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प्रश्न
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:
- P is equidistant from B and C.
- P is equidistant from AB and BC.
- Measure and record the length of PB.
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उत्तर
Steps of construction:
-
- Draw a line segment AB = 7 cm.
- Draw angle ∠ABC = 60° with the help of a compass.
- Cut off BC = 8 cm.
- Join A and C.
- The triangle ABC so formed is the required triangle.
- Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
- Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point that fulfills the condition required in i. and ii. is the intersection point of the bisector of line BC and the angular bisector of ∠ABC.
P is the required point, which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.
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संबंधित प्रश्न
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
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