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Radius of a circle whose area is of 2 other cirlces

  1. Jan 3, 2012 #1
    1. The problem statement, all variables and given/known data

    --SOLVED--

    It seems, that you just apply the pythageos thereom to 'A' and 'B' =)


    ---------------
    Lets say I have a circle 'A' with radius 6cm, and a circle 'B' with radius 6cm. These two circles would have an area of 113.097cm^2. When both are combined they would have an area of 226.194cm^2.

    Is there a way that you can calculate the new radius (or diameter) without having to get the area -> divide by PI -> take the square root? (which would equal 8.485cm^2 in this case)

    Also, how would this problem be handled with 2 different inital radiuses? eg. Circle 'A' has radius 5cm, and circle 'B' has radius 11cm.

    Thankyou

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jan 3, 2012
  2. jcsd
  3. Jan 3, 2012 #2
    It's not Pythagoras' theorem. It's simply that area is proportional to the square of a characteristic length dimension when scaling a shape. (Not restricted to circles). So if I scale a shape up by a factor of 3 in lengths, the area increases by a factor of 9.
     
  4. Jan 3, 2012 #3

    Mentallic

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    Homework Helper

    So what you're looking for is to add the area of two circles with varying radii, say, r1 radius for the first circle and r2 for the second.

    The area of each circle is

    [tex]A_1=\pi r_1^2[/tex]

    [tex]A_2=\pi r_2^2[/tex]

    and the sum of these is

    [tex]A_1+A_2=\pi (r_1^2+r_2^2)[/tex]

    You want this to be equal to a third circle A3 with unknown radius r3 so what you're looking for is to solve for r3 in the following equation:

    [tex]A_1+A_2=A_3=\pi (r_1^2+r_2^2)=\pi r_3^2[/tex]

    So, all you need to do is solve for r3 in

    [tex]\pi (r_1^2+r_2^2)=\pi r_3^2[/tex]

    and you have yourself a formula for converting the area of 2 varying circles into 1. You can also extend this idea to convert any number of circles into 1.

    Edit: I've never actually noticed it before, but yes, you're right about the pythagorean formula being applied here :smile:
     
  5. Jan 3, 2012 #4

    Curious3141

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    Homework Helper

    And the way in which you can "see" that is to think about the implication of Pythagoras' Theorem for areas of squares whose side is equal to each of the lengths of a right triangle, and draw incircles in each of those squares. The areas of the incircles are also related to one another as the the areas of the squares to one another. Each of the incircles has a diameter equal to the respective side of the right triangle. And therefore each of the incircles has an area which is related to the same constant of proportionality (1/4) to the larger circles whose radius is equal to the respective side; the proportionality constants cancel out, giving the same inter-relationship between the areas of these larger circles as the smaller circles and the squares adjoining the sides, which is specified by the Pythagorean Theorem.
     
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