CAT 2006 — QA Question 13

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)

$$\begin{gathered} =2\times \left[\left(\frac{\pi \times 1^2}{4}\right]-\left(\frac{1}{2}\times 1^2\right)\right) \\ =\frac{\pi -2}{2} \mathrm{sq}.\mathrm{units} \end{gathered}$$

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

$=\pi -\left(\frac{\pi -2}{2}\right)=\frac{\pi +2}{2} \mathrm{sq}.\mathrm{units}$

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

$=4-\left(\frac{\pi +2}{2}\right)=\frac{6-\pi }{2} \mathrm{sq}.\mathrm{units}$

∴ Proportion of sheet area that remains after punching $=\frac{\left({\displaystyle \frac{6-\pi }{2}}\right)}{4}=\frac{6-\pi }{8}$

Hence, option (b).