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27 November, 18:06

The length of a rectangle is $3x+10$ feet and its width is $x+12$ feet. If the perimeter of the rectangle is 76 feet, how many square feet are in the area of the rectangle?

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  1. 27 November, 18:31
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    352ft²

    Step-by-step explanation:

    Given the length of a rectangle = 3x+10

    Width of the rectangle = x+12

    Perimeter of the rectangle = 76ft

    For us to get the area of the rectangle we need to get the value of its length and width.

    Given the perimeter of a rectangle P = 2 (L+W) where;

    L is the length and W is the width, on substitution

    76 = 2 (3x+10+x+12)

    76 = 2 (4x+22)

    76 = 8x+44

    8x = 76-44

    8x = 32

    x = 32/8

    x = 4

    Substituting x = 4 into the function of the length and width we have;

    Length = 3 (4) + 10

    Length = 22feet

    Width = 4+12

    width = 16feet

    Area of the rectangle = Length * Width

    Area of the rectangle = 22feet * 16feet

    Area of the rectangle = 352ft²
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