# What are the dimensions of a rectangle which has a perimeter of 48 cm, if its length is 12 cm longer than twice its width?

Jan 9, 2018

Width: $4 c m$

Length: $20 c m$

#### Explanation:

Recall that the perimeter of a rectangle is twice the sum of its length and width

So we have

$P = 2 l + 2 w$

We are also given that the length of the rectangle in question is 12cm longer than twice it width

Then

$l = 12 + 2 w$

Then we can plug in

$P = 2 w + 2 l = 2 w + 2 \left(12 + 2 w\right)$

And simplify

$P = 2 w + 24 + 4 w = 24 + 6 w$

Since we know that the perimeter is 48 cm we can say

$P = 48 = 24 + 6 w$

$\iff$ subtract 24 from both sides

$24 = 6 w$

$\iff$ divide both sides by 6

$4 = w$

Then we can plug that in to find l

$l = 12 + 2 w = 12 + 2 \left(4\right) = 12 + 8 = 20$

Jan 9, 2018

Width = $4 c m$
Length = $20 c m$

#### Explanation:

Set:
$x =$width
$y =$length

Translate words into equations:
Equation 1: "length" = 2xx"width" + 12 " " → y = 2x + 12
Equation 2:" perimeter" = 2xx"width" + 2xx"length " → 48 = 2x + 2y

Insert equation 1 into equation 2
$48 = 2 x + 2 y$
$48 = 2 x + 2 \left(2 x + 12\right)$
$48 = 2 x + 4 x + 24$
$24 = 6 x$
$x = 4 c m$

Find y through equation 1
$y = 2 x + 12$
$y = 2 \times 4 + 12$
$y = 20 c m$

Jan 9, 2018

The width is $4 c m$ and the length is $20 c m$

#### Explanation:

We are told how the length of the rectangle is related to the width, so we can use one variable to define both sides.

Let the width be $x$

The the length is $2 x + 12 \text{ } \leftarrow ' 12$ more than twice the width'

The perimeter is the sum of two widths and two lengths.

$2 \left(x\right) + 2 \left(2 x + 12\right) = 48 \text{ } \leftarrow$ write an equation

$2 x + 4 x + 24 = 48$

$6 x = 48 - 24$

$6 x = 24$

$x = 4$

The width is $4 c m$ and the length is $20 c m$