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  1. Chapter 11 Class 9 Constructions
  2. Serial order wise

Transcript

Construction 11.2 : To construct the perpendicular bisector of a given line segment. Given a line segment AB, we want to construct its perpendicular bisector. Steps of Construction: Taking A as center, and radius more than 1/2 AB, draw arcs on top and bottom of AB 2. Taking B as center, and same radius as before, draw arcs on top and bottom of AB 3. Let the point where arcs intersect on top of AB be P And the point where arcs intersect at the botton be Q Join PQ. 4. Let PQ intersect AB at the Point M. Thus, Line PMQ is the required Perpendicular bisector of AB Justification We have to prove PQ is perpendicular bisector, i.e. AM = BM and ∠ PMA = ∠ PMB = 90° Join A,B to P and A, B to Q In Δ PAQ and Δ PBQ, AP = BP AQ = BQ PQ = PQ ∴ ∆ PAQ ≅ ∆PBQ So, ∠ APM = ∠ BPM Now , In Δ PAM and Δ PBM, AP = BP ∠APM = ∠BPM PM = PM ∴ ∆ PMA ≅ ∆PMB So, AM = BM & ∠ PMA = ∠ PMB Also, ∠PMA + ∠PMB = 180° ∠PMA + ∠PMA = 180° 2∠PMA = 180° ∠ PMA = (180°)/2° ∠ PMA = 90° ∴ ∠ PMB = ∠ PMA = 90° Thus, AM = BM & ∠ PMA = ∠ PMB = 90° ∴ PQ is the perpendicular bisector of AB

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.