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Math Help - Two equally big squares

  1. #1
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    Two equally big squares

    I'm kind of lost on this one. Could someone please help me.



    Two equally big squares with the sides 12 cm partly covers each other as the figure shows. One of the squares corner is in the other squares center. Decide the area of the shadowed part.


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  2. #2
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    Quote Originally Posted by kevin3000 View Post
    I'm kind of lost on this one. Could someone please help me.



    Two equally big squares with the sides 12 cm partly covers each other as the figure shows. One of the squares corner is in the other squares center. Decide the area of the shadowed part.


    If the sides of both squares were parallel to the other, then then shaded area would be 1/4 of the total square area.

    If the partially tilted square were at a 45 degree incline, then you could see that again it is 1/4 of the total shaded area.

    You need to determine an additional intermediate area to state that the area is static and does not ossilate as the squares rotate.
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  3. #3
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    thank U. Now I've succeded to solve it
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  4. #4
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    Quote Originally Posted by kevin3000 View Post
    thank U. Now I've succeded to solve it
    Since you've solved it but haven't explained it
    here is my version.

    Name the squares 'upper square' and 'lower square'.

    Draw the squares so that the sides of the squares are
    parallel to each other. The shaded region is
    in the lower-right quadrant of the upper square,
    and in the higher-left quadrant of the lower square.

    The distance from the centerpoint of the upper square
    to any side of the upper square is 6 or label it 'h'.
    Restated: the distance from the center of the upper
    square to the lower or bottom side of the square is
    distance h, which is also the distance from the center
    point to the right side of the upper square.
    That is re-stating what should be obvious.


    Cut the shaded region into two triangles.
    Cut from the center of the upper-square to
    the lower-right corner of the upper square.


    You have two shaded triangles.
    On lower-left triangle, label the vertical side
    as 'h' and label the bottom as 'b'

    On the upper-right triangle label the horizonal
    distance as 'h' and the vertical distance is the same
    as 'c'. (It is the same length as 'b'.)


    The area of the lower-left triangle is:  \frac {h \times b}{2}

    The area of the upper-right triangle is:  \frac {h \times c}{2}

    Since b=c the two triangles have the same area which
    sums to the area of the shaded square region.

    NOW.
    Rotate the lower square counter-clockwise through a small angle theta.
    We need to calculate the area lost by the lower-left triangle.

    The distance along the bottom edge of the upper-square is 'd'
     d = h\times \tan \theta

    The area lost is  \frac {h \times d}{2}



    The area gained by the upper-right triangle:
     d = h\times \tan \theta

    When rotated,
    The area lost by the lower-left triangle will always be
    equal to the area gained by the upper-right triangle,
    thus the area is constant.
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