Geometry

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For a given box, the height measures 4 meters. If the length of the rectangular base is 2 meters greater than the width of the base and the lateral area L is 96 square meters, find the dimensions of the box.

  • Geometry -

    let the width be w
    then length = x+2

    SA = 2lw + 2lh + 2 wh

    2w(w+2) + 2(w+2)(4) + 2(w)(4) = 96
    2w^2 + 4w + 8w + 16 + 8w = 96
    w^2 + 10w - 40 = 0
    completing the square
    w^2 + 10w + 25 = 40+25
    (w+5)^2 = 65
    w +5 = ±√65
    w = -5 + √65 or x = -5 -√65 , the last being negative, thus resulting in "silly answer"

    w = -5+√65 = appr. 3.06 or whatever number of decimals you want

    so the width = 3.06
    lenth = 5.06
    height = 4

    check:
    2(3.06)(5.06) + 2(3.06)(4) + 2( 5.06)(4) = 95.92...
    (not bad)

  • Geometry -

    area = 2(lh + wh)
    you know that h=4, l=w+2, area=96

    96 = 2((w+2)(4) + 4(4))
    48 = 4w + 8 + 16
    w = 6

    the box is 6x8x4

  • Geometry - PS -

    hmm. forgot the w
    96 = 2((w+2)(4) + w(4))
    48 = 4w+8+4w
    40 = 8w
    w=5

    box is 5x7x4
    Lateral area is height*perimeter = 4*2(5+7) = 4*24 = 96

  • Geometry -

    I think Steve has an error

    2(5x7) + 2(5x4) + 2(7x4) ≠ 96

  • Geometry -

    Ahem. Lateral area does not include the bases.

  • Geometry -

    Thanks Steve, didn't know that

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