Answer:
Steps of Construction:
Part I: To construct the ΔABC.
(a) Draw BC = 8·5 cm.
(b) Construct\[\angle CBX=90{}^\circ \].
(c) With centre C and radius equal to 9·5 cm, draw an arc to cut at A.
(d) Join A and C. The ΔABC is the required triangle.
Part II: Perpendicular bisectors of AB and BC.
(a) With centre at A and radius more than half of AB, draw two arcs on both sides of AB.
(b) With centre at B and the same radius as in
step (i), draw two arcs intersecting the arcs drawn in step (i) at R and S.
(c) Join R and S and extend it on both sides to P and Q. Now PQ is the perpendicular bisector of AB.
(d) Similarly, draw NM perpendicular bisector of BC.
We note that PQ and NM meet at a point D, which is at AC.
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