# How do you find the dimensions of a rectangle if the area is 63cm^2 and the perimeter is 32cm where x represents the width of the rectangle?

Apr 3, 2017

Width = $7 c m$; Length = $9 c m$

#### Explanation:

let $\textcolor{red}{x}$ be the width and $\textcolor{red}{y}$ be the length of the rectangle.

Given
$a r e a$ $= x \cdot y = 63$ -----------$\left(1\right)$

$p e r i m e t e r$ $= x + y + x + y = 2 x + 2 y = 32$ ----------$\left(2\right)$

from $\left(1\right) , x = \frac{63}{y}$

substituting this in $\left(2\right) ,$

$\frac{63}{y} + y = 16$

$\implies {y}^{2} - 16 y + 63 = 0$

$\implies {y}^{2} - 7 \cdot y - 9 \cdot y + 63 = 0$

$\implies y \left(y - 7\right) - 9 \left(y - 7\right) = 0$

$\implies \left(y - 9\right) \left(y - 7\right) = 0$

$\implies$ either $y - 9 = 0$ i.e. $y = 9$

or $y - 7 = 0$ i.e. $y = 7$

therefore $x = \frac{63}{9} = 7$ or $x = \frac{63}{7} = 9$

Hence, width = $7 c m$; length = $9 c m$