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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:

  1. 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).
  2. Let These arcs intersect each other at P and Q. Join PQ.
  3. 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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  1. Chapter 11 Class 9 Constructions
  2. Concept wise
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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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