Question

Two circles of radius 15 cm and 13 cm intersect at A and B , such that AB = 24 cm

The distance between their centers is ?

Answer 1

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Answer 2

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Please refer the attachment below:

We know that,

$\triangle AOR \cong \triangle BOR$

Because,

OR = OR (Common)

AO = OB (Radii)

AR = BR (Radii)

So,

AM = BM (CPCT)

AM = BM = 24/2 = 12 cm

We can see that AOBR is a Kite,

$\therefore \angle M = 90^o$

In ΔBOM,

By Pythagoras theorem,

$BO^2 - BM^2 = OM^2$

$15^2 - 12^2 = OM^2$

$225 - 144 = OM^2$

$81 = OM^2$

Therefore,

$OM = \sqrt 81$

$OM = 9 cm$

Similarly in ΔBMR,

$BR^2 - BM^2 = MR^2$

$13^2 - 12^2 = MR^2$

$169 - 144 = MR^2$

$25 = MR^2$

Therefore,

$MR = \sqrt 25$

$MR = 5cm$

Hence,

Distance between their centers,

$=OM + MR$

$= 9 + 5$

$= 14 cm$