CAT 2018 Question Paper | Quants Slot 2

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Question 6 : 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. 88 + 12π
2. 80 + 16π
3. 86 + 8π
4. 82 + 24π

Video Explanation

Explanatory Answer

Method of solving this CAT Question from Mensuration

Given that ABCD be the rectangle of area of 768 sq cm. A semicircular part with diameter AB and area 72π sq cm is removed
From the area given we can find the radius r as
\\frac{πr^2}{2}) = 72π
r² = 144
r = 12 cm

The diameter AB is 24 since r = 12
So we have a rectangle ABCD
AB × BC = 768
24 × BC = 768
BC = \\frac{768}{24})
BC = 32

Now we have to find the perimeter of the left over portion (shaded area)
Perimeter = AD + DC + BC + \\frac{1}{2}) perimeter of the circle
Perimeter = 32 + 24 + 32 + \\frac{2πr}{2}) (r = 12 cm)
Perimeter = 88 + 12π cm
The perimeter of the leftover portion, in cm, is 88 + 12π

The question is "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"

Hence, the answer is 88 + 12π cm

Choice A is the correct answer.