment.
Central angle: The angle formed by two radii at the centre of a circle.
The central angle is usually represented by the symbol θ.
If the minor angle is θ, then:
Major angle = 360° − θ
A sector is enclosed by two radii and an arc, while a segment is enclosed by a chord and an arc.
| Feature | Sector | Segment |
| Enclosed by | Two radii and an arc | A chord and an arc |
| Reaches the centre | Yes | Usually no |
| Shape | Pizza-slice shape | Curved cap shape |
| Main area method | Fraction of circle area | Sector area − triangle area |
| Major region method | Circle area − minor sector | Circle area − minor segment |
| Common mistake | Forgetting two radii in perimeter | Forgetting to subtract the triangle |
A sector contains the triangle formed by the two radii and the chord. A segment is the part that remains after this triangle is removed from the sector.
The main Areas Related to Circles formulas calculate circumference, circle area, arc length, sector area, sector perimeter, and segment area.
In the formulas below:
Circumference = 2 × π × radius
C = 2 × π × r
It can also be written as:
Circumference = π × diameter
C = π × d
Unit: centimetres, metres, kilometres, or another linear unit.
Area of circle = π × radius × radius
A = π × r²
Unit: square centimetres, square metres, square kilometres, or another square unit.
A semicircle is half of a complete circle.
Area of semicircle = 1/2 × π × radius × radius
The perimeter of a semicircle includes the curved part and the diameter.
Perimeter of semicircle = π × radius + 2 × radius
Students should not write only π × radius because that gives only the curved length.
A quadrant is one-fourth of a complete circle.
Area of quadrant = 1/4 × π × radius × radius
The perimeter of a quadrant includes the curved arc and two radii.
Perimeter of quadrant = 1/2 × π × radius + 2 × radius
Arc length = Central angle ÷ 360 × 2 × π × radius
The formula works because the arc represents the same fraction of the complete circumference as its central angle represents of 360°.
Unit: centimetres, metres, kilometres, or another linear unit.
Area of sector = Central angle ÷ 360 × π × radius × radius
The formula works because the sector represents the same fraction of the complete circular area as its central angle represents of 360°.
Unit: square centimetres, square metres, or another square unit.
Perimeter of sector = Arc length + 2 × radius
Therefore:
Perimeter of sector = Central angle ÷ 360 × 2 × π × radius + 2 × radius
Students often lose marks by writing only the arc length and forgetting the two radii.
Area of minor segment = Area of sector − Area of triangle
The triangle is formed by the two radii and the chord.
Area of major sector = Area of complete circle − Area of minor sector
It can also be calculated by using the major angle.
Major angle = 360° − Minor angle
Area of major segment = Area of complete circle − Area of minor segment
A circular ring is formed between two circles with the same centre.
Area of ring = Area of outer circle − Area of inner circle
Area of ring = π × outer radius × outer radius − π × inner radius × inner radius
Area of ring = π × (R² − r²)
Do not use π × (R − r)². The radii must be squared before subtraction.
| Required Quantity | Formula | Unit |
| Circumference | 2 × π × radius | Linear unit |
| Area of circle | π × radius × radius | Square unit |
| Arc length | Angle ÷ 360 × 2 × π × radius | Linear unit |
| Area of sector | Angle ÷ 360 × π × radius × radius | Square unit |
| Perimeter of sector | Arc length + 2 × radius | Linear unit |
| Area of minor segment | Sector area − triangle area | Square unit |
| Area of major sector | Circle area − minor sector area | Square unit |
| Area of major segment | Circle area − minor segment area | Square unit |
| Area of circular ring | π × (R² − r²) | Square unit |
Also Check: Class 10 Mensuration Formula Sheet
The correct formula depends on whether the question asks for a length, area, perimeter, sector, segment, or shaded region.
Use the following steps:
| Words in the Question | Formula or Method |
| Length of the arc | Angle ÷ 360 × 2 × π × radius |
| Area swept by a hand or blade | Area of sector |
| Region between two radii | Area of sector |
| Complete boundary of a sector | Arc length + 2 × radius |
| Region between chord and arc | Area of segment |
| Major sector | Circle area − minor sector |
| Major segment | Circle area − minor segment |
| Circle inside a square | Square area − circle area |
| Circular path | Outer circle area − inner circle area |
| Equal sectors | Divide 360° by the number of sectors |
Every Areas Related to Circles problem can be solved by identifying the required region before substituting values into a formula.
