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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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- Construct a ΔABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
- Construct the locus of points at a distance of 3.5 cm from A.
- Construct the locus of points equidistant from AC and BC.
- Mark 2 points X and Y which are at a distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
How will you find a point equidistant from three given points A, B, C which are not in the same straight line?
Ruler and compasses 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 a ΔABC, in which BC = 6 cm, AB = 9 cm and ∠ABC = 60°.
(ii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to ΔABC.
(iii) Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
(iv) Measure and record the length of CQ.
