NCERT Class 10 Maths Chapter 11 Areas Related to Circles Solutions MP Board
Class 10 Mathematics: Chapter 11 Areas Related to Circles (MP Board)
Welcome Class 10 MP Board students! Chapter 11, Areas Related to Circles, is a crucial unit in geometry that builds your understanding of perimeter, area, sectors, and segments of circles. Below, you will find important formulas along with step-by-step NCERT solutions to help you excel in your board exams.
Important Formulas
- Circumference of a circle: C = 2πr = πd
- Area of a circle: A = πr2
- Length of an arc of a sector with angle θ: l = (θ / 360°) × 2πr
- Area of a sector of angle θ: A = (θ / 360°) × πr2
- Area of a segment: Area of Sector - Area of Corresponding Triangle
- Area of an equilateral triangle: (√3 / 4) × side2
Exercise 11.1 Solutions
Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Given: Radius of circle (r) = 6 cm, Angle of sector (θ) = 60°.
Area of sector = (θ / 360°) × πr2
= (60° / 360°) × (22 / 7) × 6 × 6
= (1 / 6) × (22 / 7) × 36
= (22 × 6) / 7 = 132 / 7 cm2.
Answer: The area of the sector is 132/7 cm2 (or 18.86 cm2).
Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Circumference of circle = 2πr = 22 cm
2 × (22 / 7) × r = 22 ⇒ r = 7 / 2 cm.
Angle of a quadrant (θ) = 90°.
Area of quadrant = (90° / 360°) × πr2
= (1 / 4) × (22 / 7) × (7 / 2) × (7 / 2)
= 77 / 8 cm2.
Answer: The area of the quadrant is 77/8 cm2 (or 9.625 cm2).
Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Length of minute hand (r) = 14 cm.
Angle swept in 60 minutes = 360°.
Angle swept in 5 minutes (θ) = (360° / 60) × 5 = 30°.
Area swept in 5 minutes = (θ / 360°) × πr2
= (30° / 360°) × (22 / 7) × 14 × 14
= (1 / 12) × (22 / 7) × 196
= 154 / 3 cm2.
Answer: The area swept by the minute hand is 154/3 cm2 (or 51.33 cm2).
Q4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)
Solution:
Given: r = 10 cm, θ = 90°.
(i) Area of minor sector = (90° / 360°) × 3.14 × 10 × 10 = (1 / 4) × 314 = 78.5 cm2.
Area of triangle = (1 / 2) × base × height = (1 / 2) × 10 × 10 = 50 cm2.
Area of minor segment = Area of minor sector - Area of triangle = 78.5 - 50 = 28.5 cm2.
(ii) Major sector angle = 360° - 90° = 270°.
Area of major sector = (270° / 360°) × 3.14 × 10 × 10 = (3 / 4) × 314 = 235.5 cm2.
Answer: (i) Minor segment = 28.5 cm2, (ii) Major sector = 235.5 cm2.
Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) length of the arc (ii) area of the sector (iii) area of the segment.
Solution:
Given: r = 21 cm, θ = 60°.
(i) Length of arc = (θ / 360°) × 2πr = (60° / 360°) × 2 × (22 / 7) × 21 = (1 / 6) × 132 = 22 cm.
(ii) Area of sector = (θ / 360°) × πr2 = (60° / 360°) × (22 / 7) × 21 × 21 = 231 cm2.
(iii) Area of segment = Area of sector - Area of equilateral triangle
= 231 - [(√3 / 4) × 212] = (231 - 441√3 / 4) cm2.
Answer: (i) 22 cm, (ii) 231 cm2, (iii) (231 - 441√3 / 4) cm2.
Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments. (Use π = 3.14, √3 = 1.73)
Solution:
Given: r = 15 cm, θ = 60°.
Area of sector = (60° / 360°) × 3.14 × 15 × 15 = 117.75 cm2.
Area of equilateral triangle = (√3 / 4) × 152 = (1.73 × 225) / 4 = 97.3125 cm2.
Area of minor segment = 117.75 - 97.3125 = 20.4375 cm2.
Area of circle = 3.14 × 15 × 15 = 706.5 cm2.
Area of major segment = 706.5 - 20.4375 = 686.0625 cm2.
Answer: Minor segment = 20.4375 cm2, Major segment = 686.0625 cm2.
