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Explanation:

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Let O be the centre of the smaller circle.

Let the small circle touch AB at P. OQ is perpendicular to AD.

Now, AO = 60 – r and & OP = r

Now,

QO2 = AP2 = (60 – r)2 – r2              … (i)

Now, in ΔDQO,

QO2 = DO2 – DQ2

Now, DO = 60 + r and DQ = 60 – r

DQ = 60 – r

∴ OQ2 = (60 + r)2 – (60 – r)2         … (ii) 

∴ (60 – r)2 +3600 – 120r = (60 + r)2

From (i) and (ii),

∴ r = 10

Hence, option (b).

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