NCERT Solutions for Class 10 Maths Chapter 11: Areas Related to Circles (2026-27) – Free PDF Download

Class 10 Maths Chapter 11, Areas Related to Circles, has a single exercise (Exercise 11.1) in the current 2023-rationalised, 2026-27 session syllabus, combining sector/segment area and combination-of-figures problems. Below are original, step-by-step solutions to all 14 questions.

Last Updated: September 10, 2026

Why This Chapter Matters for Boards

Areas Related to Circles is a high-weightage Mensuration chapter in the CBSE Class 10 Maths board exam. Questions here test application of the sector and segment area formulas to real-life contexts (clocks, wipers, brooches, umbrellas, tables) and frequently appear as 3-5 mark application-based questions.

Exercise 11.1 Solutions

1. Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
   Area = (θ/360) × πr² = (60/360) × (22/7) × 6² = (1/6) × (22/7) × 36 = 132/7 = 18.86 cm² (approx.)
2. Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.
   2πr = 22 ⇒ r = 22 × 7 / (2 × 22) = 3.5 cm.
   Quadrant area = (90/360) × πr² = (1/4) × (22/7) × 3.5² = (1/4) × 38.5 = 9.625 cm²
3. Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by it in 5 minutes.
   Angle swept in 5 min = (5/60) × 360° = 30°.
   Area = (30/360) × (22/7) × 14² = (1/12) × 616 = 51.33 cm² (approx.)
4. Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of (i) minor segment (ii) major sector (use π = 3.14).
   (i) Sector(90°) area = (1/4) × 3.14 × 100 = 78.5 cm². Triangle area (right angle, legs = r) = (1/2) × 10 × 10 = 50 cm².
   Minor segment = 78.5 − 50 = 28.5 cm²
   (ii) Circle area = 3.14 × 100 = 314 cm². Major sector = 314 − 78.5 = 235.5 cm²
5. Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (i) length of arc (ii) area of sector (iii) area of segment.
   (i) Arc length = (60/360) × 2 × (22/7) × 21 = (1/6) × 132 = 22 cm
   (ii) Sector area = (60/360) × (22/7) × 441 = (1/6) × 1386 = 231 cm²
   (iii) Triangle area (equilateral since angle = 60°, two sides = r) = (√3/4) × 21² ≈ 190.87 cm². Segment = 231 − 190.87 = 40.13 cm² (approx.)
6. Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the minor and major segments (use π = 3.14, √3 = 1.73).
   Sector(60°) = (1/6) × 3.14 × 225 = 117.75 cm². Triangle (equilateral) = (1.73/4) × 225 = 97.31 cm².
   Minor segment = 117.75 − 97.31 = 20.44 cm²
   Circle area = 3.14 × 225 = 706.5 cm². Major segment = 706.5 − 20.44 = 686.06 cm²
7. Q7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment (use π = 3.14, √3 = 1.73).
   Sector(120°) = (1/3) × 3.14 × 144 = 150.72 cm². Triangle area = (1/2)r²sin120° = (1/2) × 144 × 0.866 = 62.35 cm².
   Segment = 150.72 − 62.35 = 88.37 cm²
8. Q8. A horse is tied to a peg at one corner of a square field of side 15 m by a 5 m rope. Find (i) the grazing area, (ii) the increase in grazing area if the rope is 10 m long (use π = 3.14). Since the peg is at a corner of a square, the horse can graze a quarter circle (90°).
   (i) Area = (1/4) × 3.14 × 5² = 19.625 m²
   (ii) With 10 m rope: (1/4) × 3.14 × 10² = 78.5 m². Increase = 78.5 − 19.625 = 58.875 m²
9. Q9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used to make 5 diameters which divide the circle into 10 equal sectors. Find (i) total length of silver wire, (ii) area of each sector.
   Radius = 17.5 mm. Circumference = πd = (22/7) × 35 = 110 mm. 5 diameters = 5 × 35 = 175 mm.
   (i) Total wire = 110 + 175 = 285 mm
   (ii) Each sector area = πr²/10 = [(22/7) × 17.5²]/10 = 962.5/10 = 96.25 mm²
10. Q10. An umbrella has 8 ribs equally spaced, with radius 45 cm. Find the area between two consecutive ribs.
    Each sector angle = 360/8 = 45°. Area = πr²/8 = [(22/7) × 45²]/8 = 6364.29/8 = 795.54 cm² (approx.)
11. Q11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through 115°. Find the total area cleaned.
    Each sweep area = (115/360) × (22/7) × 25² = 627.49 cm² (approx). Total for 2 wipers = 1254.96 cm² (approx.)
12. Q12. A ship is spreading a warning red-light signal through a sector of angle 80° over the sea, to a distance of 16.5 km. Find the area of the sea over which ships are warned (use π = 3.14).
    Area = (80/360) × 3.14 × 16.5² = (2/9) × 3.14 × 272.25 = 189.97 km² (approx.)
13. Q13. A round table cover has six equal designs made by dividing it into 6 equal segments, radius 28 cm. Find the cost of making the designs at Rs 0.35 per cm² (use √3 = 1.7).
    Each design corresponds to a 60° segment. Sector = (1/6) × (22/7) × 784 = 410.67 cm². Triangle (equilateral) = (1.7/4) × 784 = 333.2 cm².
    One segment = 410.67 − 333.2 = 77.47 cm². Six segments = 464.8 cm² (approx). Cost = 464.8 × 0.35 = ₹162.68 (approx, using √3=1.7 as specified).
14. Q14. Tick the correct answer: Area of a sector of angle P (in degrees) of a circle with radius R is:
    (A) P/180 × 2πR   (B) P/180 × πR²   (C) P/360 × 2πR   (D) P/720 × 2πR²
    Since sector area = (P/360) × πR² = (P/720) × 2πR², the answer is (D).

More on This Chapter

• NCERT Solutions – Chapter 11
• Extra Questions (HOTS) – Chapter 11
• Revision Notes – Chapter 11
• Class 10 Maths Book (Catalog Page)

FAQ

Q: Is this content updated for the 2026-27 NCERT edition?
A: Yes. Chapter 11 in the current rationalised syllabus has only Exercise 11.1 (the older separate Exercise 12.1/12.2 split from pre-2023 editions has been merged into one exercise) — verified against multiple independent sources before writing.

Q: Which formula should I remember for board exams?
A: Area of sector = (θ/360) × πr², and area of segment = area of sector − area of the triangle formed by the two radii and the chord.

Written by Satish

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