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NCERT Class 10 Mathematics Exercise 11.1 Solutions: Areas Related to Circles

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 11.1 in NCERT's Class 10 Mathematics book, chapter “Areas Related to Circles”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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Ex 11.1 Q1
Find the area of a sector of radius 6 cm and angle 60° (π = 22/7).
1. The formula for the area of a sector is (θ/360°) × πr².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
2. Substitute the given values: θ = 60°, r = 6 cm, and π = 22/7.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
3. Area = (60/360) × (22/7) × 6².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
4. Simplify the fraction 60/360 to 1/6.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
5. Area = (1/6) × (22/7) × 36.
6. Multiply 36 by 1/6 to get 6.
7. Area = (22/7) × 6.
8. Calculate the final product: 132/7.
Answer: 132/7 cm²
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Ex 11.1 Q2
Find the area of a quadrant of a circle whose circumference is 22 cm (π = 22/7).
1. Let r be the radius of the circle.
2. The circumference is given by the formula 2πr.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
3. Substitute the given values into the formula: 2 × (22/7) × r = 22.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
4. Simplify the equation to get (44/7) × r = 22.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. Solve for r by multiplying both sides by 7/44: r = 22 × (7/44).
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
6. Calculate the radius: r = 7/2 cm.
7. The area of a quadrant is one-fourth of the area of the full circle, so Area = (1/4)πr².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
8. Substitute the values of π and r: Area = (1/4) × (22/7) × (7/2)².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
9. Square the radius: (7/2)² = 49/4.
10. Multiply the terms: Area = (1/4) × (22/7) × (49/4).
11. Simplify the expression by cancelling common factors: Area = (1/4) × (22 × 7) / 4.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
12. Calculate the final area: Area = 154 / 16 = 77/8 cm².
Answer: 77/8 cm²
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Ex 11.1 Q3
A clock's minute hand is 14 cm long. Find the area it sweeps in 5 minutes (π = 22/7).
1. The minute hand of a clock completes a full circle in 60 minutes.
2. In 5 minutes, the minute hand sweeps an angle of (5/60) × 360° = 30°.
3. The area swept by the minute hand is the area of a sector with radius 14 cm and angle 30°.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
4. Area of sector = (θ/360) × π × r² = (30/360) × (22/7) × 14 × 14.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
5. Simplifying, Area = (1/12) × 22 × 28 = (1/12) × 616 = 154/3 cm².
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: 154/3 cm²
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Ex 11.1 Q4
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of (i) the minor segment (ii) the major sector (π = 3.14).
1. (i) The chord subtends a right angle at the centre, so the angle of the minor sector is 90°.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
2. Area of the minor sector = (90/360) × π × r² = (1/4) × 3.14 × 10² = 78.5 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
3. The triangle formed by the radii and the chord is a right-angled isosceles triangle with legs 10 cm.
4. Area of the triangle = (1/2) × base × height = (1/2) × 10 × 10 = 50 cm².
5. Area of the minor segment = Area of minor sector − Area of triangle = 78.5 − 50 = 28.5 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
6. (ii) The angle of the major sector is 360° − 90° = 270°.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
7. Area of the major sector = (270/360) × π × r² = (3/4) × 3.14 × 10².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
8. Area of the major sector = 0.75 × 314 = 235.5 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
Answer: (i) 28.5 cm²; (ii) 235.5 cm²
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Ex 11.1 Q5
In a circle of radius 21 cm an arc subtends 60° at the centre. Find (i) the arc length (ii) the sector area (iii) the area of the segment cut off by the chord (π = 22/7).
1. (i) The formula for arc length is l = (θ/360) × 2πr.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
2. (i) Substituting θ = 60° and r = 21 cm, we get l = (60/360) × 2 × (22/7) × 21.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
3. (i) Simplifying the expression gives l = (1/6) × 2 × 22 × 3 = 22 cm.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
4. (ii) The formula for sector area is A = (θ/360) × πr².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
5. (ii) Substituting the values, A = (60/360) × (22/7) × 21 × 21.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
6. (ii) Simplifying gives A = (1/6) × 22 × 3 × 21 = 231 cm².
