If the side of a square is doubled and then increased by 7, the new perimeter is 8 more than 3 times the old perimeter. ?

What is the side length of the original square??

4 Answers

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  • 1 month ago

    First stage, the side is s₁. The perimeter is:

    p₁ = 4.s₁

    Second stage, the side is doubled:

    s₂ = 2.s₁ and then increased by 7

    s₂ = 2.s₁ + 7

    The perimeter is:

    p₂ = 4.s₂

    p₂ = 4.(2.s₁ + 7)

    p₂ = 8.s₁ + 28

    The new perimeter is 8 more than 3 times the old perimeter.

    p₂ = 3.p₁ + 8 → recall: p₂ = 8.s₁ + 28

    8.s₁ + 28 = 3.p₁ + 8

    8.s₁ - 3.p₁ = - 20 → recall: p₁ = 4.s₁

    8.s₁ - 3.(4.s₁) = - 20

    8.s₁ - 12.s₁ = - 20

    - 4.s₁ = - 20

    s₁ = 5 ← this is the side length of the original square

  • 1 month ago

    2S + 7 so that 4(2S + 7) = 8*4S find S = ?  We have 8S + 28 = 32S; so that 28 = 24S and S = 7/6 ANS.

  • 1 month ago

    If the side of a square is doubled and then increased by 7, 

    the new perimeter is 8 more than 3 times the old perimeter. 

    The old perimeter is 4s and the new perimeter is 12s + 8

    What is the side length of the original square?

  • david
    Lv 7
    1 month ago

    old perimeter  =  4x

    new per. = 4(x + 7) = 8  + 3(4x)

     , , ,   4x + 28  =  12x + 8

     . . .   8x = 20

     . . .   x = 2.5  <<<  answer

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