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प्रश्न
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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उत्तर
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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संबंधित प्रश्न
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
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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.
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.
