Question

The area of a rectangle is 120m^2. If its length is decreased by 8m and its width is increased by 4m, the area stays the same. Find the perimeter of the original rectangle if all measurements are whole numbers, and show your working. Please answer, much appreciated.

Answer 1

$\text{Let l and b be the length and breadth the given rectangle}$

$\textbf{Given:}$

$\text{Area of the rectangle= 120\,m^2 }$

$\implies\,lb=120$........(1)

$\text{As per given data,}$

$(l-8)(b+4)=120$

$\implies\,lb+4l-8b-32=120$

$\implies\,120+4l-8b-32=120$

$\implies\,4l-8b-32=0$

$\implies\,l-2b-8=0$

$\implies\,l=2b+8$ .......(2)

$\text{Using (2) in (1), we get}$

$(2b+8)b=120$

$2\,b^2+8b-120=0$

$b^2+4b-60=0$

$(b+10)(b-6)=0$

$\implies\,b=-10,6$

$\text{Since b cannot be negative, we have b=6 }$

$\text{When b=6\,m ,}\;l=\dfrac{120}{6}=20\,m$

$\text{Perimeter of the original rectangle}$

$=2(l+b)$

$=2(20+6)$

$=2(26)$

$=52\;m$

$\therefore\textbf{Perimeter of the original rectangle is 52 m}$