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A certain rectangle is 5 times as long as it is wide. Suppose the length and width are both tripled. The perimeter of the second rectangle is how many times as large as the perimeter of the first rectangle?

Respuesta :

MrGumby
It is tripled as well. You are tripling every distance and thus tripling the sum of the distances as well. 

The perimeter of the second rectangle is 9 times as large as the perimeter of the first rectangle.

Given:

  • A certain rectangle is 5 times as long as it is wide.
  • A second rectangle is formed by tripling the length and width of previous reactangle.

To find:

The perimeter of the second rectangle is how many times as large as the perimeter of the first rectangle.

Solution:

  • Width of the rectangle-1 = w
  • Lenght of the rectangle-1 = l = 5w

The perimeter of the rectangle-1 = p

Perimeter = Width × Length

[tex]p=l\times w\\\\=5w\times w= 5w^2[/tex]

Suppose the length and width of rectangl-1 are tripled and rectangle-2 formed.

  • Width of the rectangle-2 = W = [tex]3\times w=3w[/tex]
  • Lenght of the rectangle-2 = L = [tex]3\times 5w=15w[/tex]

The perimeter of the rectangle-2 = P

Perimeter = Width × Length

[tex]P=L\times W\\\\=15w\times 3w= 45w^2[/tex]

Dividing perimeter of both recatnges:

[tex]\frac{P}{p}=\frac{45w^2}{5w^2}\\\\P=p\times 9\\\\P=9p[/tex]

The perimeter of the second rectangle is 9 times as large as the perimeter of the first rectangle.

Learn more about perimeter of rectangle here:

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