A triangle ABC is constructed with sides AB = 5.6 cm, AC = BC = 9.2 cm. What is the distance between the points which are both equidistant from AB and AC and also 2 cm from BC?
Steps of construction:
- Draw a line segment AB = 5.6 cm
- From A and B as centers and radius 9.2 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 2 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 2 cm from BC, we draw two lines parallel to BC at a distance of 2 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 2 cm from BC.
On measuring, we see that the distance between P and Q is 4.3 cm.
Steps of construction:
- Draw a line segment AB = 5.6 cm
- From A and B as centers and radius 9.2 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 2 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 2 cm from BC, we draw two lines parallel to BC at a distance of 2 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 2 cm from BC.
On measuring, we see that the distance between P and Q is 4.3 cm.
