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
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
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.
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.