Use the value of π stated in the question.
The current NCERT Exercise 11.1 instructs students to use 22/7 unless another value is stated.
Keep fractions and exact values during the working and round only at the final step.
For example:
Area of sector
= 120 ÷ 360 × 22/7 × 7 × 7
= 1/3 × 154
= 154/3 square centimetres
= 51.33 square centimetres approximately
Rounding separate values too early may produce a different final answer.
| Quantity | Correct Unit Type |
| Radius | cm, m, km |
| Diameter | cm, m, km |
| Arc length | cm, m, km |
| Circumference | cm, m, km |
| Perimeter | cm, m, km |
| Sector area | cm², m², km² |
| Segment area | cm², m², km² |
| Cost | Rupees or another currency after applying the stated rate |
The following solved examples cover the main direct, segment, shaded-region, and real-life question types.
Find the area of a 90° sector of a circle with radius 14 cm. Use π = 22/7.
Step 1: Write the formula.
Area of sector = Angle ÷ 360 × π × radius × radius
Step 2: Substitute the values.
Area of sector
= 90 ÷ 360 × 22/7 × 14 × 14
Step 3: Simplify.
Area of sector
= 1/4 × 616
= 154 square centimetres
Answer: The area of the sector is 154 cm².
Find the length of a 60° arc in a circle of radius 21 cm. Use π = 22/7.
Arc length = Angle ÷ 360 × 2 × π × radius
Arc length
= 60 ÷ 360 × 2 × 22/7 × 21
= 1/6 × 132
= 22 centimetres
Answer: The length of the arc is 22 cm.
Find the perimeter of a 120° sector with radius 7 cm. Use π = 22/7.
Step 1: Calculate the arc length.
Arc length
= 120 ÷ 360 × 2 × 22/7 × 7
= 1/3 × 44
= 44/3 centimetres
Step 2: Add the two radii.
Perimeter of sector
= Arc length + 2 × radius
= 44/3 + 14
= 44/3 + 42/3
= 86/3 centimetres
= 28.67 centimetres approximately
Answer: The perimeter of the sector is approximately 28.67 cm.
A chord in a circle of radius 14 cm forms a central angle of 60°. Find the area of the minor segment. Use π = 22/7 and √3 = 1.732.
Step 1: Calculate the sector area.
Sector area
= 60 ÷ 360 × 22/7 × 14 × 14
= 1/6 × 616
= 102.67 square centimetres approximately
Step 2: Calculate the triangle area.
The triangle is equilateral because all three angles are 60°.
Area of equilateral triangle
= √3 ÷ 4 × side × side
= 1.732 ÷ 4 × 14 × 14
= 84.87 square centimetres approximately
Step 3: Calculate the segment area.
Area of minor segment
= Sector area − Triangle area
= 102.67 − 84.87
= 17.80 square centimetres
Answer: The area of the minor segment is approximately 17.80 cm².
A chord of a circle with radius 14 cm forms a 90° angle at the centre. Find the area of the minor segment. Use π = 22/7.
Step 1: Calculate the sector area.
Sector area
= 90 ÷ 360 × 22/7 × 14 × 14
= 154 square centimetres
Step 2: Calculate the triangle area.
The two radii form a right-angled triangle.
Triangle area
= 1/2 × base × height
= 1/2 × 14 × 14
= 98 square centimetres
Step 3: Subtract.
Segment area
= 154 − 98
= 56 square centimetres
Answer: The area of the minor segment is 56 cm².
A chord of a circle with radius 14 cm forms a central angle of 120°. Find the area of the minor segment. Use π = 22/7 and √3 = 1.732.
Step 1: Calculate the sector area.
Sector area
= 120 ÷ 360 × 22/7 × 14 × 14
= 1/3 × 616
= 205.33 square centimetres approximately
Step 2: Calculate the triangle area.
Triangle area
= 1/2 × radius × radius × sin 120°
= 1/2 × 14 × 14 × √3/2
= 49 × 1.732
= 84.87 square centimetres approximately
Step 3: Calculate the segment area.
