Construct a triangle ABC, with AB= 6 cm,AC=BC=9 cm.Find a point 4 cm from A and equidistant from B and C.

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Solution

Steps of Construction:

i) Draw a line segment AB = 6 cm

ii) With A and B as centres and radius 9 cm, draw two arcs which intersect each other at C.

iii) Join AC and BC.

iv) Draw the perpendicular bisector of BC.

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

P is the required point which is equidistant from B and C and at a distance of 4 cm from A.

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Construct a triangle ABC, having given AB= 4.8 cm, AC=4 cm, and $∠ A = 75^{o}$ .Find a point P

(i) inside the triangle ABC;

(ii) outside the triangle ABC;

equidistant from B and C; and at a distance of 1.2 cm from BC.

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.

Q. Construct a triangle

A B C

with

A B = 5.5 c m, A C = 6 c m

and

∠ B A X = 105^{o}

Hence :
Construct the locus of points equidistant from

B A

and

B C

Q. Construct a triangle

A B C

with

A B = 5.5 c m, A C = 6 c m

and

∠ B A X = 105^{o}

Hence :
Construct the locus of point equidistant from

B

and

C

On constructing a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm, the distance between the points which are both equidistant from AB and AC and also 2 cm from BC is

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