Question

Construct a triangle ABC with AB = 6 cm, AC = BC = 9 cm. Then there exists a point which is 4 cm from A and equidistant from B and C, on the perpendicular bisector of the line BC.

• A. True
• B. False

Answer 1

The correct option is A

True

Steps of construction:

• Draw a line segment AB = 6 cm.
• With A and B as centres and radius 9 cm, draw two arcs which intersect each other at C.
• Join AC 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. Hence every point on the perpendicular bisector 'l' of BC is equidistant from B and C).

• With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.

Now, note that P is a point which is on the perpendicular bisector of BC and hence equidistant from B and C and also it is at a distance of 4 cm from A. Therefore P is the required point.

So, there exists a point which is 4 cm from A and equidistant from B and C, which lies on the perpendicular bisector of BC. Hence the given statement is true.