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. (iii) The area of the segment is the sector area minus the area of the equilateral triangle formed by the radii and chord.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
8. (iii) Since the central angle is 60° and radii are equal, the triangle is equilateral with side 21 cm.
9. (iii) The area of an equilateral triangle is (√3/4) × side².
10. (iii) Triangle area = (√3/4) × 21² = (√3/4) × 441 = 441√3/4 cm².
11. (iii) Segment area = 231 − 441√3/4 cm².
Answer: (i) 22 cm; (ii) 231 cm²; (iii) (231 - 441√3/4) cm²
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Ex 11.1 Q6
A chord of a circle of radius 15 cm subtends 60° at the centre. Find the areas of the minor and major segments (π = 3.14, √3 = 1.73).
1. Given radius r = 15 cm and central angle θ = 60°.
2. Area of the circle = πr² = 3.14 × 15² = 3.14 × 225 = 706.5 cm².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
3. Area of the sector = (θ/360) × πr² = (60/360) × 706.5 = 117.75 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
4. Since the central angle is 60° and the triangle is isosceles with two sides equal to the radius, the triangle formed by the radii and the chord is equilateral.
5. Area of the equilateral triangle = (√3/4) × side² = (1.73/4) × 15² = (1.73/4) × 225 = 97.3125 cm².
6. Area of the minor segment = Area of sector − Area of triangle = 117.75 − 97.3125 = 20.4375 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
7. Area of the major segment = Area of circle − Area of minor segment = 706.5 − 20.4375 = 686.0625 cm².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
Answer: minor segment 20.4375 cm², major segment 686.0625 cm²
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Ex 11.1 Q7
A chord of a circle of radius 12 cm subtends 120° at the centre. Find the area of the segment (π = 3.14, √3 = 1.73).
1. Let O be the centre (radius 12 cm) and AB the chord subtending ∠AOB = 120° at O.
2. Draw OM perpendicular to AB, which bisects AB and angle AOB, so angle AOM is 60°.
3. In right triangle AOM, sin 60° = AM/OA, so AM = 12 × (√3/2) = 6√3 cm.
  why?
  In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
4. Thus, the full chord AB = 2 × 6√3 = 12√3 cm.
5. In right triangle AOM, cos 60° = OM/OA, so OM = 12 × (1/2) = 6 cm.
  why?
  In a right triangle, sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent (cosec, sec, cot are their reciprocals). The ratio depends only on the angle, not on the size of the triangle.
6. Area of triangle AOB = (1/2) × base × height = (1/2) × 12√3 × 6 = 36√3 cm².
7. Substitute √3 = 1.73 to get area of triangle AOB = 36 × 1.73 = 62.28 cm².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
8. Area of sector AOB = (θ/360) × πr² = (120/360) × 3.14 × 12².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
9. Area of sector AOB = (1/3) × 3.14 × 144 = 3.14 × 48 = 150.72 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
10. Area of the minor segment = Area of sector − Area of triangle = 150.72 − 62.28 = 88.44 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
Answer: 88.44 cm²
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Ex 11.1 Q8
A horse is tied with a 5 m rope to a corner peg of a 15 m square field. Find (i) the area it can graze (ii) the increase if the rope were 10 m (π = 3.14).
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The horse is tied to a corner of the square field, so it can graze in a quarter circle (90° sector).
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
2. (i) For a 5 m rope, the radius r is 5 m.
3. The area of a quarter circle is (90/360) × π × r².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
4. Substituting the values, Area = (1/4) × 3.14 × 5².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
5. Area = (1/4) × 3.14 × 25 = 19.625 m².
6. (ii) If the rope is 10 m, the radius r becomes 10 m.
7. The new grazing area is (1/4) × 3.14 × 10².
8. New Area = (1/4) × 3.14 × 100 = 78.5 m².
9. The increase in area is the new area minus the original area.
10. Increase = 78.5 − 19.625 = 58.875 m².
Answer: (i) 19.625 m²; (ii) 58.875 m²
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Ex 11.1 Q9
A brooch is a silver-wire circle of diameter 35 mm with 5 diameters dividing it into 10 equal sectors. Find (i) the total wire length (ii) the area of each sector (π = 22/7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. (i) The diameter of the brooch is 35 mm, so the radius r is 35 ÷ 2 = 17.5 mm.
2. (i) The length of the circular wire is the circumference, calculated as 2 × π × r = 2 × (22/7) × 17.5 = 110 mm.
  why?
  Circumference = 2πr = πd: π is the same ratio (circumference ÷ diameter) for every circle. Take π = 22/7 or 3.14 as the question says.
3. (i) The 5 diameters are each 35 mm long, so their total length is 5 × 35 = 175 mm.
4. (i) The total wire length is the sum of the circumference and the diameters: 110 + 175 = 285 mm.
5. (ii) The area of the entire circle is π × r² = (22/7) × (17.5)² = (22/7) × 306.25 = 962.5 mm².
6. (ii) The 5 diameters divide the circle into 10 equal sectors.
