2019 Mock AMC 10B Problems/Problem 15

The area of the larger square is 144, so each side length is 12. $m + n = 12$ . Also, each of the smaller 4 triangles are right triangles, so $m^2+n^2= 108$ .

$n=12-m$

$m^2+(12-m)^2=108$

$m^2+144-24m+m^2=108$ .

Solving for $m$ gives $m= 12+\frac{\sqrt72}{2}$

$m = 6+3\sqrt{2}$

$a=6$ , $b=3$ , $c=2$ . $a+b+c=11$ . (A)

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2019 Mock AMC 10B Problems/Problem 15