Question

A punching machine is used to punch a circular hole of diameter two units form a square sheet of aluminium of width 2 units as shown in the diagram. The hole is punched such that the circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square. The proportion of the sheet area that remains after punching is:

(a) (π + 2)/8 

(b) (6 – π)/8 

(c) (4 – π)/4 

(d) (π – 2)/4 

(e) (π – 2)/6

Answer 1

Answer : (b) \(\frac{6-\pi}{8}\)

In △FAE, ∠FAE = 90º, as it is an angle of a square. EF will be the diameter of the circle as an angle subtended by a diameter of the circumference of a circle = 90º. 

So EF will pass through center O.

In △AOF and AOE, OF = OE (radii of circle) AO = OA (common) ∠OAF = ∠OAE = 45º 

∴ △AOF ~ △AOE ⇒ AF = AE

Also, area of △AFE = \(\frac{1}{2}\) × AF × AE = \(\frac{1}{2}\)AF² 

Given FE = 2

⇒ AF² + AE² = FE² = 2² = 4 ⇒ AF² = 2

∴ Area of △AFE = \(\frac{1}{2}\) × 2 = 1.

Area of shaded region 2 = 2 × (Area of sector AOE – Area of DAOE)

∴ Area of the unshaded region of the circle 

= Area of the circle – Area of shaded region 2

∴ Area of shaded region 1 

= Area of the square – area of the unshaded region of the circle

= 2² - \(\frac{\pi +2}{2} \) = \(\frac{6-\pi}{2}\)

∴ The proportion of the sheet that remains after punching 

= \(\frac{Area\, of\, region\,1}{Total\, area\, of\, the\, square}\) = \(\frac{6-\pi}{\frac{2}{4}}\)

= \(\frac{6-\pi}{8}\)