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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.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
Prove that:
F is equidistant from AB and AC.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Describe the locus for questions 1 to 13 given below:
1. The locus of a point at a distant 3 cm from a fixed point.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Describe the locus of points at distances greater than 4 cm from a given point.
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
ΔPBC and ΔQBC are two isosceles triangles on the same base. Show that the line PQ is bisector of BC and is perpendicular to BC.
In Fig. AB = AC, BD and CE are the bisectors of ∠ABC and ∠ACB respectively such that BD and CE intersect each other at O. AO produced meets BC at F. Prove that AF is the right bisector of BC.
