Study Guide

Areas Related to Circles Class 10: Sectors, Segments & Solved Problems

Your complete guide to sectors, segments, arc lengths, and combination-of-figures problems for CBSE boards!

CBSEClass 10
The SparkEd Authors (IITian & Googler)15 March 202645 min read
Circle with shaded sector and segment regions illustrating areas related to circles for Class 10 CBSE.

Why This Chapter Is a Marks Goldmine

Areas Related to Circles (NCERT Chapter 11) is one of the most scoring chapters in Class 10 Maths. It typically carries 4-6 marks in the board exam, and almost every question follows a predictable formula-based pattern.

The chapter builds on what you already know about circles — circumference and area — and introduces three key concepts:
1. Arc length — the curved distance along part of a circle.
2. Area of a sector — the pie-slice shaped region.
3. Area of a segment — the region between a chord and its arc.

Once you master these formulas and learn how to handle combination-of-figures problems (where you add and subtract areas of sectors, triangles, and circles), you can confidently tackle any board exam question.

Let's go through it all, step by step!

Recap: Circle Basics

Before we dive into sectors and segments, let's lock in the fundamentals.

For a circle with radius rr:

Circumference=2πr\text{Circumference} = 2\pi r

Area=πr2\text{Area} = \pi r^2

For a circle with diameter dd (where d=2rd = 2r):

Circumference=πd\text{Circumference} = \pi d

Area=πd24\text{Area} = \frac{\pi d^2}{4}

**What is π\pi?** It's the ratio of a circle's circumference to its diameter, approximately 3.141593.14159. In CBSE board exams, use π=227\pi = \frac{22}{7} unless the question specifies otherwise.

A semicircle has:

Perimeter=πr+2r=r(π+2)\text{Perimeter} = \pi r + 2r = r(\pi + 2)

Area=12πr2\text{Area} = \frac{1}{2}\pi r^2

A quadrant (quarter circle) has:

Perimeter=πr2+2r\text{Perimeter} = \frac{\pi r}{2} + 2r

Area=14πr2\text{Area} = \frac{1}{4}\pi r^2

Arc, Sector, and Segment: Definitions and Formulas

These three concepts are the core of this chapter. Let's define each one precisely.

Arc and Arc Length

An arc is a continuous part of the circumference of a circle. When a sector has angle θ\theta at the centre:

Length of arc=θ360∘×2πr\text{Length of arc} = \frac{\theta}{360^\circ} \times 2\pi r

For example, if θ=90∘\theta = 90^\circ, the arc length is 90360×2πr=πr2\frac{90}{360} \times 2\pi r = \frac{\pi r}{2}, which is one-quarter of the full circumference.

The arc divides the circle into two parts:
- Minor arc — the shorter arc (when θ<180∘\theta < 180^\circ)
- Major arc — the longer arc (when θ<180∘\theta < 180^\circ, the major arc subtends 360∘−θ360^\circ - \theta)

Sector and Area of Sector

A sector is the region enclosed between two radii and the arc they intercept. It looks like a slice of pizza or pie.

Area of sector=θ360∘×πr2\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2

Perimeter of sector=2r+θ360∘×2πr=2r+arc length\text{Perimeter of sector} = 2r + \frac{\theta}{360^\circ} \times 2\pi r = 2r + \text{arc length}

Minor sector: The sector corresponding to the minor arc (θ<180∘\theta < 180^\circ).
Major sector: The sector corresponding to the major arc.

Area of major sector=πr2−Area of minor sector\text{Area of major sector} = \pi r^2 - \text{Area of minor sector}

=360∘−θ360∘×πr2= \frac{360^\circ - \theta}{360^\circ} \times \pi r^2

Segment and Area of Segment

A segment is the region between a chord and the arc it cuts off. It's the area of the sector minus the area of the triangle formed by the two radii and the chord.

