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Explanation:

Consider the diagram below.

Let r be the radius of the smaller semi-circles and s be the radius of the smaller circle.

OS = 2r − s

PS = r + s

PO = r

But, ∆PSO is a right-angled triangle.

∴ PS2 = PO2 + SO2

(r + s)2 = r2 + (2r − s)2

∴ r² + s² + 2rs = r² + 4r² + s² − 4rs

 s=2r3

∴ Total area not grazed = 12 π(2r)22×12πr2+π2r32

= 2πr2 - πr249πr2

59πr2

∴ Required percentage = 59πr22πr2 × 100

59×2 × 100

≈ 28%

Hence, option (b).

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