geometry

let TC=y, and break up $\sqrt{85}$ into 2 parts: $n_1,n_2$ let BT= $n_1$ , let TA = $n_2$ .................. lets use Pythagorean theorem. both BTC and ATC are right-angled triangle.. we get $n_1^{2}+ y^{2} =9^2, n_1= \sqrt{81-y^2}$ $n_2^{2}+ y^{2} =\sqrt{40}^2, n_2= \sqrt{40-y^2}$ we get $n_1+n_2= \sqrt{81-y^2}+\sqrt{40-y^2}$ since $n_1,n_2$ is broken from $\sqrt{85}$ , $n_1+n_2=\sqrt{85}$ we get $\sqrt{85}= \sqrt{81-y^2}+\sqrt{40-y^2}$ square both sides $85 = 81 -y^2 +40-y^2 +2\sqrt{(40-y^2)(81-y^2)}$ $85 = 121-2y^2 +2\sqrt{3240-121y^2+y^4}$ $-36 = -2y^2 +2\sqrt{3240-121y^2+y^4}$ $2y^2-36 =2\sqrt{3240-121y^2+y^4}$ $\frac{2y^2-36}{2} =\sqrt{3240-121y^2+y^4}$ $y^2-18 =\sqrt{3240-121y^2+y^4}$ square both sides $y^4 -36y^2 +324=3240-121y^2+y^4$ $-36y^2 +324=3240-121y^2$ $85y^2 +324=3240$ $85y^2=2916$ $y^2 = \frac{ 2916}{85}$ $y =\sqrt{ \frac{2916}{85}}$ so, A= 2916, b= 85, A+B=2916+85= $\boxed{3001}$