Question 3 - End-of-Chapter Exercises - Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja
Last updated at August 12, 2026 by Teachoo
End-of-Chapter Exercises
Question 1
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Maths 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
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Transcript
Question 3 The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord. Let’s draw the diagram Here, we have circle with center O Since Diamter is 26 cm Radius = 26/2 cm = 13 cm Let AB be the chord And, OM be perpendicular distance to AB from O We know that Perpendicular from the center to the chord, bisects the chord So, we can write So, we can write AM = MB = 𝟏/𝟐AB = 1/2 × 24 = 12 cm We need to find OM Joining OA Since ∆ AOM is a right angled triangle By Baudhāyana–Pythagoras theorem 〖𝑂𝐴〗^2=〖𝑂𝑀〗^2+〖𝐴𝑀〗^2 Putting OA = Radius = 13 cm, AM = 12 cm 〖𝟏𝟑〗^𝟐=𝑶𝑴^𝟐+𝟏𝟐^𝟐 169=𝑂𝑀^2+144 169−144=𝑂𝑀^2 25=𝑂𝑀^2 〖𝑶𝑴〗^𝟐=𝟐𝟓 𝑂𝑀^2=5^2 𝑂𝑀=√(5^2 ) 𝑶𝑴=𝟓 cm Thus, distance from center of circle to the chord is 5 cm
