So Many Circles

Consider one of the 25 congruent right triangle $\triangle PQR$ , where $PQ=QR =1$ and $\angle PQR = 90^\circ$ . Let the center of the incircle be $O$ , the extension of the line $QO$ will meet the hypotenuse $PR$ at a right angle at $M$ . The length of $MQ=\frac {1}{\sqrt{2}}$ . Let the radius of the incircle be $r$ , then we have:

$\quad MQ = \dfrac {1}{\sqrt{2}} = \sqrt{2}r+r = (\sqrt{2}+1)r\quad \Rightarrow r = \dfrac {1}{\sqrt{2}(\sqrt{2}+1)} = \dfrac {1}{2 + \sqrt{2}}$

The required answer is $25r = \dfrac {25}{2 + \sqrt{2}} \approx \boxed {7.32}$

r = (a+b-c)/2

r = (1+1-sqrt of 2)/2

r= 0.29 multiply by number of circles (25)

r= 7.32