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Ex 5.5, 1
Last updated at May 26, 2026 by Teachoo
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
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Exercise Set 5.5
- Angle subtended by an arc
- Exercise Set 5.6
- Concyclicity of Points and Cyclic Quadrilateral
- Theorems (and their Proofs)
- End-of-Chapter Exercises
Transcript
Ex 5.5, 1 Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm. Let’s draw the diagram Here, we have circle with center O And, Radius = 7 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 Joining OA Since ∆ AOM is a right angled triangle By Baudhāyana–Pythagoras theorem 〖𝑶𝑨〗^𝟐=〖𝑶𝑴〗^𝟐+〖𝑨𝑴〗^𝟐 Putting AC = Radius = 7 cm, OM = 6 cm 𝒓^𝟐=〖𝑪𝑭〗^𝟐+〖𝑨𝑭〗^𝟐 Since AM = 𝟏/𝟐AB We can write 𝐴𝑀=1/2 𝐴𝐵 √13=1/2 𝐴𝐵 2√13=𝐴𝐵 𝑨𝑩=𝟐√𝟏𝟑 cm Thus, length of the chord is 𝟐√𝟏𝟑 cm
