The length of a rectangle is 5 cm greater than it width. The perimeter is 58 cm. What are the dimension of the rectangle?

2(x+(x+5))=58

2(2x+5)=58
4x+10=58
then subtract 10 from both side, which gives you,
4x=48
here x=12 and the length is x+5 which is 12+5=17
L:17 and B:12

perimeter=2(l+W)=2(w+5+w)

To solve this problem, we can create an equation using the given information.

Let's assume the width of the rectangle is x cm.
According to the given information, the length of the rectangle is 5 cm greater than its width. Therefore, the length can be expressed as (x + 5) cm.

The perimeter of a rectangle is given by the formula: Perimeter = 2(Length + Width).

Substituting the values, we get:
58 = 2[(x + 5) + x]

Simplifying the equation:
58 = 2(2x + 5)
58 = 4x + 10
48 = 4x
x = 12

Therefore, the width of the rectangle is 12 cm, and the length is (12 + 5) = 17 cm.

To find the dimensions of the rectangle, let's set up an equation based on the given information.

Let's say the width of the rectangle is W cm. According to the problem, the length of the rectangle is 5 cm greater than its width, so the length would be (W + 5) cm.

The perimeter of a rectangle is given by the formula: P = 2(L + W)

We are given that the perimeter is 58 cm, so we can substitute these values into the formula:

58 = 2((W + 5) + W)

Now, we can simplify and solve for W:

58 = 2(2W + 5)
58 = 4W + 10
48 = 4W
W = 12

So the width of the rectangle is 12 cm. Now, we can plug this value back into the equation to find the length:

L = W + 5
L = 12 + 5
L = 17

Therefore, the dimensions of the rectangle are 12 cm (width) and 17 cm (length).