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)
Solution:
Given: r = 12 cm, θ = 120°.
Area of sector = (120° / 360°) × 3.14 × 12 × 12 = (1 / 3) × 3.14 × 144 = 150.72 cm2.
Area of triangle = (1 / 2) × r2 × sin(120°) = (1 / 2) × 144 × (√3 / 2) = 36 × 1.73 = 62.28 cm2.
Area of segment = 150.72 - 62.28 = 88.44 cm2.
Answer: The area of the segment is 88.44 cm2.
Q8. A horse is tied to a peg at one corner of a square grass field of side 15 m by a 5 m long rope. Find: (i) grazing area (ii) increase in area if rope is 10 m long. (Use π = 3.14)
Solution:
Corner angle = 90°.
(i) For r = 5 m: Grazing area = (90° / 360°) × 3.14 × 52 = (1 / 4) × 78.5 = 19.625 m2.
(ii) For r = 10 m: New grazing area = (90° / 360°) × 3.14 × 102 = (1 / 4) × 314 = 78.5 m2.
Increase in area = 78.5 - 19.625 = 58.875 m2.
Answer: (i) 19.625 m2, (ii) Increase = 58.875 m2.
Q9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. Wire is also used for 5 diameters dividing into 10 equal sectors. Find: (i) total length of wire (ii) area of each sector.
Solution:
Diameter (d) = 35 mm, Radius (r) = 35 / 2 mm.
(i) Total length of wire = Circumference + 5 × Diameter
= πd + 5(35) = (22 / 7 × 35) + 175 = 110 + 175 = 285 mm.
(ii) Area of each sector = Total circle area / 10 = [(22 / 7) × (35 / 2)2] / 10 = (3850 / 4) / 10 = 385 / 4 mm2.
Answer: (i) Total wire = 285 mm, (ii) Area of each sector = 385/4 mm2 (or 96.25 mm2).
Q10. An umbrella has 8 ribs equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find area between two consecutive ribs.
Solution:
Total sectors = 8, Radius (r) = 45 cm.
Area between two consecutive ribs = Area of circle / 8
= [(22 / 7) × 45 × 45] / 8 = 22275 / 28 cm2.
Answer: Area between two ribs is 22275/28 cm2 (or 795.53 cm2).
Q11. A car has two wipers. Each wiper blade length is 25 cm sweeping through an angle of 115°. Find total area cleaned at each sweep.
Solution:
r = 25 cm, θ = 115°.
Area cleaned by 1 wiper = (115° / 360°) × (22 / 7) × 25 × 25.
Total area cleaned by 2 wipers = 2 × (115 / 360) × (22 / 7) × 625 = 158125 / 126 cm2.
Answer: Total area cleaned is 158125/126 cm2 (or 1254.96 cm2).
Q12. A lighthouse spreads red light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea warned. (Use π = 3.14)
Solution:
θ = 80°, r = 16.5 km.
Area = (80° / 360°) × 3.14 × (16.5)2 = (2 / 9) × 3.14 × 272.25 = 189.97 km2.
Answer: Area warned is 189.97 km2.
Q13. A round table cover has six equal designs. If radius is 28 cm, find the cost of making designs at ₹ 0.35 per cm2. (Use √3 = 1.73)
Solution:
Number of designs = 6, θ for each sector = 360° / 6 = 60°, r = 28 cm.
Area of 6 sectors = 6 × [(60° / 360°) × (22 / 7) × 282] = 2464 cm2.
Area of 6 equilateral triangles = 6 × [(√3 / 4) × 282] = 6 × [1.73 × 196] = 2034.48 cm2.
Area of 6 designs = Area of 6 sectors - Area of 6 triangles = 2464 - 2034.48 = 429.52 cm2.
Cost = 429.52 × 0.35 = ₹ 150.33.
Answer: Total cost of making the designs is ₹ 150.33.
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
Solution:
Area of sector = (p / 360) × πR2. Multiplying numerator and denominator by 2 gives (p / 720) × 2πR2.
Answer: Option (D).
We hope these detailed NCERT solutions for Class 10 Chapter 11 Areas Related to Circles help you score top marks in your MP Board exams. Practice regularly and memorize the key formulas to solve problems efficiently!
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