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

  1. $80 + 16\pi$
  2. $86+8\pi$
  3. $82+24\pi$
  4. $88+12\pi$
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Given that, area of rectangle ${ABCD} = 768 \; \text{cm}^{2}.$

Let the radius of semicircle be $`r\text{’} \; \text{cm}.$



$\text{Area of semicircle} = \dfrac{\pi r^{2}}{2},$ where $`r\text{’}$ is the radius of semicircle.

$ \Rightarrow \frac{\pi r^{2}}{2} = 72 \pi $

$ \Rightarrow r^{2} = 144 $

$ \Rightarrow \boxed{r = 12 \; \text{cm}} $

$ \Rightarrow \boxed{ AB = 2r = 24 \; \text{cm}} $

$\text{Area of rectangle} = AB \times BC $

$ \Rightarrow 24 \; BC = 768 $

$ \Rightarrow BC = \frac{768}{24} $

$ \Rightarrow \boxed{ BC = 32 \; \text{cm}} $

A semicircular part is removed from the rectangle. The figure will be :



The perimeter of the remaining portion $ = AD + DC + BC + \text{arc} (AB) $

$\quad = 32 + 24 + 32 + $ Perimeter of semicircle

$\quad = 88 + \frac{2 \pi (12)} {2} \quad [\because \text{Perimeter of semicircle} = \frac {2 \pi r}{2} = \pi r] $

$ \quad = (88 + 12 \pi) \; \text{cm}.  $

Correct Answer $: \text{D} $

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