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Math Question?

The length of a rectangular pool is to be 2 times its width, and a sidewalk 6.5 feet wide will surround the pool. If a total area of 1809 ft2 has been set aside for construction, what are the dimensions of the pool?

6 Answers

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  • 1 month ago

    Let w ft be the width of the pool, then 2w

    will be the length. 1809 ft^2 is the total area

    of the pool & the sidewalk, you should first

    write down the true length & width of the

    rectangle for this area which are

    length=2w+2(6.5)

    width=w+2(6.5)

    & then write down an equation & solve for w

    & hence get the dimensions of the pool.

    Ans. w=20.5 ft; 2w=41 ft.

  • 1 month ago

    The length of a rectangular pool is to be 2 times its width, 

    and a sidewalk 6.5 feet wide will surround the pool. 

    If a total area of 1809 ft^2 has been set aside for construction, 

    what are the dimensions of the pool?

    (2w + 13)*(w + 13) = 1809

    w = 41/2, l = 41

    The length of the pool is 41 feet and its width is 20 1/2 feet.

  • 1 month ago

    A = 1809 ft²

    s = 6.5 ft

    l = pool length

    w = pool width

    l = 2w

    A = (2w + 2s)(w + 2s)

    1809 = (2w + 2(6.5))(w + 2(6.5)

    1809 = 2w² + 13w + 26w + 13²

    2w² + 39w - 1640 = 0

    If ax² + bx + c = 0 then x = [-b ± √(b² - 4ac)] / 2a

    w = [-39 ± √(39² - 4(2)(-1640))] / 2(2)

    w = -39/4 ± 121/4

    w = 20.5, -40

    Throw out the negative value which makes no sense.

    w = 20.5 ft

    l = 2w

    l = 2(20.5)

    l = 41 ft

  • 1 month ago

    Let w be the width.

    Let 2w be the length.

    The total width of the area is w + 2(6.5) --> w + 13

    The total length of the area is 2w + 2(6.5) --> 2w + 13

    The total  area is 1809 ft²:

    (w + 13)(2w + 13) = 1809

    Expand the left side:

    2w² + 13w + 26w + 169 = 1809

    2w² + 39w - 1640 = 0

    Factor that:

    (2w - 41)(w + 40) = 0

    That leads to two potential solutions:

    2w - 41 = 0

    2w = 41

    w = 20.5

    w + 40 = 0

    w = -40

    We can ignore the negative result.

    Calculate the length (double the width):

    2w = 2(20.5) = 41

    Answer:

    The pool is 20.5 feet wide and 41 feet long.

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  • 1 month ago

    pool

    width = w

    length = 2w

    entier = pool plus border sidwalk

    width = w + 2(6.5ft) = w + 13ft

    length = 2w + 2(6.5ft) = 2w + 13ft

    Area = 1809ft^2

    1809ft^2 = (w + 13ft)(2w + 13ft)

    1809ft^2 = 2w^2 + 39ftw + 169ft^2

    2w^2 + 39ftw - 1640ft

    w = 20.5ft and/or -40ft

    ignore negative width

    pool measurements

    w = 20.5ft

    l = 41ft

    verify

    (41 + 13)(20.5 + 13) = 1809ft

  • 1 month ago

    Let's say the dimensions of the pool are l and w. imagine a "frame" of 6.5 ft added around the pool. the dimensions of the new rectangle are l + 6.5 + 6.5 (6.5 on each side) and w + 6.5 + 6.5 or in short l + 13 and w + 13. We know the total surface is 1809, so (l + 13)(w + 13) = 1809.

    We also know that the length is twice the width, so we could say l = 2w. This means (2w + 13)(w + 13) = 1809. or 2w² + 39w + 169 = 1809 or 2w² + 39w - 1640 = 0.

    Solve this equation and you'll find your dimensions.

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