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  1. Asked: In:

    Find the area of a quadrant of a circle whose circumference is 44 cm.

    Virat

    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/

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  2. Asked: In:

    The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

    Virat

    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/

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  3. Asked: In:

    Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

    Virat

    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/

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  5. Asked: In:

    In 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).

    Virat

    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/

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  6. Asked: In:

    Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB and ΔPCD) and the green region (ΔPBC and ΔPDA)?

    Virat

    Let the side length of square ABCD be s. Draw a line through point P perpendicular to AB and CD. Let the perpendicular distance from P to AB be h1 and to CD be h2. Then h1 + h2 = s. Area of red region = Area(ΔPAB) + Area(ΔPCD) = (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s². SRead more

    Let the side length of square ABCD be s.

    Draw a line through point P perpendicular to AB and CD.

    Let the perpendicular distance from P to AB be h1 and to CD be h2.

    Then h1 + h2 = s.

    Area of red region = Area(ΔPAB) + Area(ΔPCD)

    = (1/2) x s x h1 + (1/2) x s x h2 = (1/2) x s x (h1 + h2) = (1/2) x s².

    Since total area of the square is s², the green region also equals s² – s²/2 = (1/2) x s².

    Thus, the ratio of areas is 1:1.

     

    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/

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