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Math Help - Square and Circle

  1. #1
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    Square and Circle

    I'm trying to figure out the length of a side of a square:
    A square is inscribed in a circle of radius r. Express the length of the side of the square in terms of r.
    I tried to work it out this way:

    x^2+x^2=r^2 since the the lines insribed on the insides of a square are right angles right?

    2x^2=2r^2

    x^2=r^2

    x=r

    but the answer says r doesn't = x so what am I doing wrong here and how can I get the right answer?
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  2. #2
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    Quote Originally Posted by Kataangel View Post
    I'm trying to figure out the length of a side of a square:


    I tried to work it out this way:

    x^2+x^2=r^2 since the the lines insribed on the insides of a square are right angles right?

    2x^2= 2r^2

    x^2=r^2

    x=r

    but the answer says r doesn't = x so what am I doing wrong here and how can I get the right answer?
    You are close:

     x^2 + x^2 = r^2
    is equal to this:  2x^2 = r^2

    which can be changed to
     x^2 = \dfrac{r^2}{2}
    &
     x = \sqrt{ \dfrac{r^2}{2} }
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  3. #3
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    Quote Originally Posted by Kataangel View Post
    I'm trying to figure out the length of a side of a square:
    A square is inscribed in a circle of radius r. Express the length of the side of the square in terms of r.
    Take notice that the diagonals of the square have length 2r, a diameter.
    So 2s^2  = 4r^2 \, \Rightarrow \,s = r\sqrt 2
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by Kataangel View Post
    I'm trying to figure out the length of a side of a square:


    I tried to work it out this way:

    x^2+x^2=r^2 since the the lines insribed on the insides of a square are right angles right?

    2x^2=2r^2

    x^2=r^2

    x=r


    but the answer says r doesn't = x so what am I doing wrong here and how can I get the right answer?
    Hi Kataangel,

    I interpreted it this way.

    A square is inscribed in a circle of radius r. Express the length of the side x of the square in terms of r.

    Draw r from the center of the circle to a vertex on the square.

    Drop a perpendicular from the center of the circle to the side of the square.

    You have formed a 45-45-90 triangle with r as the hypotenuse.

    If x represents a side of the square, then x/2 will represent the sides of the isosceles triangle.

    Use the Pythagorean Theorem to represent r in terms of x.

    r^2=\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\r  ight)^2

    r^2=\frac{x^2}{4}+\frac{x^2}{4}

    r^2=\frac{2x^2}{4}

    r^2=\frac{x^2}{2}

    r=\frac{x}{\sqrt{2}}

    x=r\sqrt{2}
    Attached Thumbnails Attached Thumbnails Square and Circle-1.png  
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by Plato View Post
    Take notice that the diagonals of the square have length 2r, a diameter.
    So 2s^2 = 4r^2 \, \Rightarrow \,s = r\sqrt 2
    Yes, Plato, much easier your way.
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  6. #6
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    Ahh... I love this forum! =)

    I've been drooling over this prolbem for over 30 minutes, and couldn't have solved it without help from you guys. Thanks a lot!
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