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Question
Ruler and compass only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
(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.
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Solution
(i) Steps of Construction:
1. Draw a line segment BC = 8 cm.
2. Make ∠CBX = 60°
3. Set off BA = 5 cm, along BX.
4. Join CA.
Then, ΔABC is the required triangle.
(ii) We know that the locus of point equidistant from two intersecting straight lines consist of a pair of straight lines that bisect the angles between the given straight lines.
Therefore in this case is the angle bisector of angle B, It is shown in the adjoining figure.
(iii) We know that the locus of a point equidistant from two fixed points is the right bisector of the straight line joining the two fixed points.
Therefore, in this case the right bisector of side BC of ΔABC. It is shown in the given figure.
(iv) The point P, is the point in intersecting of angle bisector of ∠ABC and the right bisector of BC.
It is shown in the following figure.
(v) On measuring we find the length of PB = 3 cm.
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