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Help Me With This Problem.

  1. Dec 14, 2004 #1
    The Area Of These 2 Adjacent Squares Are 4 cm^2 and 196 cm^2.

    Find The Length Of The Segment Joining The Centers Of Their Inscribed Circled

    Attached Files:

  2. jcsd
  3. Dec 14, 2004 #2
    it's the square root of four plus the square root of 196 and all that over two
  4. Dec 14, 2004 #3
    does anyone agree/disagree?
  5. Dec 14, 2004 #4
    I disagree.

    Hint: Construct a right-angled triangle with the connecting segment as its hyp
  6. Dec 15, 2004 #5
    tell me step by step........
  7. Dec 15, 2004 #6


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    It will be the hypotenuse of a triangle whose sides are (14-2)/2 and (14+2)/2.
    Last edited: Dec 16, 2004
  8. Dec 16, 2004 #7
    make a triangle using the line as the hypotenuse

    side 1:

    Get the radius of the big circle and subtract the radius of the small circle. Giving you 7 - 1 = 6

    side 2:

    Get the radius of the big and the small circle.
    Giving you 7 + 1 = 8

    Hopefully you can solve the problem now.
    Last edited by a moderator: Dec 16, 2004
  9. Dec 16, 2004 #8
    let me see ..
    draw two lines, one from each center perpendicular to the bottom side (as in the figure ) Then draw a line parallel to the bottom line through the center of the smaller circle.. Then u will see a
    right angled triangle formed with hypotenuse being the line connecting the two centers ...
    u can now very easiy find the length of thwo sides and hence that of the hypotenuse ..

    answer = sqrt (6^2+8^2) = 10
  10. Dec 21, 2004 #9
    youre making it all more complicated, the center of an inscribed circle is righ in the center of the square so the distance from the center to one side is the half of what the whole length of the square is so you need to square root the areas given to find the lengths divide each length by two, and add them together, the answer should be nine
  11. Dec 21, 2004 #10


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    That's simple alright! It just suffers from the minor problem of being wrong.

    It would be correct IF the line through the denters were parallel to the sides- but that's not true.
  12. Dec 30, 2004 #11
    hmnn so the squares make more of a stairstep and not a pyramid.
  13. Dec 30, 2004 #12


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    What pyramid??This is plane geometry...
    My answer is
    [tex] d=\sqrt{106} [/tex]
    The question is:"how did I get that??"

  14. Dec 30, 2004 #13


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    Good method,bad answer.

    1.Wrong answer,wrong method.
    2.There's more common the expression "perimeter of a square".Similar,the"circumference of a circle".

  15. Dec 30, 2004 #14
    No your answer was wrong... Mahesh's was the right one.

    Here's where you went wrong:

    106 = 9^2 + 5^2

    wrong numbers. You forgot that the radius of the smaller circle is only 1 not 2.
    Thus 7-1 = 6 not 7-2.

    As well 7+1 = 8 not 7+2.

    I agree with Mahesh's answer...
  16. Dec 30, 2004 #15


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    Yeah,you're right,sorry,it must be because it's almost 2 am. :blushing:
    Advice:don't do math at 2 am. :tongue2:

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