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प्रश्न
Use ruler and compasses only for this question.
- Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
- Construct the locus of points inside the triangle which are equidistant from BA and BC.
- Construct the locus of points inside the triangle which are equidistant from B and C.
- Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
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उत्तर
Steps of construction:
- Draw line BC = 6 cm and an angle CBX = 60°. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
- Draw perpendicular bisector of BC and bisector of angle B.
- Bisector of angle B meets bisector of BC at P.
`=>` BP is the required length, where, PB = 3.5 cm - P is the point which is equidistant from BA and BC, also equidistant from B and C.
PB = 3.6 cm
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संबंधित प्रश्न
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
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| i. |  |
| ii. |  |
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