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Ex 5.6, 3
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, 3 (Method 1) Find 𝑥 in Fig. 5.26. Here, we will use the property Opposite angles of Cyclic Quadrilateral are supplementary Thus, ∠ D + ∠ B = 180° 100° + x = 180° x = 180° – 100° x = 80° Ex 5.6, 3 (Method 2) Find 𝑥 in Fig. 5.26. Let’s modify the figure We mark center O And join AO & OC Now, we will apply the Theorem Angle subtended by chord at center is double the angle subtended at any other point For Arc ADC Thus, Reflex angle ∠ AOC = 2 × 100° = 200° For Arc ABC Thus, ∠ AOC = 2 × x = 2x We know that Angle around the center = 360° Thus, we can write ∠ AOC + Reflex angle ∠ AOC = 360° 200° + 2x = 360° 2x = 360° – 200° 2x = 160° x = (160° )/2 x = 80°
