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Math Help - Rectangle in a square

  1. #1
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    Rectangle in a square

    The diagram shows a 1 by x rectangular plank which fits neatly inside a 10 10 square frame. What is the value of x?
    Attached Thumbnails Attached Thumbnails Rectangle in a square-2003-24.gif  
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  2. #2
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    Note there are 2 big 45-45-90triangles, 2 small triangles 45-45-90 triangles and a rectangle in the middle.
    Thus, the sum of all these areas is the area of big square which is 10x10=100.

    The area of the rectange in the middle is 1xX=X
    The area of the small triangles can easily be determined because they add up to a perfect square with diagnol of 1, thus the area is 1/2
    The area of the big triangles can easily be determined because they add up to a perfect square with diagnol of x thus thee area is (1/2)x^2

    Thus we have,
    \frac{1}{2}x^2+x+\frac{1}{2}=100
    Thus,
    x^2+2x+1=200
    Thus,
    (x+1)^2=200
    Thus,
    x=10\sqrt{2}-1
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  3. #3
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    Thanks . I had a question though, do you find those little triangles by going through this reasoning.

    We know the diagonals of these triangles is of 1 unit long then their sides are of lengths 1 / SQRT 2.

    Therefore one little triangle on its own has area 1/4 but both have area 1/2.
    Last edited by Natasha1; May 30th 2006 at 03:34 PM.
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  4. #4
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    Quote Originally Posted by Natasha1
    Thanks . I had a question though, do you find those little triangles by going through this reasoning.

    We know the diagonals of these triangles is of 1 unit long then their sides are of lengths 1 / SQRT 2.

    Therefore one little triangle on its own has area 1/4 but both have area 1/2.
    You can do that.

    I actually used a different reasoning. Since as I said the little triangles combine to give you a square. With a diagnol of 1 I used the following theorem. If the diagnols of a rhombus are d1 and d2 then is area is (1/2)d1d2 . Since a square is a rhombus and both its diagnols are equal we have its area as, \frac{1}{2}d\cdot d=(1/2)d^2
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  5. #5
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    Hello, Natasha!

    1) The diagram shows a 1 by x rectangular plank which fits neatly inside a 10 10 square frame.
    What is the value of x?
    Code:
          A   D         B
          * - * - - - - *
          | *1  *       |
        G *       * x   |
          | *       *   |
          |   *       * |
          |     *       * E
          |       *   * |
          * - - - - * - *
          D         F   C
    We see that \Delta DAG is a isosceles right triangle with hypotenuse DG = 1

    . . Then: AD = \frac{\sqrt{2}}{2}\quad\Rightarrow\quad DB = 10 - \frac{\sqrt{2}}{2}

    Since \Delta DBE is an isosceles right triangle with legs 10 - \frac{\sqrt{2}}{2}
    . . then: x\;=\;\sqrt{2}(DB) \;=\;\sqrt{2}\left(10 -\frac{\sqrt{2}}{2}\right) \;= \;10\sqrt{2} - 1
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  6. #6
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    Got it! Thanks. It works my way too
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