7. (ii) The area of each sector is the total area divided by 10: 962.5 ÷ 10 = 96.25 mm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
8. (ii) Converting 96.25 to a fraction gives 385/4 mm².
Answer: (i) 285 mm; (ii) 385/4 mm²
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Ex 11.1 Q10
An umbrella has 8 equally spaced ribs; taking it as a flat circle of radius 45 cm, find the area between two consecutive ribs (π = 22/7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The umbrella is treated as a circle with radius r = 45 cm.
2. There are 8 equally spaced ribs, so the angle between two consecutive ribs is 360° ÷ 8 = 45°.
3. The area between two consecutive ribs is the area of a sector with central angle 45°.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
4. Area of sector = (θ/360°) × π × r².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
5. Substituting values: Area = (45/360) × (22/7) × 45 × 45.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
6. Simplifying the fraction: 45/360 = 1/8.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
7. So, Area = (1/8) × (22/7) × 2025.
8. Multiplying: (22 × 2025) ÷ (8 × 7) = 44550 ÷ 56.
9. Simplifying the fraction: 44550/56 = 22275/28.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
10. Thus, the area between two consecutive ribs is 22275/28 cm².
Answer: 22275/28 cm²
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Ex 11.1 Q11
A car has two non-overlapping wipers, each with a 25 cm blade sweeping 115°. Find the total area cleaned per sweep (π = 22/7).
1. The area swept by one wiper is the area of a sector with radius 25 cm and angle 115°.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
2. Area of one sector = (115/360) × (22/7) × 25².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
3. Calculating the area for one wiper gives 31625/126 cm².
4. Since there are two identical wipers, the total area is 2 × 31625/126 cm².
5. Simplifying the total area gives 158125/126 cm².
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: 158125/126 cm²
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Ex 11.1 Q12
A lighthouse lights a sector of angle 80° to a distance of 16.5 km. Find the sea area covered (π = 3.14).
1. The area of the sector is given by the formula (θ/360) × π × r².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
2. Substitute the given values: θ = 80°, r = 16.5 km, and π = 3.14.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
3. Area = (80/360) × 3.14 × (16.5)².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
4. Simplify the fraction 80/360 to 2/9.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
5. Calculate (16.5)² which equals 272.25.
6. Multiply 2/9 by 272.25 to get 60.5.
7. Multiply 60.5 by 3.14 to find the final area.
8. The calculated area is 189.97 km².
Answer: 189.97 km²
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Ex 11.1 Q13
A round table cover of radius 28 cm has six equal designs, each the segment cut off by a side of a regular hexagon inscribed in it (Fig. 11.11). Find the cost of the designs at ₹0.35 per cm² (π = 22/7, √3 = 1.7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Each of the six designs is the segment cut off by one side of a regular hexagon inscribed in the circle, so each chord subtends 360° ÷ 6 = 60° at the centre.
2. Area of one 60° sector = (60/360) × (22/7) × 28² = (1/6) × (22/7) × 784 = 1232/3 ≈ 410.67 cm².
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
3. The triangle formed by the two radii and the chord is equilateral with side 28 cm: area = (√3/4) × 28² = (1.7/4) × 784 = 333.2 cm².
4. Area of one design = 1232/3 − 333.2 ≈ 77.47 cm².
5. Area of six designs = 6 × (1232/3 − 333.2) = 2464 − 1999.2 = 464.8 cm².
6. Cost = 464.8 × ₹0.35 = ₹162.68.
Answer: Area of the designs = 464.8 cm²; cost ₹162.68
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Ex 11.1 Q14
The area of a sector of angle p° of a circle of radius R is (A) p/180 × 2πR (B) p/180 × πR² (C) p/360 × 2πR (D) p/720 × 2πR².
1. The area of a full circle with radius R is given by the formula πR².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
2. A full circle corresponds to an angle of 360° at the center.
3. Therefore, the area of a sector with angle p° is the fraction p/360 of the total area.
  why?
  A sector of angle θ is θ/360 of the whole circle, so its area is (θ/360) × πr² and its arc length is (θ/360) × 2πr.
4. Substituting the total area, the formula becomes (p/360) × πR².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
5. We can rewrite πR² as (2πR²)/2 to match the options provided.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
6. This gives (p/360) × (2πR²)/2, which simplifies to (p/720) × 2πR².
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
7. This expression matches option (D).
Answer: (D)
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NCERT Class 10 Mathematics Exercise 11.1 Solutions: Areas Related to Circles

Ex 11.1 Q1

Ex 11.1 Q2

Ex 11.1 Q3

Ex 11.1 Q4

Ex 11.1 Q5

Ex 11.1 Q6

Ex 11.1 Q7

Ex 11.1 Q8

Ex 11.1 Q9

Ex 11.1 Q10

Ex 11.1 Q11

Ex 11.1 Q12

Ex 11.1 Q13

Ex 11.1 Q14