Question 5 - 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 5 Prove that the perpendicular bisector of a chord passes through the centre of the circle. Given: AB is a chord And PM is perpendicular bisector of AB ∴ PM ⊥ AB & M is mid-point of AB, i.e. AM = BM To Prove: PM passes through center O Proof: Join OA & OB We prove ∆OAM & ∆OBM congruent In ∆OAM & ∆OBM OA = OB OM = OM AM = BM ∴ ∆OAM ≅ ∆OBM Thus, by Corresponding parts of Congruent Triangles (CPCT) ∠ OMA = ∠ OMB Since AB is a line, by linear pair ∠ OMA + ∠ OMB = 180° From (1): ∠ OMA = ∠ OMB ∠ OMA + ∠ OMA = 180° 2 × ∠ OMA = 180° ∠ OMA = (180° )/2 ∠ OMA = 90° This proves that OM is perpendicular bisector of AB But, there can only be one unique perpendicular line passing through the midpoint M, the perpendicular bisector of chord AB must pass through the center O Hence proved
