Let $m > 1$ .

$\triangle{ABD} \sim \triangle{EFD} \implies m = \dfrac{\overline{BD}}{y_{1}} \implies \overline{BD} = my_{1}$

Using the Pythagorean theorem on $\triangle{ABD} \implies m^2y_{1}^2 = m^2 + (1 + y_{1})^2 \implies$

$(m^2 - 1)y_{1}^2 - 2y_{1} - (m^2 + 1) = 0 \implies y_{1} = \dfrac{m^2 + 1}{m^2 - 1} \implies$

$d_{1} = \overline{AD} = \dfrac{2m^2}{m^2 - 1} \implies A_{1} = \dfrac{m^3}{m^2 - 1}$ .

$d_{1} = \dfrac{2m^2}{m^2 - 1} \implies m^2 + (d_{1} + 1 + y_{2}^2)^2 = m^2y_{2}^2 \implies$

$(m^2 - 1)y_{2}^2 - 2(d_{1} + 1)y_{2} - (m^2 + (d_{1} + 1)^2) = 0 \implies$

$y_{2} = \dfrac{d_{1} + 1 + m\sqrt{(d_{1} + 1)^2 + m^2 - 1}}{m^2 - 1}$

$d_{2} = d_{1} + 1 + y_{2} = \dfrac{m}{m^2 - 1}((d_{1} + 1)m + \sqrt{(d_{1} + 1)^2 + m^2 - 1})$

Similarly $m^2 + (d_{2} + 1 + y_{3})^2 = m^2y_{3}^2 \implies$

$(m^2 - 1)y_{3}^2 - 2(d_{2} + 1)y_{3} - (m^2 + (d_{2} + 1)^2) = 0 \implies$

$y_{3} = \dfrac{d_{2} + 1 + m\sqrt{(d_{2} + 1)^2 + m^2 - 1}}{m^2 - 1} \implies$

$d_{3} = d_{2} + 1 + y_{3} = d_{2} + 1 + y_{3} = \dfrac{m}{m^2 - 1}((d_{2} + 1)m + \sqrt{(d_{2} + 1)^2 + m^2 - 1})$

In General using $\boxed{A_{1} = \dfrac{m^3}{m^2 - 1}}$

Let $\boxed{d_{1} = \dfrac{2m^2}{m^2 - 1}}$

and for $\boxed{n \geq 1}$ define:

$\boxed{d_{n + 1} = \dfrac{m}{m^2 - 1}((d_{n} + 1)m + \sqrt{(d_{n} + 1)^2 + m^2 - 1})}$ and $\boxed{A_{n + 1} = \dfrac{m}{2}d_{n + 1}}$

or if you prefer $d_{n + 1} = \dfrac{m}{m^2 - 1}((d_{n} + 1)m + \sqrt{d_{n}^2 + 2d_{n} + m^2})$ and $A_{n + 1} = \dfrac{m}{2}d_{n + 1}$

Using the above recursive formula with $m = 2$ the program below computes $A_{10}$ using python:

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import math;

d = 8/3;

for n in range(1,10):
       d = 2/3 * (2 * (d + 1) + math.sqrt(pow((d + 1),2) + 3));

print(d);

$\phantom 0$

$\implies A_{10} = \boxed{2473.00493680999}$ .