Q. 17 A3.5( 8 Votes )

# Two equal circles Let Center of the Circles are A and B. CD is a common Chord of the circle. AB is the perpendicular bisector of the chord CD.

If AB is a perpendicular bisector of CD then it should passes through both the centers.

So, AB is the distance that we need to calculate.

Given, AC = 10cm, CD = 12cm

CM CM CM = 6cm

In ACM

AC2 = AM2 + CM2

AM2 = AC2-CM2

AM2 = 100-36

AM2 = 64

AM = 8cm

In BCM

BC2 = BM2 + CM2

BM2 = BC2-CM2

BM2 = 100-36

BM2 = 64

BM = 8cm

AB = 16cm

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