The area of the outer big square is $(a+b{)}^2=100$.

The area of the inner small square is $a^2+b^2$ (from Pythagoras on the triangles).

To calculate the inner square area, you take the outer square area, and subtract the area of the triangles:

58

$(a+b{)}^2=a^2+2ab+b^2=100$
$(a+b{)}^2-2ab=a^2+b^2=100-42$

Do I get a B+? Haha

$(a+b)(a+b)=10\times 10=100$
$a^2+ba+ab+b^2=100$
$a^2+b^2+2ab=100$
$a^2+b^2+2\times 21=100$
$a^2+b^2+42=100$
$a^2+b^2=58$

I started making a multi-part multi-day puzzle to solve Fermat's last theorem for $n=3$, but couldn't decide on how to properly divide it into meaningful parts as separate daily puzzles. That's why I ended up settling on this easy quick one for now, but hope to get back to Fermat in the coming days if I can find a way to approach things pedagogically.

58

Also, thank you for posting the link to the previous puzzle. I haven't been able to keep up but I want to go back and do some of the prior ones.