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Math Help - Trig with Surds

  1. #1
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    Trig with Surds

    I have a devilish problem for you, at least it was for me.

    There is a square, the length of its sides are unknown.
    Another, smaller, square is formed within the first square, its sides are connected by points placed at the mid point of each line of the first square.

    Visually it looks like a diamond in a square.

    You are told that a side (and thus all sides) of the smaller square are root 6 in length (a surd).

    The object is to define the perimeter of the first, larger, square in surd form.

    I started by defining the length between one of the vertices of the first square and the mid point of the line to the next (where it intersects with the smaller square) as n.

    It became apparent to me that root 2 x n squared was equal to root 6.
    It was also apparent that n over root 6 was equal to the inverse sin of 45 degrees.

    In spite of this I am still no closer to defining n in surd form. Obviously 8n would be equal to the perimeter of the large square.

    Please help...
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  2. #2
    o_O
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    So you mean the sides of the inside square are \sqrt{6} units long right?

    Well, you're right. You have 4 right angled triangles all constructed the same way and if each length from vertex to midpoint is called n then: 2n^{2} = \left(\sqrt{6}\right)^{2}.

    Is it the algebra you're having trouble with? Just isolate n^{2}:
    n^{2} = \frac{\left(\sqrt{6}\right)^{2}}{2}

    Take the square root of both sides. etc. ...
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  3. #3
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    Quote Originally Posted by o_O View Post
    So you mean the sides of the inside square are \sqrt{6} units long right?

    Well, you're right. You have 4 right angled triangles all constructed the same way and if each length from vertex to midpoint is called n then: 2n^{2} = \left(\sqrt{6}\right)^{2}.

    Is it the algebra you're having trouble with? Just isolate n^{2}:
    n^{2} = \frac{\left(\sqrt{6}\right)^{2}}{2}

    Take the square root of both sides. etc. ...
    meaning that I ought to end up with n = root of root 6 over 2?

    Ive been awake to long?

    EDIT Ive cheated and read the answer, it says n = root 3 but i dont see how to get it.
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  4. #4
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    doh, root 6 sq is equal to 6. thus its n sq = 6 over 2 or 3 and then root 3 is = n

    ok i really do just need sleep.
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