Consider a square ABCD of side 60 cm. lt contains arcs BD and AC drawn with centres at A and D respectively. A circle is drawn such that it is tangent to side AB, and the arcs BD and AC. What is the radius of the circle?
Explanation:
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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