Question 4 - 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 4 A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord? Let’s draw the diagram Here, we have circle with center O And, Radius = 15 cm Let AB be the chord And, OM be perpendicular distance to AB from O ∴ OM = 9 cm We know that Perpendicular from the center to the chord, bisects the chord So, we can write AM = MB = 𝟏/𝟐AB Joining OA Since ∆ AOM is a right angled triangle By Baudhāyana–Pythagoras theorem 〖𝑂𝐴〗^2=〖𝑂𝑀〗^2+〖𝐴𝑀〗^2 Putting OA = Radius = 15 cm, OM = 9 cm 〖𝟏𝟓〗^𝟐=𝟗^𝟐+𝑨𝑴^𝟐 225=81+𝐴𝑀^2 225−81=𝐴𝑀^2 144=𝐴𝑀^2 𝑨𝑴^𝟐=𝟏𝟒𝟒 𝐴𝑀^2=12^2 𝐴𝑀=√(12^2 ) 𝑨𝑴=𝟏𝟐 cm Since AM = 𝟏/𝟐AB We can write 𝐴𝑀=1/2 𝐴𝐵 12=1/2 𝐴𝐵 2 × 12=𝐴𝐵 24=𝐴𝐵 𝑨𝑩=𝟐𝟒 cm Thus, length of the chord is 𝟐𝟒 cm
