question_answer

Construct a \[\Delta \]ABC such that BC = 8.5 cm, \[\angle \]B = \[90{}^\circ \] and AC = 9.5 cm. Draw perpendicular bisectors of AC and BC. Do they meet at AC?

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.