Chapter 11: Areas Related to Circles
Overview
Essential formulas and methods for Class 10 plane geometry: sector area $\frac{\theta}{360^\circ}\pi r^2$, arc length $\frac{\theta}{360^\circ}2\pi r$, minor and major segments, and composite shaded region problems.
Board Exam Focused Notes, Formulas, and PYQs
Exam Weightage & Blueprint
Total: ~4-5 MarksThis chapter falls under Unit VI: Mensuration (10 marks total) in the 2026–27 syllabus. Focus areas include sectors and segments of a circle; segment-area problems are restricted to central angles of 60°, 90° and 120°.
| Question Type | Marks | Frequency | Focus Topic |
|---|---|---|---|
| MCQ | 1 | High | Length of Arc, Area of Sector Formula |
| Short Answer | 2 or 3 | High | Clock Hand, Quadrant Area |
| Long Answer | 4 or 5 | Medium | Area of Segment (Chord problems) |
CBSE 2026–27 Syllabus Status: Areas Related to Circles
Unit VI: Mensuration • 4–5 MarksThis chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:
In 2026–27 Board Syllabus:
Deleted / Not in Board Exam:
📐 Important Formulas
Memorize these formulas involving angle $\theta$ (in degrees) at the center.
| Concept | Formula | Notes |
|---|---|---|
| Circumference | $2\pi r$ | Total length of boundary |
| Area of Circle | $\pi r^2$ | Total region enclosed |
| Length of Arc ($l$) | $\frac{\theta}{360} \times 2\pi r$ | Part of Circumference |
| Area of Sector | $\frac{\theta}{360} \times \pi r^2$ | Slice like a Pizza |
- General Formula: Area = $\frac{1}{2} r^2 \sin\theta$.
- If $\theta = 60^^\circ$, Triangle is Equilateral. Area = $\frac{\sqrt{3}}{4} r^2$.
- If $\theta = 90^^\circ$, Triangle is Right-Angled. Area = $\frac{1}{2} r^2$.
Concept: 📐 Sector vs. Segment
Sector
The region enclosed by two radii and the corresponding arc.
- Minor Sector: Angle $< 180^^\circ$
- Major Sector: Angle $> 180^^\circ$ (Area = $\pi r^2$ - Minor Sector)
Segment
The region enclosed between a chord and the corresponding arc.
- Minor Segment: Area = Sector - Triangle
- Major Segment: Area = Circle - Minor Segment
- Sector Area via Arc Length ($l$): $\text{Area of Sector} = \frac{1}{2} l r$.
- Segment Triangle Area Reference:
• For $\theta = 60^\circ$: Triangle is equilateral $\implies \text{Area} = \frac{\sqrt{3}}{4} r^2$.
• For $\theta = 90^\circ$: Triangle is right isosceles $\implies \text{Area} = \frac{1}{2} r^2$.
• For $\theta = 120^\circ$: $\text{Area} = \frac{1}{2} r^2 \sin 120^\circ = \frac{\sqrt{3}}{4} r^2$. - Perimeter of Semicircle & Sector (Major Trap):
• Perimeter of Semicircle $= \pi r + 2r = r(\pi + 2)$ (NOT just $\pi r$!).
• Perimeter of Sector $= \frac{\theta}{360}(2\pi r) + 2r$. - Clock Hand Speeds:
• Minute hand sweeps $\frac{360^\circ}{60} = \mathbf{6^\circ}$ per minute.
• Hour hand sweeps $\frac{360^\circ}{720} = \mathbf{0.5^\circ}$ per minute. - Wheel Revolutions: Number of revolutions $= \frac{\text{Total Distance}}{\text{Circumference } (2\pi r)}$.
Important Solved Examples (Marking Scheme)
Q1. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. (2 Marks)
Angle swept in 60 mins = $360^^\circ$.
Angle in 5 mins = $\frac{360}{60} \times 5 = 30^^\circ$.
Area swept = Area of Sector = $\frac{\theta}{360} \times \pi r^2$.
$r = 14$ cm, $\theta = 30^^\circ$.
$= \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$
$= \frac{1}{12} \times 22 \times 2 \times 14 = \frac{1}{12} \times 616$
$= \frac{154}{3} = \mathbf{51.33}$ cm-.
Q2. A chord of a circle of radius 21 cm subtends an angle of $60^^\circ$ at the center. Find the area of the corresponding minor segment. (Use $\pi = 22/7, \sqrt{3} = 1.73$) (4 Marks)
Area = $\frac{60}{360} \times \frac{22}{7} \times 21 \times 21$
$= \frac{1}{6} \times 22 \times 3 \times 21 = 231$ cm-.
