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In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
Let the perpendicular distance from the centre to the chord be 5 cm and the radius be 13 cm. The perpendicular from the centre to a chord bisects the chord. Thus, we get a right triangle with hypotenuse 13 cm and one side 5 cm. Half chord = √(13² - 5²) = √144 = 12 cm Therefore, length of the chord =Read more
Let the perpendicular distance from the centre to the chord be 5 cm and the radius be 13 cm. The perpendicular from the centre to a chord bisects the chord. Thus, we get a right triangle with hypotenuse 13 cm and one side 5 cm.
Half chord = √(13² – 5²) = √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 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 lessFind the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
Draw a perpendicular from the centre of the circle to the chord. It bisects the chord into two equal parts. Applying the Baudhāyana–Pythagoras Theorem to the right triangle, the half-chord is obtained as √13 cm. Doubling this value gives the full chord length. Therefore, the required chord length isRead more
Draw a perpendicular from the centre of the circle to the chord. It bisects the chord into two equal parts. Applying the Baudhāyana–Pythagoras Theorem to the right triangle, the half-chord is obtained as √13 cm. Doubling this value gives the full chord length. Therefore, the required chord length is 2√13 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 lessExplain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r2-d2 ).
A perpendicular drawn from the centre to a chord always bisects the chord into two equal parts. Each half forms a right triangle with the radius as the hypotenuse and the perpendicular distance as one side. By applying the Baudhāyana–Pythagoras Theorem, the half-chord equals √(r² − d²). Therefore, tRead more
A perpendicular drawn from the centre to a chord always bisects the chord into two equal parts. Each half forms a right triangle with the radius as the hypotenuse and the perpendicular distance as one side. By applying the Baudhāyana–Pythagoras Theorem, the half-chord equals √(r² − d²). Therefore, the complete chord length is 2√(r² − d²).
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 lessConsider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
Since CE and CH are perpendicular distances from the centre to the chords and CE = CH, both chords are at the same distance from the centre. According to Theorem 7, chords that are equidistant from the centre of the same circle are equal in length. Hence, the two chords are equal. Therefore, AB = GFRead more
Since CE and CH are perpendicular distances from the centre to the chords and CE = CH, both chords are at the same distance from the centre. According to Theorem 7, chords that are equidistant from the centre of the same circle are equal in length. Hence, the two chords are equal. Therefore, 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