1. Subscribertalzamir
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    10 Mar '14 20:17
    This is trivial enough, but I'll probably use it in a high school someday.

    A rectangle has circumference length of p. The diagonal has length d. What is the area of the rectangle?
  2. Subscriberjoe shmo
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    11 Mar '14 21:233 edits
    Originally posted by talzamir
    This is trivial enough, but I'll probably use it in a high school someday.

    A rectangle has circumference length of p. The diagonal has length d. What is the area of the rectangle?
    I don't know if I did this the hard way, because it didn't seem obvious to me...

    Let the sides of the rectangle be x, and y respectively.

    Perimeter = P = 2*x + 2*y ....eq (1)

    from the Pythagorean theorem

    d² = x² + y² ...eq(2)

    from eq(1)

    y = 1/2*P - x ...eq(1)'

    from eq(2)

    y = Sqrt ( d² - x² )

    Set eq(1)' = eq(2)

    1/2*P - x = Sqrt ( d² - x² )

    solve for x in terms of d, and P by completing the square

    x = Sqrt(d²/2 - 1/16*P² ) + P/4 ...eq(3)

    Sub eq(3) into eq(1)'

    y = 1/2*P - [Sqrt(d²/2 - 1/16*P² ) + P/4]

    = 1/4*P - Sqrt(d²/2 - 1/16*P² )

    then Area = A = x*y

    = (P/4 + Sqrt(d²/2 - 1/16*P² ))(P/4 - Sqrt(d²/2 - 1/16*P² ))

    which is of the form (a + b)(a - b) = a² - b² : giving after simplification

    A = P²/8 - d²/2 ***
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    11 Mar '14 22:462 edits
    Originally posted by joe shmo
    Let the sides of the rectangle be x, and y respectively.

    Perimeter = P = 2*x + 2*y ....eq (1)

    from the Pythagorean theorem

    d² = x² + y² ...eq(2)
    What you did works well.

    Starting with what I have quoted above there is a neat trick which avoids needing to complete the square.

    If you square both sides of eq(1) you get something that involves a multiple of x² + y² (which is d² ) and a multiple of xy (which is the area we want).

    Then rearrange to get your result.
  4. Subscriberjoe shmo
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    11 Mar '14 23:14
    Originally posted by Diapason
    What you did works well.

    Starting with what I have quoted above there is a neat trick which avoids needing to complete the square.

    If you square both sides of eq(1) you get something that involves a multiple of x² + y² (which is d² ) and a multiple of xy (which is the area we want).

    Then rearrange to get your result.
    Wish I would have saw that,...by far less work!
  5. Subscribertalzamir
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    12 Mar '14 07:33
    Nicely done. And the squaring trick is what I was thinking too. ๐Ÿ™‚

    2x + 2y = p => x + y = p/2 => x^2 + 2xy + y^2 = p^2/4

    x^2 + y^2 = d^2

    combine for p^2/4 + 2xy = d^2 => xy = d^2/2 - p^2/8 = A
  6. Subscribersonhouse
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    21 Mar '14 11:422 edits
    Originally posted by talzamir
    Nicely done. And the squaring trick is what I was thinking too. ๐Ÿ™‚

    2x + 2y = p => x + y = p/2 => x^2 + 2xy + y^2 = p^2/4

    x^2 + y^2 = d^2

    combine for p^2/4 + 2xy = d^2 => xy = d^2/2 - p^2/8 = A
    It's funny, the first numbers I chose showed me something about rectangles:

    I chose d=10 and p=20 where A = zero๐Ÿ™‚

    which makes sense๐Ÿ™‚

    It also means you can get negative numbers: D=10 and P=22, A = minus 60.5)

    How can you fix it so that doesn't happen?

    Exploding ink?๐Ÿ™‚
  7. Subscriberjoe shmo
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    21 Mar '14 22:121 edit
    Originally posted by sonhouse
    It's funny, the first numbers I chose showed me something about rectangles:

    I chose d=10 and p=20 where A = zero๐Ÿ™‚

    which makes sense๐Ÿ™‚

    It also means you can get negative numbers: D=10 and P=22, A = minus 60.5)

    How can you fix it so that doesn't happen?

    Exploding ink?๐Ÿ™‚
    You happened to pick your variables outside of the domain of the equation.

    In order for the equation to give valid results ( that is results that are physical ) the area of the rectangle must be strictly greater than zero. For if it were equal to zero, you don't have a rectangle.

    So;

    A > 0

    P²/8 - d²/2 > 0

    P²/8 > d²/2

    P² > 4*d²

    P > 2*d (this is the domain of the equation), which say that all perimeters must be greater than twice the diagonal.

    You picked a perimeter that was exactly equal to twice the diagonal, which is out of the equations domain. See?
  8. Subscribersonhouse
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    22 Mar '14 01:29
    Originally posted by joe shmo
    You happened to pick your variables outside of the domain of the equation.

    In order for the equation to give valid results ( that is results that are physical ) the area of the rectangle must be strictly greater than zero. For if it were equal to zero, you don't have a rectangle.

    So;

    A > 0

    P²/8 - d²/2 > 0

    P²/8 > d²/2

    P² > 4*d²

    P > 2*d ...[text shortened]... imeter that was exactly equal to twice the diagonal, which is out of the equations domain. See?
    I knew that, just pulling your chain๐Ÿ™‚
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