In a circle, chord MN ≅ chord RT. Chord RT is at a distance of 6 cm from the centre. Find the distance of the chord MN from the centre.

Given:

Chord RT is at a distance of from centre A.

Draw a figure using given condition

Let us draw a perpendicular bisector AC to MN and AB to RT and

To Show: AB = AC

Join AT and AN, we get,

Now in ∆ABT and ∆ACN

RT = MN (since, chord MN ≅ chord RT )

BT = CN

∠ABT = ∠ACN = 90^°

seg AT = seg AN (since, AT and AN are radius of a circle)

Therefore, ∆ABT ≅ ∆ACN (By RHS Congruent Rule)

Therefore, seg AC = seg AB = 6cm (By CPCT)

Alternate Method: Given: chord MN ≅ chord RT

Chord RT is at a distance of from centre A.

To Find length of seg AC

We know that, Congruent chord are equidistant from the centre.

Therefore, seg AC = seg AB = 6cm