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Question
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
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Solution
Steps of construction:
- Draw a ray BC.
- At B, draw a ray BA making an angle of 75° with BC.
- Draw a line l parallel to AB at a distance of 2 cm
- Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.
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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.