Area of minor segment=Area of sector−Area of △OAB\text{Area of minor segment} = \text{Area of sector} - \text{Area of } \triangle OAB

=θ360∘×πr2−12r2sin⁡θ= \frac{\theta}{360^\circ} \times \pi r^2 - \frac{1}{2}r^2 \sin\theta

For the common case θ=60∘\theta = 60^\circ (equilateral triangle formed):

Area of segment=60360πr2−34r2=r2(π6−34)\text{Area of segment} = \frac{60}{360}\pi r^2 - \frac{\sqrt{3}}{4}r^2 = r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)

For θ=90∘\theta = 90^\circ:

Area of segment=90360πr2−12r2=r2(π4−12)\text{Area of segment} = \frac{90}{360}\pi r^2 - \frac{1}{2}r^2 = r^2\left(\frac{\pi}{4} - \frac{1}{2}\right)

Area of major segment=πr2−Area of minor segment\text{Area of major segment} = \pi r^2 - \text{Area of minor segment}

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Solved Examples: Sector and Arc Problems

Let's work through the most common problem types you'll see in exams.

Solved Example 1: Finding Arc Length and Sector Area

Problem: Find the length of arc and area of the sector of a circle of radius 21 cm and central angle 60∘60^\circ.

Solution:
r=21r = 21 cm, θ=60∘\theta = 60^\circ.

Arc length:

=θ360∘×2πr=60360×2×227×21= \frac{\theta}{360^\circ} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21

=16×2×22×3=22 cm= \frac{1}{6} \times 2 \times 22 \times 3 = 22 \text{ cm}

Area of sector:

=θ360∘×πr2=60360×227×212= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 21^2

=16×227×441=16×22×63=231 cm2= \frac{1}{6} \times \frac{22}{7} \times 441 = \frac{1}{6} \times 22 \times 63 = 231 \text{ cm}^2

Answer: Arc length =22= 22 cm, Area of sector =231= 231 cm2^2.

Solved Example 2: Finding the Angle from Arc Length

Problem: The length of an arc of a circle of radius 14 cm is 22 cm. Find the angle subtended at the centre.

Solution:

Arc length=θ360∘×2πr\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi r

22=θ360×2×227×1422 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 14

22=θ360×8822 = \frac{\theta}{360} \times 88

θ360=2288=14\frac{\theta}{360} = \frac{22}{88} = \frac{1}{4}

θ=90∘\theta = 90^\circ

Answer: The angle subtended at the centre is 90∘90^\circ.

Solved Example 3: Perimeter of a Sector

Problem: Find the perimeter of a sector of a circle with radius 10.5 cm and central angle 60∘60^\circ.

Solution:

Perimeter=2r+θ360∘×2πr\text{Perimeter} = 2r + \frac{\theta}{360^\circ} \times 2\pi r

=2×10.5+60360×2×227×10.5= 2 \times 10.5 + \frac{60}{360} \times 2 \times \frac{22}{7} \times 10.5

=21+16×66= 21 + \frac{1}{6} \times 66

=21+11=32 cm= 21 + 11 = 32 \text{ cm}

Answer: Perimeter of the sector =32= 32 cm.

Solved Examples: Segment Problems

Segment problems combine sector area with triangle area. These are slightly trickier and appear as 3-4 mark questions.

Solved Example 4: Area of Minor Segment (60-degree)

Problem: A chord of a circle of radius 12 cm subtends an angle of 60∘60^\circ at the centre. Find the area of the minor segment. (Use π=3.14\pi = 3.14 and 3=1.73\sqrt{3} = 1.73.)

Solution:
When θ=60∘\theta = 60^\circ and both radii equal 12 cm, the triangle OABOAB is equilateral (since two sides are radii = 12 cm and the included angle is 60∘60^\circ, the triangle is isosceles with 60∘60^\circ vertex angle, so all angles are 60∘60^\circ).

