The proportion of the sheet area that remains after punching is:
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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