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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
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"
Choice A is the correct answer.
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