Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles

Important questions on Areas Related to Circles for Class 10 are provided here to help CBSE students prepare effectively for their exams. These NCERT-based practice questions from Class 10 Maths Chapter 12: Areas Related to Circles are designed to strengthen conceptual understanding and improve problem-solving skills.

Students can use these CBSE Class 10 Maths important questions to understand the exam pattern, revise key concepts, and build confidence in solving different types of polynomial problems.

Table of Contents

• Key Concepts of Class 10 Areas Related to Circles
• Important Questions on Area and Perimeter of a Circle
• Important Questions on Area of a Sector of a Circle
• Important Questions on Area of a Segment of a Circle
• Important Questions on Areas of Combinations of Plane Figures
• Case-study based Questions on Areas Related to Circles for Class 10

Key Concepts of Class 10 Areas Related to Circles

Before solving any question on areas related to circles, you need these formulas at your fingertips.

Important Questions on Area and Perimeter of a Circle

Q1. The circumference of a circle is numerically equal to its area. Find the radius.

Answer: 2πr = πr²

⇒ 2r = r²

⇒ r(r − 2) = 0 ⇒ r ≠ 0

Therefore, radius r = 2 units

Q2. If the diameter of a semicircular protractor is 14 cm, find its perimeter.

Answer: r = 7 cm

Perimeter = πr + 2r = (22/7) × 7 + 2 × 7 = 22 + 14 = 36 cm

Q3. A circular playground of area 22,176 m² needs to be fenced. Find the cost of fencing at ₹50 per metre.

Answer: πr² = 22176

(22/7) × r² = 22176

r² = 22176 × 7/22 = 7056

r = 84 m

Circumference = 2πr = 2 × (22/7) × 84 = 528 m

Cost = 528 × 50 = ₹26,400

Q4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km/h?

Answer: Speed = 66 km/h = (66 × 1000)/60 m/min = 1100 m/min

Distance in 10 minutes = 1100 × 10 = 11,000 m = 11,000 × 100 cm = 11,00,000 cm

Circumference of wheel = πd = (22/7) × 80 = 1760/7 cm

Number of revolutions = Distance / Circumference

= 1100000 ÷ (1760/7)

= 1100000 × 7/1760

= 7700000/1760

= 4375 revolutions

Important Questions on Area of a Sector of a Circle

Q1. Find the length of an arc of a circle with radius 6 cm and central angle 60°.

Answer: Length of arc = (θ/360°) × 2πr

= (60/360) × 2 × (22/7) × 6

= (1/6) × (264/7)

= 44/7 cm ≈ 6.28 cm

Q2. The minute hand of a clock is 14 cm long. Find the area swept by the minute hand in 5 minutes.

Answer: The minute hand completes 360° in 60 minutes, so in 5 minutes it sweeps:

θ = (5/60) × 360° = 30°

Area swept = (θ/360°) × πr²

= (30/360) × (22/7) × 14²

= (1/12) × (22/7) × 196

= (1/12) × 616

= 154/3 cm² ≈ 51.33 cm²

Q3. To warn ships about underwater rocks, a lighthouse spreads light over a sector of angle 80° to a distance of 16.5 km. Find the area covered. (Use π = 3.14)

Answer: Area = (θ/360°) × πr²

= (80/360) × 3.14 × (16.5)²

= (2/9) × 3.14 × 272.25

= (2/9) × 854.865

= 189.97 km² ≈ 190 km²

Important Questions on Area of a Segment of a Circle

Q1. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use π = 3.14)

Answer: Give: θ = 90°, r = 10 cm

Area of sector = (90/360) × 3.14 × 100 = (1/4) × 314 = 78.5 cm²

Area of triangle (right triangle with both legs = r = 10 cm):

= (1/2) × 10 × 10 = 50 cm²

Area of minor segment = 78.5 − 50 = 28.5 cm²

Q2. Find the area of a segment of a circle with radius 12 cm, where the central angle is 120°.

Answer: Given: θ = 120°, r = 12 cm

Area of sector = (120/360) × π × 144 = (1/3) × (22/7) × 144 = 4224/21 = 201.14 cm² (approx.)

For the triangle: with θ = 120°, the triangle has two sides of 12 cm with included angle 120°.

Area of triangle = (1/2) × r² × sin 120° = (1/2) × 144 × (√3/2) = 36√3 = 62.35 cm² (approx.)

Area of segment = 201.14 − 62.35 = 138.79 cm² (approx.)

