Punched Circular Hole in Square Sheet

CAT 2006 Slot 1 · QA · Hard · Quantitative Aptitude

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Passage / data set

A punching machine is used to punch a circular hole of diameter 2 units from a square sheet of aluminium of width 2 units. 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.

Question 1 of 2

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

  1. A.

    (pi + 2)/8

  2. B.

    (6 - pi)/8

  3. C.

    (4 - pi)/4

  4. D.

    (pi - 2)/4

  5. E.

    (14 - 3pi)/6

Answer

(6 - pi)/8

Explanation

Total area of square sheet =2×2=4= 2 \times 2 = 4. The region of the circle inside the square consists of:

  • A semicircle of radius 1 (area π/2\pi/2).
  • A right isosceles triangle of legs 1 and 1 (area 1/2×1×1=1/21/2 \times 1 \times 1 = 1/2). Total area punched out of the square =π2+1= \frac{\pi}{2} + 1. Remaining area =4−(π2+1)=3−π2=6−π2= 4 - (\frac{\pi}{2} + 1) = 3 - \frac{\pi}{2} = \frac{6 - \pi}{2}. Proportion of sheet area remaining =(6−π)/24=6−π8= \frac{(6 - \pi)/2}{4} = \frac{6 - \pi}{8}.

Question 2 of 2

Find the area of the part of the circle (round punch) falling outside the square sheet.

  1. A.

    pi/4

  2. B.

    (pi - 1)/2

  3. C.

    (pi - 1)/4

  4. D.

    (pi - 2)/2

  5. E.

    (pi - 2)/4

Answer

(pi - 2)/2

Explanation

Total area of circle =πr2=π(12)=π= \pi r^2 = \pi (1^2) = \pi. Area of circle inside square =π2+1= \frac{\pi}{2} + 1. Area outside square =π−(π2+1)=π2−1=π−22= \pi - \left(\frac{\pi}{2} + 1\right) = \frac{\pi}{2} - 1 = \frac{\pi - 2}{2}.

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