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From a rectangle $\text{ABCD}$ of area $768$ sq cm, a semicircular part with diameter $\text{AB}$ 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. $88+12\pi$

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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