What's your question?
  1. The best way is to use the perpendicular bisectors of the sides. Draw the perpendicular bisectors of any two sides of the cyclic quadrilateral. Their point of intersection is the centre of the circumcircle because the centre is equidistant from the endpoints of each chord. After locating this point,Read more

    The best way is to use the perpendicular bisectors of the sides. Draw the perpendicular bisectors of any two sides of the cyclic quadrilateral. Their point of intersection is the centre of the circumcircle because the centre is equidistant from the endpoints of each chord. After locating this point, check its position with respect to the quadrilateral. If it lies within the quadrilateral, the centre is inside; otherwise, it is outside.

     

    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/

    See less
    • 262
  2. Let equal chords AB and CD intersect at P. Since the chords are equal, their corresponding parts have equal sums. Using the intersecting-chords theorem and equality of the chords, we get AP = CP and BP = DP. Hence proved. 75 Words Answer: Let equal chords AB and CD intersect at P. Since AB = CD, weRead more

    Let equal chords AB and CD intersect at P. Since the chords are equal, their corresponding parts have equal sums. Using the intersecting-chords theorem and equality of the chords, we get AP = CP and BP = DP. Hence proved.

    75 Words Answer:
    Let equal chords AB and CD intersect at P. Since AB = CD, we have

    AP + PB = CP + PD.

    Also, by the intersecting chords theorem,

    AP × PB = CP × PD.

    Thus, the two pairs of segments have equal sum and equal product. Therefore, the corresponding segments are equal.

    Hence,

    AP = CP and PB = PD.

    Therefore, if two equal chords intersect, their corresponding line segments are equal. 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/

     

    See less
    • 246
  3. Construct a right triangle ABC with AB = 3 cm and BC = 3 cm. Then AC = 3√2 cm. Draw the perpendicular bisector of AC. Since the triangle is right-angled, the circumcentre is the midpoint of the hypotenuse AB. The perpendicular distance from this centre to AC is 3 cm, while AC is the required chord oRead more

    Construct a right triangle ABC with AB = 3 cm and BC = 3 cm. Then AC = 3√2 cm. Draw the perpendicular bisector of AC. Since the triangle is right-angled, the circumcentre is the midpoint of the hypotenuse AB. The perpendicular distance from this centre to AC is 3 cm, while AC is the required chord of length 6 cm. Thus, the circumcircle of a suitable right triangle gives the required circle.

     

    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/

     

    See less
    • 246
  4. According to the theorem on the angle subtended by an arc, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circle. Given central angle = 70° Therefore, angle at a point on the circle = 70° / 2 = 35°. Hence, the measure of the angle subtendRead more

    According to the theorem on the angle subtended by an arc, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circle.

    Given central angle = 70°

    Therefore, angle at a point on the circle = 70° / 2 = 35°.

    Hence, the measure of the angle subtended by the arc at a point on the circle is 35°.

    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/

    See less
    • 271
  5. The diameter of the circle is 26 cm, so its radius is 13 cm. The perpendicular from the centre to a chord bisects the chord. Therefore, half of the chord is 12 cm. Using the right triangle: Distance² = 13² - 12² = 169 - 144 = 25 Therefore, distance = 5 cm. Hence, the distance from the centre of theRead more

    The diameter of the circle is 26 cm, so its radius is 13 cm. The perpendicular from the centre to a chord bisects the chord. Therefore, half of the chord is 12 cm.

    Using the right triangle:

    Distance² = 13² – 12²
    = 169 – 144
    = 25

    Therefore, distance = 5 cm.

    Hence, the distance from the centre of the circle to the chord is 5 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/

    See less
    • 264