Given blade length r = 28 cm, angle of sweep θ = 120° and number of wipers = 2. Area swept by one wiper blade: Area = (θ / 360) x π x r² Area = (120 / 360) x (22/7) x 28 x 28 Area = (1/3) x 22 x 4 x 28 = 2464 / 3 cm². Total area cleaned by both wipers: Total Area = 2 x (2464 / 3) = 4928 / 3 cm². InRead more
Given blade length r = 28 cm, angle of sweep θ = 120° and number of wipers = 2.
Area swept by one wiper blade:
Area = (θ / 360) x π x r²
Area = (120 / 360) x (22/7) x 28 x 28
Area = (1/3) x 22 x 4 x 28 = 2464 / 3 cm².
Total area cleaned by both wipers:
Total Area = 2 x (2464 / 3) = 4928 / 3 cm².
In decimal form, 4928 / 3 is approximately 1642.67 cm².
Thus, total area cleaned is 4928/3 cm² (or 1642.67 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Given r = 15 cm, θ = 60°, π = 3.14 and √3 = 1.73. Total circle area = π x r² = 3.14 x 15² = 3.14 x 225 = 706.5 cm². Area of minor sector = (60 / 360) x 3.14 x 225 = (1/6) x 706.5 = 117.75 cm². The triangle formed is equilateral (angle 60°): Area of triangle = (√3 / 4) x r² = (1.73 / 4) x 225 = 97.31Read more
Given r = 15 cm, θ = 60°, π = 3.14 and √3 = 1.73.
Total circle area = π x r² = 3.14 x 15² = 3.14 x 225 = 706.5 cm².
Area of minor sector = (60 / 360) x 3.14 x 225 = (1/6) x 706.5 = 117.75 cm².
The triangle formed is equilateral (angle 60°):
Area of triangle = (√3 / 4) x r² = (1.73 / 4) x 225 = 97.3125 cm².
Minor segment = Minor sector – Triangle = 117.75 – 97.3125 = 20.4375 cm² (or 20.44 cm²).
Major segment = Circle area – Minor segment = 706.5 – 20.4375 = 686.0625 cm² (or 686.06 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Given radius r = 10 cm and π = 3.14. Total area of the circle = π x r² = 3.14 x 10² = 314 cm². (i) For the minor sector, angle θ = 90°: Area = (90 / 360) x π x r² = (1/4) x 314 = 78.5 cm². (ii) For the major sector, angle θ = 360° - 90° = 270°: Area = (270 / 360) x π x r² = (3/4) x 314 = 235.5 cm² (Read more
Given radius r = 10 cm and π = 3.14.
Total area of the circle = π x r² = 3.14 x 10² = 314 cm².
(i) For the minor sector, angle θ = 90°:
Area = (90 / 360) x π x r² = (1/4) x 314 = 78.5 cm².
(ii) For the major sector, angle θ = 360° – 90° = 270°:
Area = (270 / 360) x π x r² = (3/4) x 314 = 235.5 cm²
(or Total Area – Minor Sector Area = 314 – 78.5 = 235.5 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Given circumference of the circle = 44 cm. Circumference = 2 x π x r = 44 2 x (22/7) x r = 44 (44/7) x r = 44, so r = 7 cm. A quadrant is one-fourth of a circle (central angle θ = 90°). Area of a quadrant = (1/4) x π x r² = (1/4) x (22/7) x 7 x 7 = (1/4) x 154 = 77 / 2 = 38.5 cm². Hence, the area ofRead more
Given circumference of the circle = 44 cm.
Circumference = 2 x π x r = 44
2 x (22/7) x r = 44
(44/7) x r = 44, so r = 7 cm.
A quadrant is one-fourth of a circle (central angle θ = 90°).
Area of a quadrant = (1/4) x π x r²
= (1/4) x (22/7) x 7 x 7
= (1/4) x 154 = 77 / 2 = 38.5 cm².
Hence, the area of the quadrant is 38.5 cm².
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Length of the minute hand acts as the radius r = 7 cm. The minute hand completes a full rotation of 360° in 60 minutes. Angle swept in 1 minute = 360° / 60 = 6°. Angle swept in 10 minutes: θ = 10 x 6° = 60°. The swept region is a circular sector with r = 7 cm and θ = 60°: Area = (θ / 360) x π x r² ARead more
Length of the minute hand acts as the radius r = 7 cm.
The minute hand completes a full rotation of 360° in 60 minutes.
Angle swept in 1 minute = 360° / 60 = 6°.
Angle swept in 10 minutes:
θ = 10 x 6° = 60°.
The swept region is a circular sector with r = 7 cm and θ = 60°:
Area = (θ / 360) x π x r²
Area = (60 / 360) x (22/7) x 7 x 7 = (1/6) x 154 = 77 / 3 cm² = 25.67 cm².
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.
