Question 9 - End-of-Chapter Exercises - Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja
Last updated at May 26, 2026 by Teachoo
End-of-Chapter Exercises
Question 1
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Class 9
Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja
- Definitions
- Symmetries of Circle
- Number of Circles
- Exercise Set 5.1
- Think, Draw and Infer (Page 98)
- Angle Subtended by Chords
- Exercise Set 5.2
- Mid-points and Perpendicular Bisectors of Chords
- Exercise Set 5.3
- Distance of Chords from the Centre
- Exercise Set 5.4
- Distance of unequal chords from center
- Exercise Set 5.5
- Angle subtended by an arc
- Exercise Set 5.6
- Concyclicity of Points and Cyclic Quadrilateral
- Theorems (and their Proofs)
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End-of-Chapter Exercises
Transcript
Question 9 The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle. Let’s draw the diagram Here, we have circle with center O Let Radius = r cm Let AB be the chord And, OM be perpendicular distance to AB from O ∴ OM = 6 cm We know that Perpendicular from the center to the chord, bisects the chord So, we can write AM = MB = 𝟏/𝟐AB = 1/2 × 16 = 8 cm We need to find Radius, i.e. OA Joining OA Since ∆ AOM is a right angled triangle By Baudhāyana–Pythagoras theorem 〖𝑂𝐴〗^2=〖𝑂𝑀〗^2+〖𝐴𝑀〗^2 Putting OA = Radius = r, OM = 6 cm, AM = 8 cm 𝒓^𝟐=𝟔^𝟐+𝟖^𝟐 𝑟^2=36+64 𝑟^2=100 𝑟^2=10^2 𝑟=√(10^2 ) 𝒓=𝟏𝟎 cm Thus, radius is 10 cm
