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  1. Asked: In:

    A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?

    Piyush365

    The radius of the circle is 15 cm and the perpendicular distance from the centre to the chord is 9 cm. The perpendicular from the centre to a chord bisects the chord. Thus, half of the chord forms a right triangle. Half chord = √(15² - 9²) = √(225 - 81) = √144 = 12 cm Therefore, length of the chordRead more

    The radius of the circle is 15 cm and the perpendicular distance from the centre to the chord is 9 cm. The perpendicular from the centre to a chord bisects the chord. Thus, half of the chord forms a right triangle.

    Half chord = √(15² – 9²)
    = √(225 – 81)
    = √144
    = 12 cm

    Therefore, length of the chord = 2 × 12 = 24 cm.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/

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  2. Asked: In:

    Prove that the perpendicular bisector of a chord passes through the centre of the circle.

    Piyush365

    Let AB be a chord of a circle with centre O. Join OA and OB. Since OA and OB are radii of the same circle, OA = OB. Thus, O is equidistant from the two endpoints A and B of the chord. The locus of points equidistant from A and B is the perpendicular bisector of AB. Therefore, the perpendicular bisecRead more

    Let AB be a chord of a circle with centre O. Join OA and OB. Since OA and OB are radii of the same circle, OA = OB. Thus, O is equidistant from the two endpoints A and B of the chord. The locus of points equidistant from A and B is the perpendicular bisector of AB. Therefore, the perpendicular bisector of chord AB passes through the centre O of the circle. Hence proved.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/

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  3. Asked: In:

    The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.

    Piyush365

    AB is the diameter of the circle and C is a point on its circumference. According to the theorem, the angle subtended by a diameter at any point on the circle is 90°. Here, the diameter AB subtends ∠ACB at point C. Therefore, ∠ACB = 90°. Hence, the measure of ∠ACB is 90°.   For more NCERT SolutRead more

    AB is the diameter of the circle and C is a point on its circumference. According to the theorem, the angle subtended by a diameter at any point on the circle is 90°. Here, the diameter AB subtends ∠ACB at point C. Therefore,

    ∠ACB = 90°.

    Hence, the measure of ∠ACB is 90°.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/

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    • 257
  5. Asked: In:

    ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?

    Piyush365

    ABCD is a cyclic quadrilateral. According to the property of a cyclic quadrilateral, the sum of each pair of opposite angles is 180°. For angles A and C: ∠C = 180° - 75° = 105° For angles B and D: ∠D = 180° - 110° = 70° Therefore, ∠C = 105° and ∠D = 70°.   For more NCERT Solutions for Class 9 MRead more

    ABCD is a cyclic quadrilateral. According to the property of a cyclic quadrilateral, the sum of each pair of opposite angles is 180°.

    For angles A and C:

    ∠C = 180° – 75° = 105°

    For angles B and D:

    ∠D = 180° – 110° = 70°

    Therefore, ∠C = 105° and ∠D = 70°.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/

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  6. Asked: In:

    Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x – 20)°, find the value of x and the measures of ∠P and ∠R.

    Piyush365

    PQRS is a cyclic quadrilateral, so its opposite angles P and R are supplementary. Therefore, (2x + 10) + (3x - 20) = 180 5x - 10 = 180 5x = 190 x = 38 Now, ∠P = 2(38) + 10 = 86° ∠R = 3(38) - 20 = 94° Hence, x = 38, ∠P = 86° and ∠R = 94°.   For more NCERT Solutions for Class 9 Maths Ganita ManjaRead more

    PQRS is a cyclic quadrilateral, so its opposite angles P and R are supplementary.

    Therefore,

    (2x + 10) + (3x – 20) = 180

    5x – 10 = 180

    5x = 190

    x = 38

    Now,

    ∠P = 2(38) + 10 = 86°

    ∠R = 3(38) – 20 = 94°

    Hence, x = 38, ∠P = 86° and ∠R = 94°.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/

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