Area of sector:

=60360×πr2=16×3.14×144=75.36 cm2= \frac{60}{360} \times \pi r^2 = \frac{1}{6} \times 3.14 \times 144 = 75.36 \text{ cm}^2

**Area of equilateral △OAB\triangle OAB:**

=34×122=1.734×144=62.28 cm2= \frac{\sqrt{3}}{4} \times 12^2 = \frac{1.73}{4} \times 144 = 62.28 \text{ cm}^2

Area of minor segment:

=75.36−62.28=13.08 cm2= 75.36 - 62.28 = 13.08 \text{ cm}^2

Answer: Area of minor segment =13.08= 13.08 cm2^2.

Solved Example 5: Area of Minor Segment (90-degree)

Problem: Find the area of the minor segment of a circle of radius 14 cm when the central angle is 90∘90^\circ.

Solution:
Area of sector:

=90360×227×142=14×227×196=154 cm2= \frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = 154 \text{ cm}^2

**Area of right △OAB\triangle OAB** (right angle at centre):

=12×14×14=98 cm2= \frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2

Area of minor segment:

=154−98=56 cm2= 154 - 98 = 56 \text{ cm}^2

Answer: Area of minor segment =56= 56 cm2^2.

Areas of Combinations of Plane Figures

This is the most important and most frequently tested problem type in this chapter. You'll be given a figure made up of circles, semicircles, quadrants, and other shapes, and asked to find the shaded area.

Strategy: Break the complex figure into basic shapes whose areas you can compute, then add or subtract as needed.

Common patterns:
- Shaded area = Area of larger shape −- Area of smaller shape(s)
- Shaded area = Sum of areas of certain sectors −- Area of a polygon
- Shaded area = Area of circle −- Area of inscribed regular polygon

Solved Example 6: Circular Table Cloth on Square Table

Problem: A circular table cloth of radius 32 cm is placed on a square table of side 64 cm. Find the area of the table that is not covered by the cloth.

Solution:
Note that the diameter of the cloth =64= 64 cm == side of the square, so the cloth fits exactly touching the midpoints of the sides.

Area of square=642=4096 cm2\text{Area of square} = 64^2 = 4096 \text{ cm}^2

Area of circle=π×322=1024π=1024×227=225287≈3218.29 cm2\text{Area of circle} = \pi \times 32^2 = 1024\pi = 1024 \times \frac{22}{7} = \frac{22528}{7} \approx 3218.29 \text{ cm}^2

Uncovered area=4096−3218.29≈877.71 cm2\text{Uncovered area} = 4096 - 3218.29 \approx 877.71 \text{ cm}^2

Answer: Area not covered ≈877.71\approx 877.71 cm2^2.

Solved Example 7: Four Quadrants Inside a Square

Problem: Find the area of the shaded region in a square of side 14 cm where four quadrants of circles of radius 7 cm are drawn at each corner.

Solution:
Each quadrant has radius 77 cm and angle 90∘90^\circ.

Area of 4 quadrants=4×14πr2=π×72=49π=49×227=154 cm2\text{Area of 4 quadrants} = 4 \times \frac{1}{4}\pi r^2 = \pi \times 7^2 = 49\pi = 49 \times \frac{22}{7} = 154 \text{ cm}^2

Area of square=142=196 cm2\text{Area of square} = 14^2 = 196 \text{ cm}^2

Shaded area=196−154=42 cm2\text{Shaded area} = 196 - 154 = 42 \text{ cm}^2

Answer: Shaded area =42= 42 cm2^2.

Solved Example 8: Two Semicircles on a Diameter

Problem: In a circle of radius 21 cm, two semicircles of radius 10.5 cm are drawn on the two halves of a diameter as shown. Find the area of the shaded region.

Solution:
The shaded region consists of one semicircle of the larger circle plus the difference of the two smaller semicircles.

