Problem on triangle and square

Add another 3 right-angled triangles, as shown. Then we see that $AF=\frac{1}{2}AH =\frac{1}{2} \times 7\sqrt{2}\approx 4.9497$ . Hence $\displaystyle\left\lfloor 1000x \right\rfloor= \boxed{4949}$ .

Since quadrilateral $ABFC$ is cyclic, we have by Ptolemy's theorem in cyclic quad. ABFC

$AF\times BC = AB \times CF + AC \times BF$

$\Rightarrow AF \times 5 = 3 \times \dfrac{5}{\sqrt{2}} + 4 \times \dfrac{5}{\sqrt{2}}$

$\therefore AF = \dfrac{7}{\sqrt2} \approx4.949.$

$\boxed{Q.E.D}$

Since, $ABC$ and. $BCF$ are right triangles and share common hypotenuse $BC,$ they can be inscribed in a Circle as shown. Let, $A= (-a, -b) , B= (-\dfrac {BC}{2} , 0) , C =(\dfrac {BC}{2},0) , F = (0, \dfrac {BC}{2})$

$A$ will lie in $3^{rd}$ quadrant if you plot this on a Cartesian plane. $BC$ will lie along X-axis and $FM$ along Y-axis. $M$ will be at Origin.

$Ar(ABC) = \dfrac {1}{2}\times AB \times AC = \dfrac {1}{2}\times BC \times AN$

$AC = 4 , AB = 3 , BC = \sqrt{AB^{2} + AC^{2}} = 5$

We get, $AN = 2.4 = b$

Since, $A$ lies on Circle it can be parametrically written as $(-a, -b) = (-2.5\cosβ,-2.5\sinβ)$

Therefore, $\sinβ = \dfrac {2.4}{2.5} \Rightarrow \cos β = \sqrt{1 - \sin^{2}β} = \dfrac {0.7}{2.5}$

$\Rightarrow a = 0.7$

$A = (-0.7 , -2.4)$ and $F = (0 , 2.5)$

So, $AF = \sqrt{(0.7 - 0)^{2} + (-2.4 - 2.5)^{2}}=\boxed{ 4.94974746831}$

Finally we get , $ANSWER:\boxed{4949}$