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A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.
A regular hexagon divides the circle into six equal sectors. Therefore, the central angle corresponding to each side is 60°. The triangle formed by two radii and one side has two sides equal to r and included angle 60°, so it is equilateral. Hence, each side of the hexagon is r. The perpendicular diRead more
A regular hexagon divides the circle into six equal sectors. Therefore, the central angle corresponding to each side is 60°. The triangle formed by two radii and one side has two sides equal to r and included angle 60°, so it is equilateral. Hence, each side of the hexagon is r.
The perpendicular distance from the centre to a side is the altitude of this equilateral triangle.
Therefore, distance = (√3/2)r.
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 lessTwo parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
The half-lengths of the chords are 5 cm and 12 cm. Since the longer chord is closer to the centre, let its distance from the centre be x. Then the distance of the 10 cm chord is x + 7. Using right triangles: r² = x² + 12² r² = (x + 7)² + 5² Therefore, x² + 144 = x² + 14x + 49 + 25 14x = 70 x = 5 ThuRead more
The half-lengths of the chords are 5 cm and 12 cm. Since the longer chord is closer to the centre, let its distance from the centre be x. Then the distance of the 10 cm chord is x + 7.
Using right triangles:
r² = x² + 12²
r² = (x + 7)² + 5²
Therefore,
x² + 144 = x² + 14x + 49 + 25
14x = 70
x = 5
Thus, r² = 25 + 144 = 169, so r = 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 lessIn a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.
Let AB and AC be congruent chords of a circle with centre O. Equal chords are at equal distances from the centre. Therefore, the perpendicular distances from O to AB and AC are equal. Alternatively, in triangles AOB and AOC, OA is common, OB = OC because they are radii andAB = AC because the chordsRead more
Let AB and AC be congruent chords of a circle with centre O. Equal chords are at equal distances from the centre. Therefore, the perpendicular distances from O to AB and AC are equal. Alternatively, in triangles AOB and AOC, OA is common, OB = OC because they are radii andAB = AC because the chords are congruent. Thus, the triangles are congruent. Hence, ∠BAO = ∠OAC. Therefore, AO bisects ∠BAC.
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 all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Consider chords of a fixed length in a circle with centre O and radius r. The perpendicular from O to each chord bisects it. Since all chords have the same length, each half-chord has the same length. Hence, by the right triangle relation, the distance of every chord's midpoint from O is the same. TRead more
Consider chords of a fixed length in a circle with centre O and radius r. The perpendicular from O to each chord bisects it. Since all chords have the same length, each half-chord has the same length. Hence, by the right triangle relation, the distance of every chord’s midpoint from O is the same. Therefore, all the midpoints lie on a circle having O as its centre. Thus, they form a concentric 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 if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
Let ABCD be a rectangle inscribed in a circle. Its diagonals AC and BD are equal and bisect each other. Since AC and BD are equal chords passing through the centre, they are diameters of the circle. The diagonals of the rectangle intersect at their common midpoint. The centre of a circle is the midpRead more
Let ABCD be a rectangle inscribed in a circle. Its diagonals AC and BD are equal and bisect each other. Since AC and BD are equal chords passing through the centre, they are diameters of the circle. The diagonals of the rectangle intersect at their common midpoint. The centre of a circle is the midpoint of every diameter. Therefore, the point where the diagonals of the rectangle intersect must be 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 less