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The 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 lessConsider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?
The best way is to use the perpendicular bisectors of the sides. Draw the perpendicular bisectors of any two sides of the cyclic quadrilateral. Their point of intersection is the centre of the circumcircle because the centre is equidistant from the endpoints of each chord. After locating this point,Read more
The best way is to use the perpendicular bisectors of the sides. Draw the perpendicular bisectors of any two sides of the cyclic quadrilateral. Their point of intersection is the centre of the circumcircle because the centre is equidistant from the endpoints of each chord. After locating this point, check its position with respect to the quadrilateral. If it lies within the quadrilateral, the centre is inside; otherwise, it is outside.
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 lessWhen two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
Let equal chords AB and CD intersect at P. Since the chords are equal, their corresponding parts have equal sums. Using the intersecting-chords theorem and equality of the chords, we get AP = CP and BP = DP. Hence proved. 75 Words Answer: Let equal chords AB and CD intersect at P. Since AB = CD, weRead more
Let equal chords AB and CD intersect at P. Since the chords are equal, their corresponding parts have equal sums. Using the intersecting-chords theorem and equality of the chords, we get AP = CP and BP = DP. Hence proved.
75 Words Answer:
Let equal chords AB and CD intersect at P. Since AB = CD, we have
AP + PB = CP + PD.
Also, by the intersecting chords theorem,
AP × PB = CP × PD.
Thus, the two pairs of segments have equal sum and equal product. Therefore, the corresponding segments are equal.
Hence,
AP = CP and PB = PD.
Therefore, if two equal chords intersect, their corresponding line segments are equal. 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 lessDraw 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 less