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  1. 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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  2. 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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  3. 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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    • 236
  4. OA and OB are radii of the circle, so OA = OB = 12 cm. Therefore, triangle AOB is isosceles. Since ∠AOB = 60°, the remaining two angles together are 120° and each is 60°. Hence, triangle AOB is an equilateral triangle. All its sides are equal. Therefore, AB = OA = OB = 12 cm. So, the length of chordRead more

    OA and OB are radii of the circle, so OA = OB = 12 cm. Therefore, triangle AOB is isosceles. Since ∠AOB = 60°, the remaining two angles together are 120° and each is 60°. Hence, triangle AOB is an equilateral triangle. All its sides are equal. Therefore, AB = OA = OB = 12 cm. So, the length of chord AB is 12 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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  5. No. There cannot be such points X and Y on the same side of AB for which ∠AXB is different from ∠AYB. According to Theorem 9, an arc subtends the same angle at every point on the remaining part of the circle. Therefore, when X and Y lie on the same side of AB, both angles subtend the same chord AB aRead more

    No. There cannot be such points X and Y on the same side of AB for which ∠AXB is different from ∠AYB. According to Theorem 9, an arc subtends the same angle at every point on the remaining part of the circle. Therefore, when X and Y lie on the same side of AB, both angles subtend the same chord AB and are equal.

     

    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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