Do you know how to solve a system of one quadratic and one linear equation?

I literally just spent 1 hour writing the solution and clicked "Have the solution checked by a pro" and it took me to the Brilliant^2 site and the solution is gone. I'm too lazy to write it all again, so I'll just write the main things:

From the given hint, you notice that the legs of the two triangles are proportional.

Solving the proportion you'll conclude that 12(a+b)=ab.

From the Pythagorean theorem, we know that a^2+b^2=1225. We also know that $a^{2}+b^{2}=(a+b)^{2}-2ab$ so we can re-write it as (a+b)^2-2ab=1225

That gives us a system/set of two equations. We add replacements for easier solving: ab=x and a+b=y

Now the set of equations looks like this: $12y=x$ $y^{2}-2x-1225=0$

Plugging in the x from the first one into the second one, we get $y^{2}-24y-1225=0$ Solving this quadratic equation, we get that the only positive solution for y is 49. That makes x=588. That gives us a new set of equations. $a \times b=588$ $a+b=49$ Solving this, we get b=28 and a=21 (because b>a). Finally, we have: $b-a=28-21$ $\boxed{ b-a=7.00}$