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प्रश्न
Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
II. Construct the locus of points 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 the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.
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उत्तर
Steps of constructions:
1) Draw a line BC = 6 cm and CBX = 60°. Cut off AB = 3.5 cm. Join AC, ΔABC is the required triangle.
2) Draw perpendicular bisector of BC and bisector of ∠B.
3) The bisector of ∠B meets bisector of BC at P, therefore BP is the required length,
Where BP = 3.5 cm
4) P is the point which is equidistant from BA and BC, also equidistant from B and C.
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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.
