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A 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 lessFind the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Given radius r = 7 cm and sector angle θ = 60°. The formula for the area of a sector is: Area = (θ / 360) x π x r² Substitute the given values using π = 22/7: Area = (60 / 360) x (22/7) x 7 x 7 Area = (1/6) x 22 x 7 Area = 154 / 6 = 77 / 3 cm². In decimal form, 77 / 3 is approximately 25.67 cm². TheRead more
Given radius r = 7 cm and sector angle θ = 60°.
The formula for the area of a sector is:
Area = (θ / 360) x π x r²
Substitute the given values using π = 22/7:
Area = (60 / 360) x (22/7) x 7 x 7
Area = (1/6) x 22 x 7
Area = 154 / 6 = 77 / 3 cm².
In decimal form, 77 / 3 is approximately 25.67 cm².
Therefore, the area of the sector is 77/3 cm² (or 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 lessIn ΔABC, D is the midpoint of AB. P is any point on BC and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that Area (ΔBPQ) = 1/2 Area (ΔABC).
Given CQ is parallel to PD. Triangles DPQ and DPC lie on the same base PD and between the same parallel lines PD and CQ. Therefore, their areas are equal: Area(ΔDPQ) = Area(ΔDPC). Now, consider triangle BPQ: Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ) Substitute Area(ΔDPC) for Area(ΔDPQ): Area(ΔBPQ) = AreaRead more
Given CQ is parallel to PD.
Triangles DPQ and DPC lie on the same base PD and between the same parallel lines PD and CQ.
Therefore, their areas are equal:
Area(ΔDPQ) = Area(ΔDPC).
Now, consider triangle BPQ:
Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ)
Substitute Area(ΔDPC) for Area(ΔDPQ):
Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPC) = Area(ΔBDC).
Since D is the midpoint of AB, CD is a median of triangle ABC.
A median divides the triangle into two equal halves:
Area(ΔBDC) = (1/2) x Area(ΔABC).
Hence, Area(ΔBPQ) = (1/2) x Area(ΔABC).
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