1. ## Perimeter

How would I solve this?

"The perimeter of a rectangle is 40 cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm. Find the dimensions of the original rectangle."

2. Originally Posted by largebabies
How would I solve this?

"The perimeter of a rectangle is 40 cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm. Find the dimensions of the original rectangle."
Let l denote the length of the rectangle and w the width. The you know:

$2l + 2w = 40$
$2\cdot (2l)+2\cdot \left(\frac12w\right) = 40+16$

That means you have to solve for l and w:

$l+w=20$
$4l+w=56$

I've got l = 12 and w = 8

3. $P=2l+2w,$where l = length and w = width
$40cm=2l+2w$
and
$56cm=2*2l+\frac{2w}{2}$

$56cm=4l+w$

$40cm-2w=2l$

$20cm-w=l$

$56cm=4(20cm-w)+w$

$56cm=80cm-4w+w$

$-24cm=-3w$

$w=8cm$

now plug $w=8cm$ into the original equation

$40cm=2(8cm)+2w$

$24cm=2w$

$w=12cm$