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Consider 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 lessShow that rectangle is the only parallelogram that can be inscribed in a circle.
Let ABCD be a parallelogram inscribed in a circle. Since it is cyclic, its opposite angles are supplementary. Also, opposite angles of a parallelogram are equal. Therefore, ∠A + ∠C = 180° and ∠A = ∠C. Thus, 2∠A = 180°, so ∠A = 90°. Similarly, all four angles are 90°. Hence, the parallelogram is a reRead more
Let ABCD be a parallelogram inscribed in a circle. Since it is cyclic, its opposite angles are supplementary.
Also, opposite angles of a parallelogram are equal.
Therefore,
∠A + ∠C = 180°
and ∠A = ∠C.
Thus, 2∠A = 180°, so ∠A = 90°. Similarly, all four angles are 90°.
Hence, the parallelogram is a rectangle. Therefore, a rectangle is the only parallelogram that can be inscribed in a 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 lessThe distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
The chord is 16 cm long, so its half is 8 cm. The perpendicular distance from the centre to the chord is 6 cm. The perpendicular from the centre to a chord bisects the chord. Thus, we get a right triangle with sides 8 cm and 6 cm. Radius² = 8² + 6² = 64 + 36 = 100 Therefore, radius = 10 cm. Hence, tRead more
The chord is 16 cm long, so its half is 8 cm. The perpendicular distance from the centre to the chord is 6 cm. The perpendicular from the centre to a chord bisects the chord. Thus, we get a right triangle with sides 8 cm and 6 cm.
Radius² = 8² + 6²
= 64 + 36
= 100
Therefore, radius = 10 cm.
Hence, the radius of the circle is 10 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 cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
For a cyclic quadrilateral, we use Brahmagupta's formula: Area = √((s - a)(s - b)(s - c)(s - d)) Here, the sides are 5, 5, 12 and 12 units. s = (5 + 5 + 12 + 12) / 2 = 17 Therefore, Area = √(12 × 12 × 5 × 5) = √3600 = 60 square units. Hence, the area of the cyclic quadrilateral is 60 square units.Read more
For a cyclic quadrilateral, we use Brahmagupta’s formula:
Area = √((s – a)(s – b)(s – c)(s – d))
Here, the sides are 5, 5, 12 and 12 units.
s = (5 + 5 + 12 + 12) / 2 = 17
Therefore,
Area = √(12 × 12 × 5 × 5)
= √3600
= 60 square units.
Hence, the area of the cyclic quadrilateral is 60 square units.
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