pre cal

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A rectangular parcel of land has an area of 4,000 ft2. A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. What are the dimensions of the land, correct to the nearest foot?

  • pre cal -

    x^2 + y^2 = diagonal^2 = d^2
    x y = 4000 so y = 4000/x
    d = x+10 so d^2 = x^2 + 20 x + 100

    x^2 + 16*10^6/x^2 = x^2 + 20 x + 100

    16*10^6 = 20 x^3 + 100 x^2

    5 x^3 + 25 x^2 - 4*10^6 = 0

    I do not see an easy way to solve that. Use iteration, x and 5x^3+25 x^2
    x = 10^2 --> 5,250,000 too big
    x = 80 --> 2,720,000 too small
    x = 90 --> 3,847,500 getting close
    x = 92 --> 4,333,568 very close
    x = 91 --> 3,974,880 between 91 and 92
    x = 91.5-> 4,039,610 between 91 and 91.5
    so 91 to nearest foot
    91^2 + y^2 = (91+10))^2
    8281 +y^2 = 10201
    y = 43.8 or 44
    so
    91 by 44

  • pre cal -

    let one side be x
    then the diagonal is x+10
    let the third side by y
    x^2 + y^2 = (x+10)^2
    x^2 + y^2 = x^2 + 20x + 100
    y^2 = 20x + 100
    y = √(20x+100)

    so the area of the triangle is half the rectangle area

    (1/2)xy =(1/2)(4000)
    xy = 4000
    x√(20x+100) = 4000
    x^2(20x+100) = 16000000
    20x^3 + 100x^2 - 16000000 = 0
    x^3 + 5x^2 - 800000 = 0

    I must admit that I used Wolfram to solve this
    http://www.wolframalpha.com/input/?i=x%5E3+%2B+5x%5E2+-+800000+%3D+0

    and I got x = 91.1947
    then y = 43.8622
    and the diagonal would be 101.1947



    check:
    area of rectangle = xy = (91.1947)(43.8622) = 4000.00017 (close enough)

    I will let you check if Pythagoras also works out, it does.

    I don't know what method you would have to solve that cubic, since it does not factor, and therefore does not have rational roots.
    Do you know Newton's Method ?

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