NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.5 – Part 1

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Perpendicular bisector

Give perpendicular bisector at triangle ABC ,D is a mid point of AB,E is a mid point of BC and F is mid point of AC

Give Perpendicular Bisector at Triangle ABC

Give perpendicular bisector at triangle ABC ,D is a mid point of AB,E is a mid point of BC and F is mid point of AC

To find this point, you will construct three perpendicular bisectors, one for each side of the triangle. The point where all three perpendicular bisectors intersect is called the Circumcenter. Using this center point, we can draw a circle that passes through all three vertices.

1. ABC is a triangle. Locate a point the interior of ABC which is equidistant from all the vertices of ABC

Solution:

interior of △ABC perpendicular bisector PQ and RS of side AB and BC respectively of triangle ABC.PQ bisects AB at M and bisects BC at point N

Interior of △ABC

interior of △ABC perpendicular bisector PQ and RS of side AB and BC respectively of triangle ABC.PQ bisects AB at M and bisects BC at point N

  • Draw perpendicular bisectors PQ and RS of sides AB and BC respectively of triangle ABC. Let PQ bisects AB at M and RS bisects BC at point N.

  • Let PQ and RS intersect at point O.Join OA, OB and OC.

  • Now in AOMandBOM,

    AM=MB [By construction]

    AMO=BMO=90° [By construction]

    OM=OM [Common]

    AOMBOM [By SAS congruency]

    OA=OB [By Coorresponding Parts of Congruent Triangles] …..equation(1)

  • Similarly, BONCON

    OB=OC [By Corresponding Parts of Congruent Triangles.] …..equation(2)

  • From eq. (1) and (2),

    OA=OB=OC

  • Hence O, the point of intersection of perpendicular bisectors of any two sides of ABC equidistant from its vertices.

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