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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.
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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
Describe the locus of a point in space, which is always at a distance of 4 cm from a fixed point.
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
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- Measure and record the length PA in cm.
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Draw and describe the locus in the following case:
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Draw and describe the lorus in the following cases:
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Using only ruler and compasses, construct a triangle ABC 1 with AB = 5 cm, BC = 3.5 cm and AC= 4 cm. Mark a point P, which is equidistant from AB, BC and also from Band C. Measure the length of PB.
Using a ruler and compass only:
(i) Construct a triangle ABC with BC = 6 cm, ∠ABC = 120° and AB = 3.5 cm.
(ii) In the above figure, draw a circle with BC as diameter. Find a point 'P' on the circumference of the circle which is equidistant from Ab and BC.
Measure ∠BCP.
Ruler and compass 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.
(i) Construct Δ ABC, in which BC = 8 cm, AB = 5 cm, ∠ ABC = 60°.
(ii) Construct the locus of point inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark as P, the point which is equidistant from AB, BC and also equidistant from B and C.
(v) Measure and record the length of PB.
