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Math Help - another difficult problem~~proving~~so hard >.<

  1. #1
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    another difficult problem~~proving~~so hard >.<

    a rectangle ABCD is cut into eight squares as shown on the pix. find the perimeter of the rectangle ABCD if the perimeter of the smallest square is 24cm


    btw the required pix is attached

    thanks guys!!! u rok
    Attached Thumbnails Attached Thumbnails another difficult problem~~proving~~so hard &gt;.&lt;-mathenrichpix.jpg  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fring
    a rectangle ABCD is cut into eight squares as shown on the pix. find the perimeter of the rectangle ABCD if the perimeter of the smallest square is 24cm


    btw the required pix is attached

    thanks guys!!! u rok
    The diagram does not show eight smaller squares it shows eight rectangles.


    However if we suppose this is just badly drawn and that they should be
    squares, then the bottom right square is 3x3 of the smallest square. The
    one above it is 4x4 of the smallest sqare, so the height of the rectangle
    id 7 times the height of the smallest square.

    Similarly the length of the rectangle is four times twice the length of a side
    of the smallest square plus three times the side of the smallest square.

    Hence your rectangle has a height of seven times the side of the smallest
    square, and a length of eleven times the side of the smallest square.

    Therefore the perimiter of the rectangle is 2x7+2x11 times the side of the
    smallest square. Since you are told the perimiter of the smallest square you
    can find its side, and so the final answer.

    RonL
    Last edited by CaptainBlack; August 1st 2006 at 02:26 AM.
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  3. #3
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    Hello, fring!

    A rectangle ABCD is cut into eight squares as shown on the pix.
    Find the perimeter of the rectangle ABCD
    if the perimeter of the smallest square is 24cm
    Code:
                                     (1) 
                        a             ↓       b
          o - - - - - - - - - - - - - o - - - - - - - o
          |                           |               |
          |                           |               |
          |                           |               |
          |                           |b              |b
         a|                           |               |
          |                           |               |
          |                           | 6       c     |
          |                           o - o - - - - - o ← (2)
          |                          6|   |6          |
    (4) → o - - - o - - - o - - - o - o - o           |
          |       |       |       |       |           | c
         d|       |       |       |       |d          |
          |       |       |       |       |           |
          o - - - o - - - o - - - o - - - o - - - - - o
              d       d       d       d   ↑     c
                                         (3)

    At (1) we see that: . a\:=\:b+6

    At (2) we see that: . b\:=\:c+6

    At (3) we see that: . c\:=\:d + 6

    . . Hence: . a\:=\:d+18 [1]


    At (4) we see that: . 4d\:=\:a + 6\quad\Rightarrow\quad a\:=\:4d - 6 [2]


    Equate [1] and [2]: . d + 18\:=\:4d - 6\quad\Rightarrow\quad \boxed{d = 8}

    Now you can determine a,\,b,\,c and the perimeter of the rectangle.

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  4. #4
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    8 Squares

    Let,
    length of each side of the smallest square = a
    " " " " " " 3rd largest square = b
    Then, we have
    length of each side of the 2nd largest square = a+b
    " " " " " " 2nd smallest " = b-a
    " " " " " " largest " = 4b-5a
    Now, since ABCD is a rectangle,
    therefore, AD=BC
    =>(4b-5a)+(b-a) = (a+b)+b
    =>b = 7a/3------------------I
    Given,
    perimeter of the smallest square = 24cm.
    =>4a = 24cm.
    =>a = 6cm.
    Putting the value of a in eq.I, we have
    b = 14cm.
    Now, we can calculate the perimeter of ABCD.
    It comes out to be 160cm.
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