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प्रश्न
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
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उत्तर
Steps of oonstruction:
(i) Draw a line segment AB of 9 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
Proof:
(i) Take any point on LM say P.
(ii) Join PA and PB.
Since, Plies on the right bisector of line AB.
Therefore, Pis equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.
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संबंधित प्रश्न
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Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
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(ii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to ΔABC.
(iii) Mark the point Q, in your construction, which would make ΔQBC equal in area to ΔABC, and isosceles.
(iv) Measure and record the length of CQ.