[Using exact values: (48π − 36√3) cm²]

Important Questions on Areas of Combinations of Plane Figures

Q1. A square is inscribed in a circle of diameter p cm. What is the area of the square?

Answer: The diagonal of the square = diameter of circle = p cm.

Using diagonal² = 2 × side²: side² = p²/2

Therefore, Area of square = p²/2 cm²

Q2. A race track is in the form of a ring. The inner circumference is 352 m and the outer circumference is 396 m. Find the width of the track and its area.

Answer: Let inner radius = r and outer radius = R.

2πr = 352 ⇒ r = 352/(2 × 22/7) = 352 × 7/44 = 56 m

2πR = 396 ⇒ R = 396 × 7/44 = 63 m

Width of track = R − r = 63 − 56 = 7 m

Area of track = π(R² − r²) = (22/7)(63² − 56²)

= (22/7)(3969 − 3136)

= (22/7) × 833

= 22 × 119

= 2618 m²

Q3. The area of an equilateral triangle is 49√3 cm². Taking each vertex as the centre, circles are drawn with radius equal to half the side of the triangle. Find the area of the part of the triangle not covered by the circles.

Answer: Given: Area of equilateral triangle = (√3/4) × a² = 49√3

So a² = 196 ⇒ a = 14 cm

Radius of each circle = 14/2 = 7 cm

Each interior angle of an equilateral triangle = 60°

Area of each sector (inside triangle) = (60/360) × π × 7² = (1/6) × (22/7) × 49 = 77/6 cm²

⇒ Total area of 3 sectors = 3 × 77/6 = 77/2 = 38.5 cm²

⇒ Area not covered = 49√3 − 38.5 = 49 × 1.732 − 38.5 ≈ 84.87 − 38.5 = ≈ 46.37 cm²

[Exact: (49√3 − 77/2) cm²]

Case study-based Questions on Areas Related to Circles for Class 10

CBSE has been introducing case study questions and assertion based questions in Board exams since 2021. Here are two fully solved repeated questions on Areas Related to Circles.

Case Study 1: 

A school decided to create a beautiful circular garden in its front yard. The garden has a circular centre piece of radius 7 m. Around this centrepiece, a ring-shaped path of uniform width 3.5 m has been laid. The garden also has a decorative sector-shaped region with a central angle of 120°, which is painted red.

Based on the above information, answer the following questions:

Q1. What is the area of the circular centrepiece?

Answer: πr² = (22/7) × 49 = 154 m²

Q2. What is the outer radius of the ring-shaped path?

Answer: Outer radius = 7 + 3.5 = 10.5 m

Q3. What is the area of the ring-shaped path?

Answer: Area = π(R² − r²) = (22/7)(10.5² − 7²)

= (22/7)(110.25 − 49)

= (22/7) × 61.25

= 192.5 m²

Q4. What is the area of the 120° sector of the centrepiece?

Answer: Area = (120/360) × (22/7) × 49 = (1/3) × 154 = 51.33 m² (approx.)

Q5. What is the perimeter of the sector with 120° and radius 7 m?

Answer:

Arc length = (120/360) × 2 × (22/7) × 7 = (1/3) × 44 = 44/3 m

Perimeter = 2r + arc = 14 + 44/3 = (42 + 44)/3 = 86/3 m ≈ 28.67 m

Assertion–Reason Questions

Directions: In each question, an Assertion (A) and Reason (R) are given. Choose the correct option:

(A) Both A and R are true, and R is the correct explanation of A.

(B) Both A and R are true, but R is not the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

Q1. Assertion (A): The area of a sector of a circle of radius 7 cm with central angle 90° is 38.5 cm².

Reason (R): Area of a sector = (θ/360°) × πr².

Answer: (A) Both are true and R correctly explains A.

Verification: (90/360) × (22/7) × 49 = (1/4) × 154 = 38.5 cm²

Q2. Assertion (A): A linear polynomial always has exactly one zero.

Reason (R): The zero of the minor segment is always less than the area of the sector.

Answer: (A) is True; (R) is True but not the correct explanation ⇒ (B)

The area of the minor segment = Area of sector − Area of triangle, so it is always less than the area of the sector. But R doesn't explain A.

Q3. Assertion (A): The area of the major segment is always greater than the area of the minor segment.

Reason (R): Area of major segment + Area of minor segment = Area of circle.

Answer: (A) Both are true and R correctly explains why A is true (since the total is πr², the major segment must account for more than half).