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Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre. (Hint: Is it a circumcircle of a suitable triangle?)
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 lessAn arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
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 lessThe diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
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 lessA circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
The radius of the circle is 15 cm and the perpendicular distance from the centre to the chord is 9 cm. The perpendicular from the centre to a chord bisects the chord. Thus, half of the chord forms a right triangle. Half chord = √(15² - 9²) = √(225 - 81) = √144 = 12 cm Therefore, length of the chordRead more
The radius of the circle is 15 cm and the perpendicular distance from the centre to the chord is 9 cm. The perpendicular from the centre to a chord bisects the chord. Thus, half of the chord forms a right triangle.
Half chord = √(15² – 9²)
= √(225 – 81)
= √144
= 12 cm
Therefore, length of the chord = 2 × 12 = 24 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 lessProve 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 less