Word Problem??

Question

The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52cm, find the dimensions fo the rectangle.

Answer 1

Another system of equations problem...

The length is 2 cm more than twice the width, so
L=2w + 2

The perimeter is 52 cm, so
52 = 2L + 2w

Solve by using your favorite method. I'll use substitution.

52 = 2(2w + 2) + 2w....(substitute)
52 = 4w + 4 + 2w........(distribute)
52 = 6w + 4................(combine like terms)
48 = 6w.....................(subtract 4 from both sides)
8 = w........................(divide both sides by 6)

If the width is 8cm, then the length is 18cm (using the first equation). 2(8) + 2(18) = 16 + 36 = 52, so we're good!

Answer 2

You just need to re-express what is being said in words to mathematical expressions.

We know the perimeter of a rectangle is
P = 2l + 2w (length and width)

We are told that the length l is 2 cm more than the width w so
l -2 = w

We are told the total perimeter is 52 so
52 = 2l +2w = we knopw what w equals, so
52 = 2l + 2x(l-2)

you can expand out and solve for l, then substitute back and solve for w.

Answer 3

This problem can be answered by using substituion. We know that L=2+2W, when L is length and W is width. If the perimeter is 52, we also know that 2L+2W=52. So now we can substitute all of the L's in that equation with 2+2W, because of our first equation. So our new equation looks like this 2(2+2W)+2W=52. When you solve this, you get the Width to be 8. We know L=2+2W, so we know that the Length is 18.

Answer 4

width = w
length = 2w + 2

2(w + 2w + 2) = 52
2(3w + 2) = 52
6w + 4 = 52
6w = 48
w = 8 cm
2w + 2 = 18 cm

18 + 18 + 8 + 8 = 52 cm

Answer 5

l = 2w + 2

p = 2l + 2w = 4w + 4 + 2w = 6w+4 = 52
w = 8, l = 2(8) + 2 = 18

Answer 6

2l + 2w = p

x = width.................2x + 2 = length

x + 2x+2 = 26
3x = 24

x = 8 length = 18