Question

# On constructing a triangle ABC, with AB = 4.3 cm, AC = BC = 5.6 cm, the distance between the points which are both equidistant from AB and AC and also 1 cm from BC is 1.1 cm 1.7 cm 2.1 cm 2.8 cm

Solution

## The correct option is C 2.1 cm Steps of construction: Draw a line segment AB = 4.3 cm From A and B as centres and radius 5.6 cm, make two arcs which intersect each other at C. Join CA and CB. Draw two lines n and m, parallel to  BC, at a distance of 1 cm from it. (We know that the locus of a point which is at a given distance from a given line is a pair of lines parallel to the given line and at a given distance from it. Hence to find the points which are 1 cm from BC, we draw two lines parallel to BC at a distance of 1 cm from it). Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively. (We know that the locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the given lines. Hence to find the points which are equidistant from AB and AC, we draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively). Thus P and Q are the required points which are both equidistant from AB and AC and 1 cm from BC. On measuring, we see that the distance between P and Q is 2.1 cm.

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