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  1. Chapter 11 Class 9 Constructions
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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 and B as centres and radius more than 1/2 AB, draw arcs on both sides of line segment AB (to intersect each other). Let These arcs intersect each other at P and Q. Join PQ. Let PQ intersect AB at the Point M. Then 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           PM = PM             ∠APM = ∠BPM   ∴ ∆ PMA ≅ ∆PMB        So,  AM = BM & ∠ PMA = ∠ PMB Also,   ∠PMA + ∠PMB = 180°   ∠PMA + ∠PMA = 180°   2∠PMA = 180°   ∠ PMA = 90° Thus, BM & ∠ PMA = ∠ PMB = 90° ∴ PQ is the perpendicular bisector of AB

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CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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