Let a chord through A meet the circle at B and C. The perpendicular distance of this chord from O determines its length. Since A lies inside the circle, the greatest possible distance of a chord through A from O is OA itself. This occurs when the chord is perpendicular to OA. A chord farther from thRead more
Let a chord through A meet the circle at B and C. The perpendicular distance of this chord from O determines its length. Since A lies inside the circle, the greatest possible distance of a chord through A from O is OA itself. This occurs when the chord is perpendicular to OA. A chord farther from the centre is shorter. Therefore, among all chords passing through A, the shortest chord is the one perpendicular to OA. 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):
Let the radius of the circle be r. The diameter has length 2r and passes through the centre. For any other chord, the perpendicular from the centre bisects the chord. If its distance from the centre is d, then half the chord is less than r whenever d is greater than zero. Thus, its complete length iRead more
Let the radius of the circle be r. The diameter has length 2r and passes through the centre. For any other chord, the perpendicular from the centre bisects the chord. If its distance from the centre is d, then half the chord is less than r whenever d is greater than zero. Thus, its complete length is less than 2r. Therefore, the diameter is the longest chord, and no chord can be longer than it.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Let ABCD be a cyclic quadrilateral and let CD be extended to E. Then ∠CDE and ∠CDA form a linear pair, so their sum is 180°. Also, opposite angles of a cyclic quadrilateral are supplementary. Therefore, ∠ABC + ∠CDA = 180° and ∠CDE + ∠CDA = 180°. Hence, ∠CDE = ∠ABC. Thus, the exterior angle of a cyclRead more
Let ABCD be a cyclic quadrilateral and let CD be extended to E. Then ∠CDE and ∠CDA form a linear pair, so their sum is 180°. Also, opposite angles of a cyclic quadrilateral are supplementary.
Therefore,
∠ABC + ∠CDA = 180°
and
∠CDE + ∠CDA = 180°.
Hence,
∠CDE = ∠ABC.
Thus, the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
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):
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):
Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
Let a chord through A meet the circle at B and C. The perpendicular distance of this chord from O determines its length. Since A lies inside the circle, the greatest possible distance of a chord through A from O is OA itself. This occurs when the chord is perpendicular to OA. A chord farther from thRead more
Let a chord through A meet the circle at B and C. The perpendicular distance of this chord from O determines its length. Since A lies inside the circle, the greatest possible distance of a chord through A from O is OA itself. This occurs when the chord is perpendicular to OA. A chord farther from the centre is shorter. Therefore, among all chords passing through A, the shortest chord is the one perpendicular to OA. 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 lessThere is no chord of a circle that is longer than its diameter. How do you justify this statement?
Let the radius of the circle be r. The diameter has length 2r and passes through the centre. For any other chord, the perpendicular from the centre bisects the chord. If its distance from the centre is d, then half the chord is less than r whenever d is greater than zero. Thus, its complete length iRead more
Let the radius of the circle be r. The diameter has length 2r and passes through the centre. For any other chord, the perpendicular from the centre bisects the chord. If its distance from the centre is d, then half the chord is less than r whenever d is greater than zero. Thus, its complete length is less than 2r. Therefore, the diameter is the longest chord, and no chord can be longer than it.
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 lessLet ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
Let ABCD be a cyclic quadrilateral and let CD be extended to E. Then ∠CDE and ∠CDA form a linear pair, so their sum is 180°. Also, opposite angles of a cyclic quadrilateral are supplementary. Therefore, ∠ABC + ∠CDA = 180° and ∠CDE + ∠CDA = 180°. Hence, ∠CDE = ∠ABC. Thus, the exterior angle of a cyclRead more
Let ABCD be a cyclic quadrilateral and let CD be extended to E. Then ∠CDE and ∠CDA form a linear pair, so their sum is 180°. Also, opposite angles of a cyclic quadrilateral are supplementary.
Therefore,
∠ABC + ∠CDA = 180°
and
∠CDE + ∠CDA = 180°.
Hence,
∠CDE = ∠ABC.
Thus, the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.
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 lessIn a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
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/
See lessLet A and B be two points on a circle with centre O. (i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
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/
See less