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Question
Constuct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
- Construct the locus of point equdistant from BA and BC.
- Construct the locus of points equidistant from B and C.
- Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Solution
Steps of construction:
- Draw AB = 5.5 cm.
- Construct ∠BAR = 105°.
- With centre A and radius 6 cm, cut off arc on AR at C.
- Join BC. ABC is the required triangle.
- Draw angle bisector BD of ∠ABC, which is the locus of points equidistant from BA and BC.
- Draw perpendicular bisector EF of BC, which is the loucs of point equidistant from B and C.
- BD and EF intersect each other at point P. Thus, P satisfies the above two loci. By measurement, PC = 4.8 cm.
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