Actually, when two semicircles of radius r/2r/2 are drawn on the diameter of a larger circle of radius rr, one on each side, the shaded region (lune-shaped areas) can be computed as:

Area of upper semicircle (large)=12π(21)2=441π2\text{Area of upper semicircle (large)} = \frac{1}{2}\pi (21)^2 = \frac{441\pi}{2}

The two smaller semicircles: one adds area (on the same side) and one subtracts area (on the opposite side). The net effect is that the shaded area equals the area of the larger semicircle:

Shaded area=12×227×441=22×632=693 cm2\text{Shaded area} = \frac{1}{2} \times \frac{22}{7} \times 441 = \frac{22 \times 63}{2} = 693 \text{ cm}^2

Answer: Shaded area =693= 693 cm2^2.

Common Mistakes Students Make

These errors show up again and again in board exams. Avoid them and you're already ahead!

1. Forgetting to Convert Diameter to Radius:
* Mistake: Plugging the diameter directly into πr2\pi r^2.
* Fix: Always halve the diameter first. Write r=d/2r = d/2 as your first step.

2. Confusing Sector with Segment:
* Mistake: Using the sector area formula when the question asks for segment area.
* Fix: Sector = pie slice (includes the triangle). Segment = the region between the chord and arc (sector minus triangle).

3. Wrong Triangle Area Formula:
* Mistake: Using 12×base×height\frac{1}{2} \times \text{base} \times \text{height} incorrectly for the triangle in a segment problem. When θ=60∘\theta = 60^\circ, the triangle is equilateral; when θ=90∘\theta = 90^\circ, it's a right isosceles triangle.
* Fix: For a 60∘60^\circ sector, use 34r2\frac{\sqrt{3}}{4}r^2. For a 90∘90^\circ sector, use 12r2\frac{1}{2}r^2.

4. Mixing Up Major and Minor:
* Mistake: Calculating the minor sector area when the question asks for the major sector.
* Fix: Major sector area =πr2−= \pi r^2 - minor sector area. Always check which region is shaded.

5. Forgetting the Perimeter of a Sector Includes Two Radii:
* Mistake: Giving only the arc length as the perimeter.
* Fix: Perimeter of sector =2r+= 2r + arc length.

6. **Arithmetic Errors with π=22/7\pi = 22/7:**
* Mistake: Calculation errors when working with fractions.
* Fix: Simplify before multiplying. Cancel common factors between numerator and denominator early.

7. Not Reading the Shading Carefully:
* Mistake: Finding the wrong area in combination problems because you didn't identify the shaded region correctly.
* Fix: Before computing anything, clearly identify what the shaded region is and express it as a sum/difference of known areas.

Board Exam Strategy

Weightage: Areas Related to Circles typically carries 4-6 marks in the CBSE board exam.

Question Patterns:

* 1-2 Marks (MCQ/VSA): Direct formula questions — find the area of a sector, length of an arc, or area of a semicircle.

* 3 Marks (SA): Area of a minor segment (requires computing both sector area and triangle area), or a simple combination-of-figures problem.

* 4-5 Marks (LA): Complex combination problems with shaded regions involving multiple sectors, semicircles, and polygons. These require careful identification of which areas to add and subtract.

Time-Saving Tips:

1. Memorize common angle values:
- θ=60∘\theta = 60^\circ: arc =πr3= \frac{\pi r}{3}, sector area =πr26= \frac{\pi r^2}{6}
- θ=90∘\theta = 90^\circ: arc =πr2= \frac{\pi r}{2}, sector area =πr24= \frac{\pi r^2}{4}
- θ=120∘\theta = 120^\circ: arc =2πr3= \frac{2\pi r}{3}, sector area =πr23= \frac{\pi r^2}{3}

2. Factor out common terms. Many expressions contain πr2\pi r^2 or r2r^2, so factor them out for cleaner calculation.

3. For segment area with standard angles, memorize:
- 60∘60^\circ segment: r2(π6−34)r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)
- 90∘90^\circ segment: r2(π4−12)r^2\left(\frac{\pi}{4} - \frac{1}{2}\right)

4. Always mention units in your final answer (cm2\text{cm}^2 for area, cm for length).

Practice on SparkEd's Areas Related to Circles page to build speed and confidence.

