# Find the length of a rectangle lot with a perimeter of 82 meters if the length is 7 meters more then the width?

May 8, 2018

24 meters

#### Explanation:

If the short side of the rectangle = s, then the long side would be s+7.

The total perimeter then would be the sum of all 4 sides, where the sids are equal in pairs, i.e.
Perimeter = s+(s+7)+s+(s+7) = 4s+14 = 82
Therefore 4s=82-14 = 68
so s=17
As this is the short side, the long side is s+7=17+7=24
The length of the rectangle, therefore, is 24 meters

May 8, 2018

The length of the rectangle lot is $\textcolor{red}{24}$ meters.

#### Explanation:

To start, I'll set up an equation to help solve this...

$2 l + 2 w = p$, meaning the length twice plus the width twice equals the perimeter. We can fill this in with the known values:

$2 l + 2 \left(l - 7\right) = 82$

$w$ can be replaced by $\left(l - 7\right)$ because the problem states that the width is $7$ meters shorter than the length.

Distribute the $2$ to the $\left(l - 7\right)$ to get:

$2 l + 2 l - 14 = 82$

And combine like terms to get:

$4 l = 96$

Finally divide both sides by $4$ to finish with:

$l = 24$, meaning the length is $\textcolor{red}{24}$ meters.

I'll check my work by plugging this info back in to the original equation to see if it works:

$2 l + 2 w = p$
$2 l + 2 w = 82$
$2 \left(24\right) + 2 \left(24 - 7\right) = 82$
$2 \left(24\right) + 2 \left(17\right) = 82$
$\textcolor{g r e e n}{48 + 34 = 82}$

The equation works, meaning the length is $24$ m and the width is $17$ m.