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प्रश्न
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
- Complete the rectangle ABCD such that:
- P is equidistant from AB and BC.
- P is equidistant from C and D.
- Measure and record the length of AB.
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उत्तर
- Steps of construction:
- Draw a line segment BC = 5 cm
- B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
- Join PC.
- B and C as centers, draw two perpendiculars to BC.
- P as centre and radius PC, cut an arc on the perpendicular on C at D.
- D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
- On measuring AB = 5.7 cm
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संबंधित प्रश्न
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°
Hence:
1) Construct the locus of points equidistant from BA and BC
2) Construct the locus of points equidistant from B and C.
3) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
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