2008 AMC 10B Problem 15

From $a^2+b^2=(b+1)^2$ we get $a^2=2b+1,$ so $a$ is odd and $a^2$ is an odd perfect square. Since $b\lt 100,$ we need $a^2=2b+1\lt 201,$ and $a^2\ge 9$ for $b\ge 4.$ The odd squares $9,25,49,81,121,169$ give $a=3,5,7,9,11,13,$ which is $6$ triangles. Thus, the correct answer is A.