Question

Angle ABC = 60${}^{\circ }$ and BA = BC = 8 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, 4 cm from M and 4 cm from N. Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP. Then BMPN is a

• A. Rectangle
• B. Square
• C. Rhombus
• D. Trapezium

Answer 1

The correct option is C

Rhombus

Steps of construction:

• Draw an angle of 60${}^{\circ }$ with AB = BC = 8 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 4 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.

• Join MP and NP.

BMPN is a rhombus since BM = MP = PN = NB = 4 cm.