In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give: p + v = 90° and similarly, q + u = 90°. Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the oppRead more
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give:
p + v = 90°
and similarly,
q + u = 90°.
Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the opposite angles together give 180°. Thus, the sum of the opposite angles of a cyclic quadrilateral is 180°.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
There appears to be a problem in the question as shown in the screenshot. If the two chords CC' and DD' are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M' of these chordsRead more
There appears to be a problem in the question as shown in the screenshot. If the two chords CC’ and DD’ are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M’ of these chords lie on AB. Therefore, the segment MM’ joining the midpoints lies along AB. Hence, MM’ is parallel to AB, not perpendicular to AB. So the printed statement cannot be proved as written.
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 endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° - 2a and 180° - 2b. SRead more
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° – 2a and 180° – 2b. Since BOC is a straight angle, a + b = 90°. Hence, ∠BAC = 90°.
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 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):
How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give: p + v = 90° and similarly, q + u = 90°. Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the oppRead more
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give:
p + v = 90°
and similarly,
q + u = 90°.
Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the opposite angles together give 180°. Thus, the sum of the opposite angles of a cyclic quadrilateral is 180°.
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, two chords CC’ and DD’ are drawn perpendicular to a diameter AB. Prove that the segment MM’ joining the midpoints of the chords CD and C’D’ is perpendicular to AB.
There appears to be a problem in the question as shown in the screenshot. If the two chords CC' and DD' are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M' of these chordsRead more
There appears to be a problem in the question as shown in the screenshot. If the two chords CC’ and DD’ are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M’ of these chords lie on AB. Therefore, the segment MM’ joining the midpoints lies along AB. Hence, MM’ is parallel to AB, not perpendicular to AB. So the printed statement cannot be proved as written.
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 lessHow would you use the following figure to justify the statement that the angle in a semicircle is 90°?
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° - 2a and 180° - 2b. SRead more
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° – 2a and 180° – 2b. Since BOC is a straight angle, a + b = 90°. Hence, ∠BAC = 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 lessLet 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 less