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):
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):
Consider two equal chords in the same circle. Draw perpendiculars from the centre to each chord. Since the radii are equal and half of each equal chord is also equal, the Baudhāyana–Pythagoras Theorem shows that the perpendicular distances from the centre must be equal. Therefore, equal chords are aRead more
Consider two equal chords in the same circle. Draw perpendiculars from the centre to each chord. Since the radii are equal and half of each equal chord is also equal, the Baudhāyana–Pythagoras Theorem shows that the perpendicular distances from the centre must be equal. Therefore, equal chords are always equidistant from the centre of the circle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Let AB be a chord and O be the centre of the circle. Join OA and OB. Since OA and OB are radii of the same circle, they are equal in length. A triangle having two equal sides is an isosceles triangle. Therefore, the triangle formed by the chord and the centre of the circle is an isosceles triangle.Read more
Let AB be a chord and O be the centre of the circle. Join OA and OB. Since OA and OB are radii of the same circle, they are equal in length. A triangle having two equal sides is an isosceles triangle. Therefore, the triangle formed by the chord and the centre of the circle is an isosceles triangle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Consider two triangles formed by equal chords and the centre of the same circle. Each triangle has two sides equal to the radii of the circle, and the given base lengths are also equal. Thus, all three corresponding sides of the triangles are equal. Therefore, the triangles are congruent by the SSSRead more
Consider two triangles formed by equal chords and the centre of the same circle. Each triangle has two sides equal to the radii of the circle, and the given base lengths are also equal. Thus, all three corresponding sides of the triangles are equal. Therefore, the triangles are congruent by the SSS Congruence Criterion.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Explain 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 lessUse the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.
Consider two equal chords in the same circle. Draw perpendiculars from the centre to each chord. Since the radii are equal and half of each equal chord is also equal, the Baudhāyana–Pythagoras Theorem shows that the perpendicular distances from the centre must be equal. Therefore, equal chords are aRead more
Consider two equal chords in the same circle. Draw perpendiculars from the centre to each chord. Since the radii are equal and half of each equal chord is also equal, the Baudhāyana–Pythagoras Theorem shows that the perpendicular distances from the centre must be equal. Therefore, equal chords are always equidistant from the centre of the 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 lessShow that the triangle formed by a chord and the centre of the circle is isosceles.
Let AB be a chord and O be the centre of the circle. Join OA and OB. Since OA and OB are radii of the same circle, they are equal in length. A triangle having two equal sides is an isosceles triangle. Therefore, the triangle formed by the chord and the centre of the circle is an isosceles triangle.Read more
Let AB be a chord and O be the centre of the circle. Join OA and OB. Since OA and OB are radii of the same circle, they are equal in length. A triangle having two equal sides is an isosceles triangle. Therefore, the triangle formed by the chord and the centre of the circle is an isosceles triangle.
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 lessShow that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Consider two triangles formed by equal chords and the centre of the same circle. Each triangle has two sides equal to the radii of the circle, and the given base lengths are also equal. Thus, all three corresponding sides of the triangles are equal. Therefore, the triangles are congruent by the SSSRead more
Consider two triangles formed by equal chords and the centre of the same circle. Each triangle has two sides equal to the radii of the circle, and the given base lengths are also equal. Thus, all three corresponding sides of the triangles are equal. Therefore, the triangles are congruent by the SSS Congruence Criterion.
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