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Perimeter of leftover portion

CAT 2018 Slot 2 · Quantitative Ability · Medium · Mensuration

This is a medium Quantitative Ability question from the CAT 2018 Slot 2 paper. It tests Mensuration. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π72\pi sq cm is removed. The perimeter of the leftover portion, in cm, is

  1. A.

    88 + 12π

  2. B.

    80 + 16π

  3. C.

    86 + 8π

  4. D.

    82 + 24π

Answer

A

Explanation

Area of semicircle with diameter AB=2rAB = 2r is 12πr2=72π  ⟹  r2=144  ⟹  r=12 cm\frac{1}{2}\pi r^2 = 72\pi \implies r^2 = 144 \implies r = 12\text{ cm}. So AB=2r=24 cmAB = 2r = 24\text{ cm}.

Area of rectangle ABCD=AB×BC=768 sq cmABCD = AB \times BC = 768\text{ sq cm}. 24×BC=768  ⟹  BC=32 cm24 \times BC = 768 \implies BC = 32\text{ cm}

The leftover portion consists of sides BCBC, CDCD, DADA, and the arc ABAB of the semicircle. BC=32,CD=24,DA=32BC = 32, CD = 24, DA = 32. Length of semicircular arc AB=πr=12πAB = \pi r = 12\pi.

Perimeter of leftover portion =BC+CD+DA+arc AB=32+24+32+12π=88+12π= BC + CD + DA + \text{arc } AB = 32 + 24 + 32 + 12\pi = 88 + 12\pi.

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