Question

Angle ABC = 50${}^{\circ }$ and BA = BC = 5 cm. The mid points of BA and BC are M and N respectively. Draw the locus of a point which is equidistant from BA and BC, 2.5 cm from M and 2.5 cm from N. Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Then the length of PC is

• A. 1.2 cm
• B. 1.7 cm
• C. 2.1 cm
• D. 2.9 cm

Answer 1

The correct option is C

2.1 cm

Steps of construction:

• Draw an angle of 50${}^{\circ }$ with AB = BC = 5 cm.
• Draw the angle bisector BX of angle B.

(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. So in order to find the locus of a point which is equidistant from BA and BC, we draw angle bisector of angle B).

• With centres at M and N, draw circles of radius 2.5 cm, which intersect each other at P.

(This gives us a point which is 4 cm away from both M and N)

Pis the required point.

Now, on measuring, we get, PC = 2.1 cm