Segment area
= 205.33 − 84.87
= 120.46 square centimetres
Answer: The area of the minor segment is approximately 120.46 cm².
The area of a minor segment in a circle of radius 14 cm is 56 cm². Find the area of the major segment. Use π = 22/7.
Step 1: Calculate the complete circle area.
Circle area
= 22/7 × 14 × 14
= 616 square centimetres
Step 2: Subtract the minor segment.
Major segment area
= Complete circle area − Minor segment area
= 616 − 56
= 560 square centimetres
Answer: The area of the major segment is 560 cm².
The circumference of a circle is 44 cm. Find the area of one quadrant. Use π = 22/7.
Step 1: Find the radius.
Circumference = 2 × π × radius
44 = 2 × 22/7 × radius
Radius = 7 cm
Step 2: Find the quadrant area.
Area of quadrant
= 1/4 × 22/7 × 7 × 7
= 38.5 square centimetres
Answer: The area of the quadrant is 38.5 cm².
Two non-overlapping wipers each have a blade length of 21 cm and sweep through 120°. Find their total cleaned area. Use π = 22/7.
Step 1: Find the area cleaned by one wiper.
Area of one sector
= 120 ÷ 360 × 22/7 × 21 × 21
= 462 square centimetres
Step 2: Multiply by two.
Total cleaned area
= 2 × 462
= 924 square centimetres
Answer: The total cleaned area is 924 cm².
A circular garden has a radius of 7 m. Grass costs ₹12 per square metre. Find the total cost of covering the garden.
Step 1: Calculate the garden area.
Area
= 22/7 × 7 × 7
= 154 square metres
Step 2: Calculate the cost.
Cost
= 154 × ₹12
= ₹1,848
Answer: The total cost is ₹1,848.
Shaded-region questions are solved by dividing the diagram into known shapes and then adding or subtracting their areas.
Use this pattern when the shaded region lies between a chord and an arc.
Shaded segment area = Sector area − Triangle area
When a circle is inscribed in a square, the diameter of the circle equals the side of the square.
Shaded corner area = Area of square − Area of circle
When all four vertices of a square touch the circle, the diagonal of the square equals the diameter of the circle.
Diameter of circle = Diagonal of square
Find the side of the square before calculating its area.
Calculate each semicircle and check whether any parts overlap.
Area of one semicircle = 1/2 × π × radius × radius
Total semicircle area = Number of semicircles × Area of one semicircle
Four quadrants of equal radius form one complete circle.
Area of four quadrants = π × radius × radius
This shortcut is useful when equal quadrants are removed from the corners of a square.
The region between two circles with the same centre is called a circular ring.
Area of path = π × (Outer radius² − Inner radius²)
Use the following method:
A rotating hand, blade, beam, or sprinkler forms a sector.
Swept area = Angle ÷ 360 × π × radius × radius
| Diagram Pattern | Begin With | Main Operation |
| Segment | Sector | Subtract triangle |
| Circle inside square | Square | Subtract circle |
| Square inside circle | Circle | Subtract square |
| Corner quadrants | Square | Subtract quadrants |
| Circular ring | Outer circle | Subtract inner circle |
| Equal sectors | Complete circle | Divide or multiply |
| Overlapping figures | Sum of regions | Subtract overlap |
| Swept area | Sector | Multiply for equal blades |
The current NCERT Exercise 11.1 contains 14 questions covering direct formulas, segments, real-life applications, and composite designs.
| Question Number | Main Skill Tested |
| 1 | Area of a 60° sector |
| 2 | Finding quadrant area from circumference |
| 3 | Area swept by a clock hand |
| 4 | Minor segment and major sector |
| 5 | Arc length, sector area, and segment |
| 6 | Minor and major segments |
| 7 | Area of a 120° segment |
| 8 | Quarter-circle grazing area |
| 9 | Wire length and equal sectors |
| 10 | Equal sectors in an umbrella |
| 11 | Area swept by car wipers |
| 12 | Lighthouse warning sector |
| 13 | Composite design and cost |
| 14 | Multiple-choice sector formula |
The exercise includes practical situations involving a clock, field, brooch, umbrella, car wipers, lighthouse, and table-cover design.
