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Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC.

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Question

Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained. 

Sum
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Solution

 
Steps of construction:

  1. Draw a line segment AB = 5.6 cm
  2. From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
  3. Join CA and CB.
  4. Draw two lines n and m parallel to BC at a distance of 2 cm
  5. Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
    P and Q are the required points which are equidistant from AB and AC.
    On measuring the distance between P and Q is 4.3 cm.
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Chapter 16: Loci (Locus and Its Constructions) - Exercise 16 (B) [Page 241]

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Selina Concise Mathematics [English] Class 10 ICSE
Chapter 16 Loci (Locus and Its Constructions)
Exercise 16 (B) | Q 23. | Page 241

Video TutorialsVIEW ALL [1]

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