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

(i) Construct the locus of points equidistant from BA and BC.

(ii) Construct the locus of points equidistant from B and C.

(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.

Steps of construction :

1. Draw a line segment AB = 5.5 cm.

2. Draw a ray AX such that ∠XAB = 105°.

3. With A as center and radius = 6 cm draw an arc intersecting AX at C.

4. Join BC. ABC is the required triangle.

5. Draw BY, angle bisector of B.

6. Draw MN perpendicular bisector of BC.

7. Mark point P as the intersection of BY and MN. Measure PC.

(i) Hence, BY is the locus of points equidistant from BA and BC.

(ii) Hence, MN is the locus of points equidistant from B and C.

(iii) Hence, PC = 4.8 cm.