Question

Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and $\angle BAC={105}^{\circ }$. Mark the point of intersection of the locus of points equidistant from BA and BC and the locus of points equidistant from B and C, as P. Then the length of PC is ___ cm.

• A. 4.8

Answer 1

The correct option is A 4.8

Steps of construction:

• Draw a line segment AB of length 5.5 cm.
• With A as the centre and radius 6 cm, draw an arc at an angle of ${105}^{\circ }$ from AB and mark it as C.
• Join BC.

$\Delta$ ABC is the required triangle.

• Draw a bisector of angle B, to meet AC at R.

(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. We draw the bisector of the angle B so as to get the locus of the point which is equidistant from BA and BC).

• Draw the perpendicular bisector 'l' of BC.

(We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points. We draw the perpendicular bisector of BC so as to get the locus of the point which is equidistant from B and C).

From the construction, we see that locus of points equidistant from BA and BC and the locus of points equidistant from B and C intersect at P.

On measuring, we get, PC = 4.8 cm