Our editorial classification of the current exercise shows:
The most important exam questions cover sector formulas, arc length, segment calculations, shaded regions, and real-life applications.
CBSE assigns 10 marks to the complete Mensuration unit, not to Areas Related to Circles alone. The Mensuration unit also includes Surface Areas and Volumes.
A chord of a circle of radius r forms an angle of 120° at the centre. Find the areas of the minor and major segments.
A circular path surrounds a garden with inner radius r and outer radius R.
Also Check: CBSE Class 10 Maths Sample Papers | Class 10 Maths Previous Year Questions
Most mistakes in this chapter occur because students identify the wrong region, use the wrong formula, or forget the correct unit.
| Common Mistake | Why It Is Wrong | Correct Check |
| Using diameter as radius | Formulas require radius | Divide diameter by 2 |
| Using radius instead of radius² | Area is two-dimensional | Check for square units |
| Using sector area for arc length | One is area and one is length | Check the required unit |
| Giving only arc length as sector perimeter | Two radii are missing | Trace the full boundary |
| Treating a segment as a sector | Triangle has not been removed | Look for the chord |
| Subtracting sector from triangle | Sector must be larger | Subtract triangle from sector |
| Using minor angle for major sector | Calculates the smaller region | Use 360° − minor angle |
| Rounding too early | Changes the final answer | Round only at the end |
| Mixing 3.14 and 22/7 | Makes the calculation inconsistent | Use one value throughout |
| Writing cm instead of cm² | Area requires square units | Check the type of quantity |
| Using π × (R − r)² for a ring | Radii must be squared first | Use π × (R² − r²) |
| Claiming the chapter carries 10 marks | The full Mensuration unit carries 10 marks | Name the complete unit |
These MCQs test definitions, formulas, and basic calculations.
A. 77/3 cm²
B. 77 cm²
C. 154 cm²
D. 44 cm²
Answer: A. 77/3 cm²
A. 11 cm
B. 22 cm
C. 44 cm
D. 154 cm
Answer: B. 22 cm
A. Only the arc
B. Only two radii
C. The arc and two radii
D. The complete circumference
Answer: C. The arc and two radii
A. Triangle area − sector area
B. Sector area − triangle area
C. Circle area − sector area
D. Arc length − radius
Answer: B. Sector area − triangle area
A. 110°
B. 180°
C. 250°
D. 360°
Answer: C. 250°
A. 3.5 cm
B. 7 cm
C. 14 cm
D. 22 cm
Answer: B. 7 cm
A. 38.5 cm²
B. 77 cm²
C. 154 cm²
D. 616 cm²
Answer: C. 154 cm²
A. cm
B. cm²
C. cm³
D. Degrees
Answer: B. cm²
A. 154 cm²
B. 196 cm²
C. 308 cm²
D. 616 cm²
Answer: C. 308 cm²
A. The radius becomes shorter.
B. The central angle changes.
C. The triangle between the radii is removed.
D. The circumference is divided by two.
Answer: C. The triangle between the radii is removed.
Also Check: Areas Related to Circles MCQ Questions]
The following questions progress from direct formula use to multi-step applications.
Find the area of a 120° sector with radius 10 cm. Use π = 3.14.
Answer:
Area
= 120 ÷ 360 × 3.14 × 10 × 10
= 104.67 cm² approximately
Find the length of a 90° arc with radius 12 cm. Use π = 3.14.
Answer:
Arc length
= 90 ÷ 360 × 2 × 3.14 × 12
= 18.84 cm
A chord in a circle of radius 10 cm forms a 90° angle at the centre. Find the minor-segment area. Use π = 3.14.
Answer:
Sector area
= 90 ÷ 360 × 3.14 × 10 × 10
= 78.5 cm²
Triangle area
= 1/2 × 10 × 10
= 50 cm²
Segment area
= 78.5 − 50
= 28.5 cm²
Using the result above, find the major-segment area.
Answer:
Complete circle area
= 3.14 × 10 × 10
= 314 cm²
Major segment area
= 314 − 28.5
= 285.5 cm²
Find the area of a circular path with outer radius 10 m and inner radius 6 m. Use π = 3.14.
