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Question
Construct a triangle ABC, such that AB= 6 cm, BC= 7.3 cm and CA= 5.2 cm. Locate a point which is equidistant from A, B and C.
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Solution
Steps of construction:
(i) Draw a line segment BC= 7.3 cm.
(ii) With Bas centre and radius 6 cm draw an arc.
(iii) With C as centre and radius 5.2 cm draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
(v) Draw perpendicuIar bisector of BC , AB and AC.
In triangIe ABC, P is the point of intersection of AB , AC and BC.
Therefore, PA = PB, PB = PC, PC = PA.
Thus, circum-centre of a triangle is the point which is equidistant from all its vertices.
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(i) Construct Δ ABC, in which BC = 8 cm, AB = 5 cm, ∠ ABC = 60°.
(ii) Construct the locus of point 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 as P, the point which is equidistant from AB, BC and also equidistant from B and C.
(v) Measure and record the length of PB.
