I4.3 In Figure $1, S_{1}$ and $S_{2}$ are two different inscribed squares of the right-angled triangle $A B C$. If the area of $S_{1}$ is $40 b+1$, the area of $S_{2}$ is $40 b$ and $A C+C B=c$, find the value of $c$.
Add the label $D, E, F, P, Q, R, S$ as shown. $C D E F, P Q R S$ are squares.
Let $D E=x, S R=y$, then $x=\sqrt{441}=21, y=\sqrt{440}$. Let $B C=a, A C=b, A B=c=\sqrt{a^{2}+b^{2}}$.
Let the height of the triangle drawn from $C$ onto $A B$ be $h$, then $a b=c h=2$ area of $\Delta$. $\qquad$
$\triangle A F E \sim \triangle A C B: \quad \frac{b-x}{b}=\frac{x}{a} \Rightarrow x=\frac{a b}{a+b}=21$ $\qquad$
$\triangle C S P \sim \triangle C A B: \frac{\text { height of } \triangle \mathrm{CSP} \text { from } \mathrm{C}}{S P}=\frac{h}{c} \Rightarrow \frac{h-y}{y}=\frac{h}{c} \Rightarrow y=\frac{c h}{c+h}=\sqrt{440}$
$\operatorname{By}(*), \frac{a b}{c+\frac{a b}{c}}=\sqrt{440} \Rightarrow \sqrt{440}=\frac{a b \sqrt{a^{2}+b^{2}}}{a^{2}+a b+b^{2}}$
From (1) $a b=21(a+b)$ $\qquad$ (3), sub. (3) into (2):
$$
\sqrt{440}=\frac{21(a+b) \sqrt{(a+b)^{2}-2 a b}}{(a+b)^{2}-a b}=\frac{21(a+b) \sqrt{(a+b)^{2}-42(a+b)}}{(a+b)^{2}-21(a+b)}=\frac{21 \sqrt{(a+b)^{2}-42(a+b)}}{(a+b)-21}
$$

Cross multiplying and squaring both sides:
$$
\begin{array}{l}
440\left[(a+b)^{2}-42(a+b)+441\right]=441\left[(a+b)^{2}-42(a+b)\right] \\
(a+b)^{2}-42(a+b)-440 \times 441=0 \\
(a+b-462)(a+b+420)=0 \\
A C+C B=a+b=462
\end{array}
$$