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प्रश्न
Draw an ∠ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
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उत्तर
Steps of construction:
- Draw a line segment BC = 5 cm
- At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
- Draw the angle bisector of ∠ABC.
- Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.
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.
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.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
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 a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
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.