Quick Reference: All Formulas at a Glance

Circle: C=2πrC = 2\pi r, A=πr2A = \pi r^2

Arc Length: θ360∘×2πr\frac{\theta}{360^\circ} \times 2\pi r

Sector Area: θ360∘×πr2\frac{\theta}{360^\circ} \times \pi r^2

Perimeter of Sector: 2r+θ360∘×2πr2r + \frac{\theta}{360^\circ} \times 2\pi r

Segment Area: Sector area −- Triangle area =θ360∘πr2−12r2sin⁡θ= \frac{\theta}{360^\circ}\pi r^2 - \frac{1}{2}r^2\sin\theta

Semicircle: Area =πr22= \frac{\pi r^2}{2}, Perimeter =πr+2r= \pi r + 2r

Quadrant: Area =πr24= \frac{\pi r^2}{4}, Perimeter =πr2+2r= \frac{\pi r}{2} + 2r

Triangle areas for common angles:
- θ=60∘\theta = 60^\circ (equilateral): 34r2\frac{\sqrt{3}}{4}r^2
- θ=90∘\theta = 90^\circ (right isosceles): 12r2\frac{1}{2}r^2
- θ=120∘\theta = 120^\circ: 34r2\frac{\sqrt{3}}{4}r^2

Key identity: Area of major sector + Area of minor sector =πr2= \pi r^2

More Practice Problems

Try these additional problems to solidify your understanding.

Solved Example 9: Wipers on a Windshield

Problem: The minute hand of a clock is 14 cm long. Find the area swept by the minute hand in 5 minutes.

Solution:
In 60 minutes, the minute hand sweeps 360∘360^\circ.

In 5 minutes, it sweeps 560×360∘=30∘\frac{5}{60} \times 360^\circ = 30^\circ.

Area swept=30360×πr2=112×227×142\text{Area swept} = \frac{30}{360} \times \pi r^2 = \frac{1}{12} \times \frac{22}{7} \times 14^2

=112×227×196=112×616=1543≈51.33 cm2= \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{1}{12} \times 616 = \frac{154}{3} \approx 51.33 \text{ cm}^2

Answer: Area swept =1543≈51.33= \frac{154}{3} \approx 51.33 cm2^2.

Solved Example 10: Ring-Shaped Region

Problem: Two concentric circles have radii 7 cm and 14 cm. Find the area of the ring-shaped region between them, and find the area of the sector of the ring with angle 30∘30^\circ.

Solution:

Area of ring=πR2−πr2=π(R2−r2)=227(196−49)=227×147=462 cm2\text{Area of ring} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \frac{22}{7}(196 - 49) = \frac{22}{7} \times 147 = 462 \text{ cm}^2

Area of ring sector=30360(πR2−πr2)=112×462=38.5 cm2\text{Area of ring sector} = \frac{30}{360}(\pi R^2 - \pi r^2) = \frac{1}{12} \times 462 = 38.5 \text{ cm}^2

Answer: Area of ring =462= 462 cm2^2, Area of ring sector =38.5= 38.5 cm2^2.

Level Up Your Practice with SparkEd

Areas Related to Circles is a chapter where formula mastery + problem practice = full marks. The more varied problems you solve, the faster you'll recognize which areas to add and subtract.

SparkEd is built to help you do exactly that:

* Structured Practice: Our Areas Related to Circles page has problems at every difficulty level, from basic sector calculations to complex shaded-region questions.

* AI Math Solver: Struggling with a tricky combination problem? Paste it into the AI Solver and get a clear, step-by-step breakdown of which areas to compute and how to combine them.

* AI Coach: Your personal study assistant that identifies which problem types give you trouble and recommends targeted practice.

* Connected Topics: Since this chapter uses concepts from Circles (tangents, chords) and feeds into Surface Areas & Volumes (circular cross-sections), practice all three together for a complete geometry preparation.

Visit sparkedmaths.com and turn your formula knowledge into board exam marks!

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