Question 24 - 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 24 How would you use the following figure to justify the statement that the angle in a semicircle is 90^∘? Let’s label the points at the end as M & N We need to prove ∠ MAN = 90° Since OM, OA, ON are radius, they are equal OM = ON = OA In ∆ OAM Since OM = OA And, we know that Angles opposite to equal sides are equal ∴ ∠ OAM = ∠ OMA = a In ∆ OAN Since ON = OA And, we know that Angles opposite to equal sides are equal ∴ ∠ OAN = ∠ ONA = b Thus, ∠ MAN = ∠ OAM + ∠ OAN = a + b In ∆ MAN By Angle sum property of triangle ∠ AMN + ∠ MAN + ∠ ANM = 180° a + (a + b) + b = 180° 2a + 2b = 180° 2 × (a + b) = 180° a + b = (180°)/2 a + b = 90° Therefore, ∠ MAN = 90° Thus, we proved angle in a semicircle is a right angle Hence proved
