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):
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):
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):
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):
Draw perpendiculars from the centre to both chords. The radii are equal, and the perpendicular distances are also equal. Using the Baudhāyana–Pythagoras Theorem, the half-lengths of both chords are equal. Doubling these equal half-chords gives equal full chord lengths. Therefore, the two chords areRead more
Draw perpendiculars from the centre to both chords. The radii are equal, and the perpendicular distances are also equal. Using the Baudhāyana–Pythagoras Theorem, the half-lengths of both chords are equal. Doubling these equal half-chords gives equal full chord lengths. Therefore, the two chords are equal, proving that AB = GF.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Prove that the perpendicular bisector of a chord passes through the centre of the circle.
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/
See lessThe diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.
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/
See lessABCD 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?
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/
See lessQuadrilateral 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.
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/
See lessSolve the previous question using the Baudhāyana–Pythagoras theorem.
Draw perpendiculars from the centre to both chords. The radii are equal, and the perpendicular distances are also equal. Using the Baudhāyana–Pythagoras Theorem, the half-lengths of both chords are equal. Doubling these equal half-chords gives equal full chord lengths. Therefore, the two chords areRead more
Draw perpendiculars from the centre to both chords. The radii are equal, and the perpendicular distances are also equal. Using the Baudhāyana–Pythagoras Theorem, the half-lengths of both chords are equal. Doubling these equal half-chords gives equal full chord lengths. Therefore, the two chords are equal, proving that AB = GF.
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