Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
II. Construct the locus of points inside the triangle which are equidistant from BA and BC.
III. Construct the locus of points inside the triangle which are equidistant from B and C.
IV. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.
Solution
Steps of constructions:
1) Draw a line BC = 6 cm and CBX = 60°. Cut off AB = 3.5 cm. Join AC, ΔABC is the required triangle.
2) Draw perpendicular bisector of BC and bisector of ∠B.
3) The bisector of ∠B meets bisector of BC at P, therefore BP is the required length,
Where BP = 3.5 cm
4) P is the point which is equidistant from BA and BC, also equidistant from B and C.
