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Question

On constructing a triangle ABC, with AB = 5 cm, AC = BC = 9 cm, the distance between the points which are both equidistant from AB and AC and also 3 cm from BC is

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Solution

The correct option is **C**

6.4 cm

Steps of construction:

- Draw a line segment AB = 5 cm
- From A and B as centres and radius 9 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 3 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 3 cm from BC, we draw two lines parallel to BC at a distance of 3 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 3 cm from BC.

On measuring, we see that the distance between P and Q is 6.4 cm.

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