Answer: 462

Small Hint:

Let the legs be $a$ and $b$ and use the first square to relate $ab$ to $a+b$

Big Hint:

For the second square, use the altitude to the hypotenuse and similar cross-sections

Solution:

Put $p=a+b,$ $q=ab,$ and let the hypotenuse be $c.$ Since $S_1$ has side $21,$ the standard leg-aligned-square relation gives $21=\frac{ab}{a+b},$ so $q=21p.$ Hence $c^2=p^2-2q=p(p-42).$

The altitude to the hypotenuse is $h=\frac{q}{c}.$ If the side of $S_2$ is $t,$ similarity gives $t=\frac{ch}{c+h}.$ Substitution simplifies this to $t=\frac{21c}{p-21}.$ Therefore $$440=\frac{441p(p-42)}{(p-21)^2}.$$ Since $p(p-42)=(p-21)^2-441,$ this becomes $$440=441\left(1-\frac{441}{(p-21)^2}\right).$$ Thus $(p-21)^2=441^2,$ and $p>42$ gives $p=462.$