# mathThe width of a rectangle is 45% of the length. If the perimeter of the rectangle is 58 cm, what is its width?

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### 3 Answers

Let W be the width and let L be the length. We can set up the following two equations:

.45W = L

2W + 2L = 58

Now substitute the value for L in the first equation into the second.

2W + 2(.45W) = 58

2W + .9W = 58

2.9W = 58

W = 20

To find the length, we can plug this value for W back in to the first equation and find that

.45*20 = L

L = 9

**So, the width of the rectangle is 20 cm and the length is 9 cm.**

The width of a rectangle is 45% of the length.

Let the length of the rectangle be L, the width of the rectangle is 0.45*L

The perimeter of a rectangle with sides W and L is 2*(W + L)

Here, the perimeter is 2*(L + 0.45*L) = 58

2*1.45L = 58

2.9*L = 58

L = 20

The width of the rectangle is 0.45*20 = 9 cm.

We'll consider x as the length of the rectangle.

We'll note the width of the rectangle as w. We know, from enunciation that the width is:

w = 45*x/100

Now, we'll write the formula of the perimeter of the rectangle:

P = x + x + w + w

P = 2x + 2w

We'll substitute w by the ratio 45*x/100:

P = 2(x + 45*x/100)

We also know, from enunciation, that the perimeter of the rectangle is of 58 cm.

58 = 2(x + 45*x/100)

We'll divide by 2:

29 = x + 0.45x

29 = 1.45x

We'll divide by 1.45:

x = 29/1.45

x = 20 cm

The length of the rectangle is of 20 cm and the width is:

w = 0.45*20 cm

w = 9 cm