Given blade length r = 28 cm, angle of sweep θ = 120° and number of wipers = 2. Area swept by one wiper blade: Area = (θ / 360) x π x r² Area = (120 / 360) x (22/7) x 28 x 28 Area = (1/3) x 22 x 4 x 28 = 2464 / 3 cm². Total area cleaned by both wipers: Total Area = 2 x (2464 / 3) = 4928 / 3 cm². InRead more
Given blade length r = 28 cm, angle of sweep θ = 120° and number of wipers = 2.
Area swept by one wiper blade:
Area = (θ / 360) x π x r²
Area = (120 / 360) x (22/7) x 28 x 28
Area = (1/3) x 22 x 4 x 28 = 2464 / 3 cm².
Total area cleaned by both wipers:
Total Area = 2 x (2464 / 3) = 4928 / 3 cm².
In decimal form, 4928 / 3 is approximately 1642.67 cm².
Thus, total area cleaned is 4928/3 cm² (or 1642.67 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessA chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and √3 ≈ 1.73.)
Given r = 15 cm, θ = 60°, π = 3.14 and √3 = 1.73. Total circle area = π x r² = 3.14 x 15² = 3.14 x 225 = 706.5 cm². Area of minor sector = (60 / 360) x 3.14 x 225 = (1/6) x 706.5 = 117.75 cm². The triangle formed is equilateral (angle 60°): Area of triangle = (√3 / 4) x r² = (1.73 / 4) x 225 = 97.31Read more
Given r = 15 cm, θ = 60°, π = 3.14 and √3 = 1.73.
Total circle area = π x r² = 3.14 x 15² = 3.14 x 225 = 706.5 cm².
Area of minor sector = (60 / 360) x 3.14 x 225 = (1/6) x 706.5 = 117.75 cm².
The triangle formed is equilateral (angle 60°):
Area of triangle = (√3 / 4) x r² = (1.73 / 4) x 225 = 97.3125 cm².
Minor segment = Minor sector – Triangle = 117.75 – 97.3125 = 20.4375 cm² (or 20.44 cm²).
Major segment = Circle area – Minor segment = 706.5 – 20.4375 = 686.0625 cm² (or 686.06 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessA chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector (that subtends 90° at the centre) and (ii) major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)
Given radius r = 10 cm and π = 3.14. Total area of the circle = π x r² = 3.14 x 10² = 314 cm². (i) For the minor sector, angle θ = 90°: Area = (90 / 360) x π x r² = (1/4) x 314 = 78.5 cm². (ii) For the major sector, angle θ = 360° - 90° = 270°: Area = (270 / 360) x π x r² = (3/4) x 314 = 235.5 cm² (Read more
Given radius r = 10 cm and π = 3.14.
Total area of the circle = π x r² = 3.14 x 10² = 314 cm².
(i) For the minor sector, angle θ = 90°:
Area = (90 / 360) x π x r² = (1/4) x 314 = 78.5 cm².
(ii) For the major sector, angle θ = 360° – 90° = 270°:
Area = (270 / 360) x π x r² = (3/4) x 314 = 235.5 cm²
(or Total Area – Minor Sector Area = 314 – 78.5 = 235.5 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessFind the area of a quadrant of a circle whose circumference is 44 cm.
Given circumference of the circle = 44 cm. Circumference = 2 x π x r = 44 2 x (22/7) x r = 44 (44/7) x r = 44, so r = 7 cm. A quadrant is one-fourth of a circle (central angle θ = 90°). Area of a quadrant = (1/4) x π x r² = (1/4) x (22/7) x 7 x 7 = (1/4) x 154 = 77 / 2 = 38.5 cm². Hence, the area ofRead more
Given circumference of the circle = 44 cm.
Circumference = 2 x π x r = 44
2 x (22/7) x r = 44
(44/7) x r = 44, so r = 7 cm.
A quadrant is one-fourth of a circle (central angle θ = 90°).
Area of a quadrant = (1/4) x π x r²
= (1/4) x (22/7) x 7 x 7
= (1/4) x 154 = 77 / 2 = 38.5 cm².
Hence, the area of the quadrant is 38.5 cm².
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessThe length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Length of the minute hand acts as the radius r = 7 cm. The minute hand completes a full rotation of 360° in 60 minutes. Angle swept in 1 minute = 360° / 60 = 6°. Angle swept in 10 minutes: θ = 10 x 6° = 60°. The swept region is a circular sector with r = 7 cm and θ = 60°: Area = (θ / 360) x π x r² ARead more
Length of the minute hand acts as the radius r = 7 cm.
The minute hand completes a full rotation of 360° in 60 minutes.
Angle swept in 1 minute = 360° / 60 = 6°.
Angle swept in 10 minutes:
θ = 10 x 6° = 60°.
The swept region is a circular sector with r = 7 cm and θ = 60°:
Area = (θ / 360) x π x r²
Area = (60 / 360) x (22/7) x 7 x 7 = (1/6) x 154 = 77 / 3 cm² = 25.67 cm².
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See less