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:

a^2 + b^2 = (a + b)^2 - 2(a * b) = 100 - \frac{4*21}{2} = 58

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\times10=100 a^2 + ba + ab + b^2 = 100 a^2 + b^2 + 2ab = 100 a^2 + b^2 + 2\times21 = 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.