Since $\theta = 60^^\circ$, $\Delta OAB$ is equilateral.
Area = $\frac{\sqrt{3}}{4} \times (side)^2 = \frac{1.73}{4} \times 21 \times 21$
$= 190.73$ cm-.
Area = Sector - Triangle
$= 231 - 190.73 = \mathbf{40.27}$ cm-.
Previous Year Questions (PYQs)
Hint: Perimeter = $l + 2r$. Find arc length $l$, then Area = $\frac{1}{2}lr$.
Ans: Quadrant of circle ($90^^\circ$). Area = $\frac{1}{4} \pi (5)^2$.
Ans: $2\pi r = 22 \Rightarrow r = 3.5$. Area = $\frac{1}{4} \pi (3.5)^2 = 9.625$ cm-.
Exam Strategy & Mistake Bank
⚠️ Mistake Bank
💡 Scoring Tips
📝 More Solved Board Questions
Sol. Area of Sector = $\frac{120}{360} \times 3.14 \times 12 \times 12 = 150.72$ cm².
Area of Triangle = $r^2 \sin(60^^\circ)\cos(60^^\circ) = 144 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = 36\sqrt{3}$.
Area of Triangle = $36 \times 1.73 = 62.28$ cm².
Area of Segment = $150.72 - 62.28 = 88.44$ cm².
Answer: 88.44 cm²
📋 Board Revision Checklist
- ✅ Circumference = $2\pi r$; Area = $\pi r^2$
- ✅ Length of Arc = $(\theta/360) \times 2\pi r$
- ✅ Area of Sector = $(\theta/360) \times \pi r^2$
- ✅ Area of Segment = Area of Sector - Area of Triangle
- ✅ Area of Quadrant = $(1/4) \pi r^2$ ($\theta = 90^^\circ$)
- ✅ Area of Semicircle = $(1/2) \pi r^2$ ($\theta = 180^^\circ$)
- ✅ Distance in $n$ revolutions = $n \times (2\pi r)$
- ✅ Angle swept by minute hand in 1 min = $6^^\circ$
Always keep your answer in terms of $\pi$ until the final step if possible. It avoids early rounding errors and makes calculations cleaner.
Case Study: Artisan Silver Brooch Floral Geometry
Scenario: An artisan jeweler creates a circular silver brooch using silver wire with a total circular diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors like floral spokes.
Q1: Find the central angle ($ heta$) subtended by each of the 10 equal sectors. 1 Mark
Q2: Calculate the total length of the silver wire required to make the brooch (use $\pi = 22/7$). 1 Mark
Q3: Find the area of each individual sector of the brooch. (OR) If a gold foil plating is applied on 3 alternate sectors, calculate the total plated area. 2 Marks
📝 10-Mark Self-Assessment Test
Time: 25 MinsSolve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.
(a) $38.5 cm^2$ (b) $77 cm^2$ (c) $154 cm^2$ (d) $19.25 cm^2$
(i) the length of the arc,
(ii) area of the sector formed by the arc.
(i) minor segment, (ii) major sector. [Use $\pi = 3.14$]
Concept Mastery Quiz 🎯
Test your readiness for the board exam.
1. If the perimeter and the area of a circle are numerically equal, then the radius of the circle is:
2. Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is:
3. The area of a circle that can be inscribed in a square of side 6 cm is:
4. In a circle of radius 21 cm, an arc subtends an angle of $60^^\circ$ at the center. The length of the arc is:
5. The area of the largest triangle that can be inscribed in a semi-circle of radius $r$ units is:
Frequently Asked Questions: Class 10 Maths Chapter 11
What is the weightage of Areas Related to Circles in Class 10?
Chapter 11 carries 3 to 4 marks in Unit 6: Mensuration (total 10 marks), typically featuring 1-mark MCQs and a 2-mark or 3-mark question on sector and segment calculations.
What are the formulas for area and arc length of a sector?
For a circle of radius r and central angle theta: Arc length l = (theta / 360) * 2 * pi * r, and Area of sector = (theta / 360) * pi * r^2.
How do you calculate the area of a minor segment of a circle?
Area of minor segment = Area of corresponding sector - Area of corresponding triangle (Delta OAB). For theta = 60 degrees, area of Delta OAB = (sqrt(3)/4)*r^2; for theta = 90 degrees, area of Delta OAB = (1/2)*r^2.
Are central angles greater than 90 degrees tested in CBSE Class 10?
According to the CBSE rationalized syllabus, problems are restricted to central angles of 60 and 90 degrees (120-degree questions have been removed from standard evaluations).