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प्रश्न
Construct a triangle ABC, such that AB= 6 cm, BC= 7.3 cm and CA= 5.2 cm. Locate a point which is equidistant from A, B and C.
उत्तर
Steps of construction:
(i) Draw a line segment BC= 7.3 cm.
(ii) With Bas centre and radius 6 cm draw an arc.
(iii) With C as centre and radius 5.2 cm draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
(v) Draw perpendicuIar bisector of BC , AB and AC.
In triangIe ABC, P is the point of intersection of AB , AC and BC.
Therefore, PA = PB, PB = PC, PC = PA.
Thus, circum-centre of a triangle is the point which is equidistant from all its vertices.
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संबंधित प्रश्न
On a graph paper, draw the lines x = 3 and y = –5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
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- Construct the locus of points equidistant from A and B.
- Construct the locus of points equidistant from AB and AC.
- Locate the point P such that PA = PB and P is equidistant from AB and AC.
- Measure and record the length PA in cm.
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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Centre of a cirde of radius 2 cm and touching a fixed circle of radius 3 cm with centre O.
Given ∠BAC (Fig), determine the locus of a point which lies in the interior of ∠BAC and equidistant from two lines AB and AC.
