CBSE Class 10 Maths — Areas Related to Circles

Chapter Overview & Weightage

Areas Related to Circles is Chapter 12 of CBSE Class 10 Maths. It regularly carries 6–8 marks in the board exam, typically as one 3-mark question and one 4-mark question, or a single 5-mark problem.

The most common question types are: (1) area of a sector or segment, (2) area of a figure formed by combining a circle with a rectangle/triangle, and (3) area of the shaded region between two geometric figures. These appear almost every year in CBSE boards and are considered scoring questions once the formulas are clear.

Year	Question type	Marks
2024	Sector + triangle combination	4 marks
2023	Shaded region (square – circle)	3 marks
2022	Area of segment	4 marks
2021	Horse tethered to a post — sector area	3 marks

Key Concepts You Must Know

Sector: A “pie slice” of a circle — the region bounded by two radii and an arc.
Minor sector: The smaller sector (angle < 180°).
Major sector: The larger sector (angle > 180°).
Segment: The region between a chord and the arc it cuts off.
Minor segment: The smaller region (the “cap”).
Major segment: The larger region.
Angle at centre ( $\theta$ ): The angle subtended by the arc at the centre. Always given in degrees for CBSE Class 10.
Perimeter of a sector: Two radii + arc length = $2r + \frac{\theta}{360} \times 2\pi r$ .
Perimeter of a segment: Chord length + arc length.

Important Formulas

\text{Arc Length} = \frac{\theta}{360°} \times 2\pi r

\text{Area of Sector} = \frac{\theta}{360°} \times \pi r^2

Tip: Think of the fraction $\theta/360°$ as the “share” of the full circle that this sector represents.

\text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle}

= \frac{\theta}{360°} \times \pi r^2 - \frac{1}{2}r^2 \sin\theta

The triangle is formed by the two radii and the chord.

For a sector with angle $\theta$ and radius $r$ , the triangle formed by the two radii:

\text{Area of triangle} = \frac{1}{2}r^2 \sin\theta

Special cases:

$\theta = 60°$ : equilateral triangle, area $= \frac{\sqrt{3}}{4}r^2$
$\theta = 90°$ : right-angled triangle, area $= \frac{1}{2}r^2$

Solved Previous Year Questions

PYQ 1 — Sector area (3 marks)

Q: Find the area of a sector of a circle with radius 6 cm and angle 60°.

Solution:

A = \frac{60}{360} \times \pi \times 6^2 = \frac{1}{6} \times \pi \times 36 = 6\pi \approx 18.86 \text{ cm}^2

Use $\pi = 22/7$ or $3.14$ as directed. Answer: $6\pi$ cm² (leave in terms of $\pi$ if not instructed otherwise).

PYQ 2 — Segment area (4 marks)

Q: A chord PQ of a circle with radius 10 cm subtends an angle of 90° at the centre. Find the area of the minor segment.

Solution:

Area of sector $= \frac{90}{360} \times \pi \times 10^2 = \frac{1}{4} \times 100\pi = 25\pi$ cm²

Area of right triangle (90° angle, both sides = radius = 10 cm): $= \frac{1}{2} \times 10 \times 10 = 50$ cm²

Area of minor segment $= 25\pi - 50 = 25(3.14) - 50 = 78.5 - 50 = \mathbf{28.5 \text{ cm}^2}$

PYQ 3 — Combination figure (5 marks)

Q: A horse is tethered to a corner of a square field of side 14 m by a rope 7 m long. Find the area it can graze.

Solution: The horse is at a corner of the square. The angle available at a corner = 90°. The grazing area is a sector with $r = 7$ m, $\theta = 90°$ :

A = \frac{90}{360} \times \frac{22}{7} \times 7^2 = \frac{1}{4} \times \frac{22}{7} \times 49 = \frac{22 \times 7}{4} = \frac{154}{4} = \mathbf{38.5 \text{ m}^2}

Difficulty Distribution

Level	% of questions	Type
Easy	30%	Direct formula: find area of sector
Medium	50%	Segment area or perimeter
Hard	20%	Combination figures, shaded regions

Expert Strategy

The most important habit: draw a clear figure for every question. Shade the region you need to find. Then identify which formulas apply.

For shaded region problems, the usual approach is: bigger area − smaller area = shaded area. Identify the simpler shapes that overlap or are subtracted.

Memorise the two “special” cases of the triangle formula because they appear frequently:

90° sector: triangle area = $r^2/2$ , so segment = $\frac{\pi r^2}{4} - \frac{r^2}{2} = r^2(\pi/4 - 1/2)$
60° sector: triangle is equilateral with side = $r$ , area = $\frac{\sqrt{3}}{4}r^2$

For perimeter questions, students often forget to add the chord length (for a segment) or two radii (for a sector). Always write out all components of the perimeter before calculating.

Common Traps

Trap 1: Forgetting to subtract the triangle area when finding segment area. Segment = Sector − Triangle. This step is missed about 40% of the time in board exams.

Trap 2: Using diameter instead of radius in formulas. All circle formulas use $r$ (radius). If the problem gives diameter, halve it first.

Trap 3: Confusing arc length with sector area. Arc length = $\frac{\theta}{360} \times 2\pi r$ (one-dimensional). Sector area = $\frac{\theta}{360} \times \pi r^2$ (two-dimensional). The $r$ in arc length is outside the fraction; in sector area it is squared.

Trap 4: Wrong value of $\pi$ . CBSE sometimes says “use $\pi = 22/7$ ” and sometimes “use $\pi = 3.14$ ”. If the radius is a multiple of 7, use 22/7 for clean arithmetic. Read the question carefully.