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Explanation:

Area of the sheet left unpainted is two-thirds of painted area i.e., area of the circles will be three-fifth of the total area.

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∴ Area of the circle = 3/5 × 135 = 54 sq. in.

Let the radius of the circle be r and l be the length of the rectangle. Width of the rectangle will be 2r.

⇒ πr2 = 81

⇒ r = 9π

Area of rectangle = 135 = 2r × l

⇒ l = 1352r = 1352×9π = 15π2

∴ Perimeter of the rectangle = 2(l + r) = 215π2+18π =  3π5 +12π

Hence, option (c).

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