question_answer 1)

Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?

Answer:

                Here, we will use the following steps of construction: Step I Mark a point O on the paper and draw a circle of radius of 4 cm with O as centre. Step II Draw any two chords \[{{l}_{1}},{{l}_{2}}\] and \[{{l}_{3}}\] of this circle. Step III Now, with A as centre, using compasses, draw an arc of radius more than half of the length of AB. Step IV With the same radius and with B as centre, draw another arc using compasses. Let it cut the previous arc at P and Q. Step V Join \[{{l}_{1}},{{l}_{2}},{{l}_{3}}\]It cuts \[{{l}_{4}}\] at M. Therefore, PQ is the perpendicular bisector of \[{{l}_{1}}\]. Step VI Now, with C as centre and radius more than half of the length of CD draw an arc. Step VII With D as centre and same radius draw another arc which intersects previous arc at R and S. Step VIII Join RS. It cuts CD at N. Therefore, RS is the perpendicular bisector of CD. From the above figure, it is clear that these perpendicular bisectors also passes through O, the centre of the circle. Hence, the perpendicular bisectors of these chords meet each other at centre of the circle.