Discussion

Explanation:

Let PQRS be the square sheet and let the hole have centre O.

As P lies on the circumference of the circle and as m ∠APC = 90°, AC is a diameter.

∵ BP is a diameter, m ∠PAB = m ∠BCP = 90°.

∵ BP = AC, ABCP is a square.

∴ m ∠POC = 90° and OP = OC = 1 unit

The area of part of the circle falling outside the square sheet

= 2 × (Area of sector OPC – Area of ∆ OPC)

=2×π×124-12×12=π-22 sq.units

Area of part of hole on sheet = Area of hole − Area of part of the circle falling outside the square sheet

=π-π-22=π+22 sq.units

Part of square remaining after punching = Area of square − Area of part of hole on sheet

=4-π+22=6-π2 sq.units

∴ Proportion of sheet area that remains after punching =6-π24=6-π8

Hence, option (b).

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