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प्रश्न
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
- the locus of the centres of all circles which touch AB and AC,
- the locus of the centres of all the circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC .
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उत्तर
Steps of construction:
- Draw a line segment BC = 4.5 cm
- With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
- Join AB and AC.
ABC is the required triangle. - Draw the angle bisector of ∠BAC
- Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
- With centre O and radius 2 cm, draw a circle which touches AB and AC.
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संबंधित प्रश्न
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
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Prove that:
F is equidistant from AB and AC.
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