Class 10 Areas of Circles Notes

Exam Weightage & Blueprint

Total: ~4-5 Marks

This chapter falls under Unit VI: Mensuration (10 marks total). As per the latest syllabus, focus is on: Area of sectors and segments of a circle. Problems restricted to central angles of 60°, 90° and 120° only.

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)

📐 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

Area of Segment = Area of Sector - Area of Corresponding Triangle

Area of Triangle Tips:
- 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

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)

Step 1: Find Angle $\theta$ (0.5 Mark)

Angle swept in 60 mins = $360^\circ$.

Angle in 5 mins = $\frac{360}{60} \times 5 = 30^\circ$.

Step 2: Apply Formula (0.5 Mark)

Area swept = Area of Sector = $\frac{\theta}{360} \times \pi r^2$.

$r = 14$ cm, $\theta = 30^\circ$.

Step 3: Calculate (1 Mark)

$= \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$

$= \frac{1}{12} \times 22 \times 2 \times 14 = \frac{1}{12} \times 616$

$= \frac{154}{3} = \mathbf{51.33}$ 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)

Step 1: Area of Sector (1.5 Marks)

Area = $\frac{60}{360} \times \frac{22}{7} \times 21 \times 21$

$= \frac{1}{6} \times 22 \times 3 \times 21 = 231$ cm-.

Step 2: Area of Triangle (1.5 Marks)

Since $\theta = 60^\circ$, $\Delta OAB$ is equilateral.

Area = $\frac{\sqrt{3}}{4} \times (\text{side})^2 = \frac{1.73}{4} \times 21 \times 21$

$= 190.73$ cm-.

Step 3: Segment Area (1 Mark)

Area = Sector - Triangle

$= 231 - 190.73 = \mathbf{40.27}$ cm-.

Previous Year Questions (PYQs)

2023 (1 Mark): The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector.
Hint: Perimeter = $l + 2r$. Find arc length $l$, then Area = $\frac{1}{2}lr$.

2020 (3 Marks): A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find the area of that part of the field in which the horse can graze.
Ans: Quadrant of circle ($90^\circ$). Area = $\frac{1}{4} \pi (5)^2$.

2019 (2 Marks): Find the area of a quadrant of a circle whose circumference is 22 cm.
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

Radius vs Diameter: Always check if diameter is given. Divide by 2 immediately! Using diameter in $\pi r^2$ is a common error.

Perimeter of Semicircle: It is $\pi r + 2r$, not just $\pi r$. Don't forget the boundary diameter line.

Calculation: Do not approximate $\pi$ as 3.14 unless specified. Use 22/7 for better cancellation with multiples of 7 (like radius 7, 14, 21).

💡 Scoring Tips

Identify Triangle Type: Check the central angle $\theta$. If $60^\circ$ -> Equilateral. If $90^\circ$ -> Right Angled. This simplifies area calculation.

Units: Areas are in $cm^2$ or $m^2$. Lengths in $cm$ or $m$. Check units before final answer.

Visualise: For complex figures, label the "Shaded Region" = Area of Larger Shape - Area of Unshaded Shape.

📝 More Solved Board Questions

Q3. Find the area of the segment of a circle of radius 12 cm if the central angle is $120^\circ$. (Use $\pi = 3.14, \sqrt{3} = 1.73$) 4 Marks

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 Pattern (2018–2025): For $\theta = 120^\circ$, you can use the formula Area of Triangle = $r^2 \sin(60^\circ)\cos(60^\circ)$ or split the triangle into two right-angled triangles to find base and height. This is a very common "Section D" question.

📋 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 = $6^\circ$

💡 Exam Tip:
Always keep your answer in terms of $\pi$ until the final step if possible. It avoids early rounding errors and makes calculations cleaner.

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:

Chapter 11: Areas Related to Circles

Overview