The width of a rectangle is fixed at 28 cm. What lengths will make the perimeter greater than 72 cm?

Oct 16, 2015

$l > \text{8 cm}$

Explanation:

Start by writing down the formula for the perimeter of a rectangle

$\text{perimeter" = P = 2 xx (l + w)" }$, where

$l$ - the length of the rectangle;
$w$ - its width.

In your case, you know that the width of the rectangle is set at $\text{28 cm}$. In order to find which lengths would make the perimeter greater than $\text{72 cm}$, determine what exact length will make the perimeter exactly $\text{72 cm}$.

$P = 2 \times \left(l + 28\right) = 72$

$l + 28 = \frac{72}{2}$

$l = 36 - 28 = \text{8 cm}$

This means that for any length that exceeds $\text{8 cm}$, the perimeter of the rectangle will be greater than $\text{72 cm}$.

$2 x \left(l + 28\right) > 72$

$l + 28 > 36$

$l > \text{8 cm}$

SIDE NOTE Don't be confused by the fact that the length turned out to be "shorter" than the width, that happens in some cases.