Answer:
Area of path
= 3.14 × (10² − 6²)
= 3.14 × (100 − 36)
= 3.14 × 64
= 200.96 m²
Competency-based questions require students to understand the situation before selecting a formula.
A sprinkler is fixed at one corner of a square garden. It waters a quarter-circle region of radius 14 m. Find the watered area using π = 22/7.
Answer:
The corner creates a 90° sector.
Watered area
= 90 ÷ 360 × 22/7 × 14 × 14
= 154 m²
A decorative clock has a minute hand 21 cm long. Find the area swept by the hand in 20 minutes.
Answer:
Angle covered in 20 minutes
= 20 ÷ 60 × 360
= 120°
Area swept
= 120 ÷ 360 × 22/7 × 21 × 21
= 462 cm²
A square logo has a side of 28 cm. A circle of radius 14 cm is printed inside it. Find the area outside the circle but inside the square.
Answer:
Square area
= 28 × 28
= 784 cm²
Circle area
= 22/7 × 14 × 14
= 616 cm²
Required area
= 784 − 616
= 168 cm²
A safety beam reaches 30 m and rotates through 72°. Find the covered area using π = 3.14.
Answer:
Covered area
= 72 ÷ 360 × 3.14 × 30 × 30
= 1/5 × 2,826
= 565.2 m²
A student calculates the area of a 90° sector with radius 14 cm using:
90 ÷ 360 × 2 × π × radius
The student writes the answer as 22 cm².
Answer:
The student used the arc-length formula but wrote an area unit.
Correct sector area
= 90 ÷ 360 × 22/7 × 14 × 14
= 154 cm²
The quick revision summary contains the formulas, definitions, and checks students should review before an examination.
Arc: A part of the circumference.
Sector: A region enclosed by two radii and an arc.
Segment: A region enclosed by a chord and an arc.
Minor region: The smaller region.
Major region: The larger remaining region.
Central angle: The angle formed by two radii at the centre.
Circumference = 2 × π × radius
Area of circle = π × radius × radius
Arc length = Angle ÷ 360 × 2 × π × radius
Area of sector = Angle ÷ 360 × π × radius × radius
Perimeter of sector = Arc length + 2 × radius
Area of minor segment = Sector area − Triangle area
Area of major sector = Circle area − Minor sector area
Area of major segment = Circle area − Minor segment area
Area of circular ring = π × (Outer radius² − Inner radius²)
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Areas Related to Circles is Chapter 11 in the current NCERT textbook. Older editions and many older online resources call it Chapter 12.
Yes. The chapter includes sector and segment areas and related perimeter and circumference problems.
The current NCERT chapter contains Exercise 11.1 with 14 numbered questions. Older textbooks may show a different exercise structure.
The main formulas cover circumference, circle area, arc length, sector area, sector perimeter, and minor and major segment areas.
A sector is enclosed by two radii and an arc. A segment is enclosed by a chord and an arc.
Sector area is calculated by multiplying the circle area by Angle ÷ 360. Sector perimeter is calculated by adding the two radii to the arc length.
Subtract the triangle area from the sector area to obtain the minor segment. Subtract the minor segment from the complete circle to obtain the major segment.
Divide the diagram into familiar shapes, calculate each area, and then add or subtract according to the shaded portion.
Use the value stated in the question. The current NCERT exercise generally asks students to use 22/7 unless another value is given.
Keep the exact fraction during the working and convert it into a decimal only when required. Round only the final answer.
Yes, this formula correctly gives the area of the triangle formed by two radii and their included angle. The complete working should still be shown clearly.
The current CBSE syllabus restricts segment-area problems to central angles of 60°, 90°, and 120°.
NCERT is the essential starting point because it defines the syllabus and includes direct, segment, and real-life questions. Students should then practise Exemplar, sample-paper, and previous-year questions.
Students should prioritise arc length, sector perimeter, 60°/90°/120° segments, major regions, shaded figures, swept areas, and cost-based applications.
CBSE assigns 10 marks to the complete Mensuration unit, which includes Areas Related to Circles and Surface Areas and Volumes. The chapter alone does not have a guaranteed 10-mark weightage.
The complete notes PDF, formula sheet, and worksheet can be downloaded from Infinity Learn website.
