Algebra

What is the length of the original rectangle?

The perimeter of a rectangle is equal to 40. If the length is halved and the width is divided by 3, the new perimeter is decreased by 24.

Select one:
a. 12
b. 8
c. 4
d. 16

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  1. P = Original perimeter = 40

    P = 2 w + 2 l = 2 ( w + l )

    40 = 2 ( w + l )

    Divide both sides by 2

    20 = w + l

    w + l = 20

    Subtract w to both sides

    w + l - w = 20 - w

    l = 20 - w


    If the length is halved and the width is divided by 3 mean:

    New lenth l1 = l / 2 = ( 20 - w ) / 2 = 10 - w / 2

    New width w1 = w / 3

    The new perimeter is decreased by 24 mean:

    P1 = New perimeter = 40 - 24 = 16

    P1 = 2 w1 + 2 l1 = 2 ( w1 + l1 )

    16 = 2 ( w1 + l1 )

    Divide both sides by 2

    8 = w1 + l1

    w1 + l1 = 8

    w / 3 + 10 - w / 2 = 8

    Subtract 10 to both sides

    w / 3 + 10 - w / 2 - 10 = 8 - 10

    w / 3 - w / 2 = - 2

    2 w / 6 - 3 w / 6 = - 2

    - w / 6 = - 2

    Multiply both sides by - 6

    ( - 6 ) ∙ ( - w / 6 ) = ( - 2 ) ∙ ( - 6 )

    w = 12

    l = 20 - w = 20 - 12 = 8


    Proof:

    Original perimeter:

    P = 2 w + 2 l = 2 ( w + l ) = 2 ∙ ( 12 + 8 ) = 2 ∙ 20 = 40

    New lenth:

    l1 = l / 2 = 8 / 2 = 4

    New width:

    w1 = w / 3 = 12 / 3 = 4

    New rectangle will be the square. ( the square is a special case of the rectangle )

    New perimeter:

    P1 = 2 w1 + 2 l1 = 2 ( 4 + 4 ) = 2 ∙ 8 = 16


    The length of the original rectangle:

    l = 8

    Answer b

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    posted by Bosnian

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