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
- 80 + 16π
- 86 + 8π
- 82 + 24π
- 88 + 12π
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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