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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. |  |
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.
Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Describe the locus of points at distances less than 3 cm from a given point.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
Find the locus of points which are equidistant from three non-collinear points.
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.
Use ruler and compasses for the following question taking a scale of 10 m = 1 cm. A park in a city is bounded by straight fences AB, BC, CD and DA. Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD. Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.
