Chords equidistant from the centre of a circle are equal

  

Topic: Chords equidistant from the centre of a circle are equal

Objective: To prove that the chords of a circle which are equidistant from the centre of the circle are equal.

Pre-requisite knowledge :

(1) To draw a line perpendicular to a line from a point outside it (by paper folding).

(ii) To draw a line perpendicular to a given line at a given point on it

(by paper folding).

Materials required :

(i) Tracing papers

(ii) Geometry box

(iii) Coloured ball point pens.

To perform the activity :

Steps:

1. Draw a circle of any radius with centre O on a tracing paper as shown in fig. 7.1.

2. Draw a chord AB of the circle.

3. From O, draw OM I AB (by paper folding).

4. Mark a point N on the tracing paper such ON = OM. It may be noted that the point N will lie in the interior of the circle (why!). Join O and N.

5. Through N, draw a line perpendicular to ON (by paper folding). Let this line meet the circle at points C and D, then CD is a chord of the circle such that ON LCD as shown in fig. 7.1. Thus the chords AB and CD are equidistant from the centre O of the circle.

6. Fold the figure/tracing paper such that M falls on N and form a crease as shown dotted in fig. 7.2.

Result :

We observe that A falls on C and B falls on D i.e. AB exactly covers CD AB = CD

Hence the chords of a circle that are equidistant from the centre of the circle are equal.

Note:

The above result is true in case of congruent circles also Le. chords in congruent circles that are equidistant from their centres are equal.

To see this :

1. Draw two congruent circles with centres O and O' on tracing papers as shown in fig. 7.3.

2. Draw a chord AB of the circle with centre O.

3. From O, draw OM I AB (by paper folding).

4. Mark a point N in the interior of the circle with centre O' such that O'N OM. Join O' and N.

5. Through N, draw a line perpendicular to O'N (by paper folding). Let this

line meet the circle at points C and D, then CD is chord of this circle such

that O'N L CD as shown in fig. 7.3. As O'N = OM, chords AB and CD

of the two circles are equidistant from their centres.

6. Superimpose the circle with centre O on the circle with centre O' such that O falls on O' and M falls on N as shown in fig. 7.4.

Result:

We observe that A falls on C and B falls on D i.e. AB exactly covers CD.

AB= CD

Hence chords in congruent circles that are equidistant from their centres are equal.

Thus we have proved:

Chords of a circle (or of congruent circles) that are equidistant from the centre (or centres) are equal.