The rectangular solid below has surface area of 270 in². What is the value of x? ?

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  • 1 month ago

    SA = 2bh + 2bw + 2hw

    270 = 2(11)x + 2(11)x + 2x^2

    270 = 22x + 22x + 2x^2

    270 = 44x + 2x^2

    2x^2 + 44x - 270 = 0

    x^2 + 22x - 135 = 0

    (x - 5)(x + 27) = 0

    x = 5 , x = - 27

    The length of x is x = 5 in

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  • Philip
    Lv 6
    1 month ago

    Notation: rs = rectangular solid.

    Surface area of rs = [4(11x) + 2x^2] [in]^2 = 270 [in]^2.

    2x^2 +44x -270 = 0.

    x^2 +22x -135 = 0.

    (x-5)(x+27) = 0.

    Clearly, x = 5 [in] since x must be > 0.

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  • ?
    Lv 7
    1 month ago

    See below for work and solution.

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  • 1 month ago

     The surface area of the rectangular solid = 270 in^2

     2x^2 + 4*11x = 270 

     2x^2 + 44x - 270 = 0

     2(x^2 + 22x - 135) = 0

     2(x - 5)(x + 27) = 0

     x = 5

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  • Ian H
    Lv 7
    1 month ago

    2x^2 + 44x - 270 = 0     div by 2 and factor

    (x + 27)(X - 5) = 0 

    Check using x = 5, A = 50 + 44*5 = 270

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  • Anonymous
    1 month ago

    Sweet mother of god.  Hint: there are six sides.

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  • 1 month ago

    If you look at this as a rectangular solid with a square base, the surface area equation becomes:

    SA = 2s² + 4sh

    Using SA = 270, s = x and h = 11, we get:

    270 = 2x² + 4x(11)

    270 = 2x² + 44x

    0 = 2x² + 44x - 270

    0 = x² + 22x - 135

    This factors:

    0 = (x + 27)(x - 5)

    x = -27 and 5

    We can't have a negative side length, so throwing that out:

    x = 5

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  • fcas80
    Lv 7
    1 month ago

    The 6 sides have four areas of 11x, and two areas of x^2.

    4(11x) + 2x^2 = 270

    2x^2 + 44x - 270 = 0

    x^2 + 22x - 135 = 0

    x = - 27 (reject; can't be negative), x = 5

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