CAT 2018 [SLOT 2] Quant Question with Solution 30

Question:
From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is

  1. 80 + 16π
  2. 86 + 8π
  3. 82 + 24π
  4. 88 + 12π
Show Answer
Correct Answer: 4

Area of the semicircle with AB as a diameter = \(\frac{1}{2}×\pi ×(\frac{{A{B^2}}}{4})\)

\( \Rightarrow \) \(\frac{1}{2}×\pi ×(\frac{{A{B^2}}}{4})\) = \(72×\pi \)

\( \Rightarrow \) \(AB = 24cm\)

It is also know that the area of the rectangle ABCD = 768 sq.cm

\( \Rightarrow \) AB×BC = 768

\( \Rightarrow \) BC = 32 cm

Observe that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

\( \Rightarrow \) 32+24+32+\(\pi ×24/2\)

\( \Rightarrow \) \(88 + 12\pi \)


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