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# 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

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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