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Ex 5.6, 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
- Exercise Set 5.5
- Angle subtended by an arc
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Exercise Set 5.6
- Concyclicity of Points and Cyclic Quadrilateral
- Theorems (and their Proofs)
- End-of-Chapter Exercises
Transcript
Ex 5.6, 1 In a circle with centre O, the central angle AOB is 60^∘. If the radius of the circle is 12 cm, what is the length of the chord AB? Let’s draw a figure Now, in ∆ OAB OA = OB We know that angles opposite equal sides are equal ∴ ∠ OAB = ∠ OBA Now, in ∆ AOB By Angle Sum Property ∠ AOB + ∠ OAB + ∠ OBA = 180° Given that ∠ AOB = 60°, putting in equation 60° + ∠ OAB + ∠ OBA = 180° ∠ OAB + ∠ OBA = 180° – 60° ∠ OAB + ∠ OBA = 120° Putting ∠ OAB = ∠ OBA from (1) ∠ OAB + ∠ OAB = 120° 2 × ∠ OAB = 120° ∠ OAB = (120° )/2 ∠ OAB = 60° Thus, ∠ OAB = ∠ OBA = 60° So, in ∆ OAB, all angles are 60° And, we know that if all angles are 60°, it is an equilateral triangle Thus, all sides of ∆ OAB are equal ∴ AB = OA = OB = Radius Thus, AB = Radius = 12 cm
