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-   -   Are of square, inside of a circle (https://www.askmehelpdesk.com/showthread.php?t=173476)

  • Jan 17, 2008, 11:00 AM
    princessgloomy
    Are of square, inside of a circle
    A square is inscribed in a circle of radius 3(square root sign)2. Find the area of the square.


    First off, Idk how to put a square root symbol up on this question haha.

    Second, I get the radius. I just did 18x3.14

    But I don't know how to get the area of the square if the only measurement given is the radius of the circle. :eek:
  • Jan 17, 2008, 12:29 PM
    Capuchin
    The radius of the circle is equal to half the diagonal of the square. Use trigonometry to find the length of the side of the square. From there it's easy. Or you canjust use the area of a triangle once you have the length of the diagonal.
  • Jan 17, 2008, 01:32 PM
    princessgloomy
    So the diagonal of this square is 6?
  • Jan 17, 2008, 01:44 PM
    Capuchin
    no it's
  • Jan 17, 2008, 01:59 PM
    inthebox
    pythagorean

    A squared + B squared = C squared

    C = 6,square root of 2


    figure it out from there
  • Jul 29, 2008, 02:49 AM
    antra786
    I do agree with this answer and it is absolutely right answer.
    _________________________
    Antra
    Wide Circles
  • Jul 30, 2008, 02:20 AM
    bert501
    The problem is interesting. I tried to solve it. But I got confused that whether the diagonal has any relationship with the side of the square. So if you try to solve this I may proceed with the problem.
    ------------------------
    Bert

    Wide Circles
  • Aug 2, 2008, 03:11 AM
    Unknown008
    Or, without the pythagoras theorem... a square is composed of two triangles if you cut it through the diagonal, in two. You have the base, , and perpendicular height. Using half base times height, you get one triangle, and multiplying by two, you have the area of the square.

    Or simpler, if you 'mix' the two steps, you have the area of the square as: diameter of circle x radius of circle.
  • Jul 30, 2010, 02:53 AM
    kinjalpanwala
    A square is inscribed in a circle of radius 3(square root sign)2. Find the area of the square.
    radius r = 3√2
    using the trigonometry you can find out the area of square as following:
    for squares,
    a² + b² = c²
    but a = b (for squares)
    so 2a² = c²

    but c = 2r
    so 2a² = (2r)²
    so a² = 2r²

    but r = 3√2
    so a² = 36.
  • Jul 30, 2010, 05:18 AM
    galactus

    Why would you answer a question that is 2 years old? It is likely a moot point by now.
  • Dec 15, 2010, 08:37 AM
    bamal
    galactus - I was searching for help with a similar problem when I came across this forum so I'm glad that kinjalpanwala answered a 2 year old question!
  • Oct 1, 2011, 09:01 AM
    twingleton
    Well, galactus, since my son has the same question TODAY, it's not moot to me! 8) His question is: the area of the square is 50, what's the area of the circle? We get 12.5 times pi, but the book says it's 25 times pi.
  • Oct 1, 2011, 10:46 AM
    Unknown008
    twingleton, just start another thread to ask your question please.

    But as it is your first time here, I'll try to help you.

    The area of the square is 50, the length of one side is therefore

    The radius of the circle is then

    This agrees with your answer. Or maybe your misread the question?

    If the square is inscribed inside the circle instead of the other way round, then, you have the radius of the circle as

    And there goes the 25pi.
  • Oct 26, 2011, 08:33 AM
    kb5uew
    This page was helpful in solving the problem even though 2 years old.

    A REAL practical application: I have a container with a diameter of 2.5" used to etch a circuit board. What is the maximum size of square circuit board that I can fit into the container? (1.78")
  • Oct 26, 2011, 08:48 AM
    Unknown008
    Is the height 1.78"?

    If so, you already have know that the square board will have to have a length of 1.78" or shorter. The diameter is even larger that that, so I see no problem for the board to fit in, nor the need to bend the board.

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