Question

Constuct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence: 

1. Construct the locus of point equdistant from BA and BC.
2. Construct the locus of points equidistant from B and C.
3. Mark the point which satisfies the above two loci as P. Measure and write the length of PC.

Answer 1

Steps of construction:

1. Draw AB = 5.5 cm.
2. Construct ∠BAR = 105°.
3. With centre A and radius 6 cm, cut off arc on AR at C. 
4. Join BC. ABC is the required triangle.

1.  Draw angle bisector BD of ∠ABC, which is the locus of points equidistant from BA and BC. 
2. Draw perpendicular bisector EF of BC, which is the loucs of point equidistant from B and C. 
3. BD and EF intersect each other at point P. Thus, P satisfies the above two loci. By measurement, PC = 4.8 cm.