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25 October, 03:19

The length of a rectangle is 1 ft more than twice the width, and the area of the rectangle is 66ft. Find the dimensions of the rectangle

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Answers (2)
  1. 25 October, 03:30
    0
    12 ft long by 5½ ft wide

    Step-by-step explanation:

    1. Set up an expression for the area.

    Let l = the length of the rectangle

    and w = the width. Then

    2w = twice the width and

    2w + 1 = 1 more than twice the width. Then

    l = 2w + 1

    The formula for the area of a rectangle is

    A = length * width

    A = lw

    66 = (2w + 1) w

    66 = 2w² + w

    2w² + w - 66 = 0

    2. Solve the quadratic for w

    2w² + w - 66 = 0

    (a) Multiply the first and last terms

    2 * (-66) = - 132

    (b) List all the factors of 132

    1 132

    2 66

    3 42

    4 33

    6 22

    11 12

    (c) Find a pair of factors whose product is - 132 and whose sum is 1.

    After some trial and error, you will choose - 11 and + 12,

    -11 * 12 = - 132 and - 11 + 12 = 1.

    (d) Rewrite w as - 11w + 12w

    2w² - 11w + 12w - 66 = 0

    (e) Factor by grouping

    w (2w - 11) + 6 (2w - 11) = 0

    (w + 6) (2w - 11) = 0

    (f) Use the zero product theorem

    w + 6 = 0 2w - 11 = 0

    w = - 6 2w = 11

    w = 5½

    We reject the negative answer, so w = 5½ ft

    3. Calculate l

    l = 2w + 1 = 2 * 5½ + 1 = 11 + 1 = 12 ft

    The rectangle is 12 ft long and 5½ ft wide.
  2. 25 October, 03:43
    0
    This is how I solve this problem
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