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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.
संबंधित प्रश्न
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 A and B.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of the door handle, as the door opens.
Describe the locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
- AB and BC;
- B and D.
Describe the locus of points at distances less than or equal to 2.5 cm from a given point.
In a quadrilateral PQRS, if the bisectors of ∠ SPQ and ∠ PQR meet at O, prove that O is equidistant from PS and QR.
Show that the locus of the centres of all circles passing through two given points A and B, is the perpendicular bisector of the line segment AB.
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.
