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i need to determine position of circle inside square,let us suppose that we have following picture

enter image description here

we have following informations:

1.$ABCD$ is square

2.all small figures ,$KMCE$,$PKEF$,$NPFD$ are square as well

3.diamter of small circle is equal to $6$ cm

problem:

we have to find area of $ABCD$ square

problem what i have is that i dont know how to express small sides of small squares in terms of diameter or radius or even find relationship between big square and circle.clearly it is not obvious that center of circle is located on intersection points of diagonals,please help me ,maybe it is very simple and i dont see key fact,but please give me a hint and i will try to get point of main idea of solution of this problem.thanks in advance

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  • $\begingroup$ You should try to express the side of the small squares as a function of the side of the big square, then you can deduce $BM$, which is equal to the diameter of the circle. $\endgroup$ – zuggg Jul 17 '13 at 5:49
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Call the sidelength of the small square $a$. Then, $|CD| = 3a$. This implies $|AD| = 3a$, and $|AN| = |AD|-|DN| = 2a$. The diameter of the small circle is thus $2a$. Now, solve $2a = 6$ to find $a$, and ...

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  • $\begingroup$ ui so simple,i think it is morning and brain has not working yet properly :D $\endgroup$ – dato datuashvili Jul 17 '13 at 5:58
  • $\begingroup$ I recommend coffee :) $\endgroup$ – Lord Soth Jul 17 '13 at 5:59
  • $\begingroup$ :D yes you are right,better Jacobs Monarch :D $\endgroup$ – dato datuashvili Jul 17 '13 at 6:00
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Area of Square is 81 cm*cm

Solution-->

 suppose ND=m then PF=KE=CM=ND   (because all PF, KE, CM, ND all four line inside parallel line MN and CD)

now we can say

 AD=6+m   ------eq(1)
 CB=6+m   ------eq(2)

now 

  CD=m+m+m
  CD=3m    ------eq(3)

we know that

 AD=CD      (lines of square)

6+m=3m
3m-m=6
2m=6
m=3

now put the value of m in eq(1)

AD=6+m=6+3=9
AD=9
AD=CD=BC=AB=9

Area of Square is 
   AD*AD =9*9=81 cm*cm
hence area of give square is 81 cm*cm
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