This is one of my all-time favourite maths challenge problems - can you solve it?

This is such a nice maths challenge problem - Q25 of the Junior Maths Challenge from 2001. Can you solve it?
This video includes questions from past UKMT maths challenge papers with the express permission of UK Mathematics Trust. The questions may not be copied or made available elsewhere without the permission of the UK Mathematics Trust except in accordance with their policy on the use of intellectual property which may be found here: bit.ly/UKMTIP
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[@ZachTheHuman]

This is interesting because I hit the answer in a different way I think. I looked at the picture and realized that if you push a black square further into a grey square the white overlap would grow meaning the grey square's area and the black square's area would both go down at the same rate making the difference between them constant. So you can just slide the black square with a side length of 9 into the 11 grey square and subtract those two areas and then do the same with the 5 square and the 7 square. (11^2-9^2)+(7^2-5^2)=64

[@MrToby11]

Cool question!

[@jickey6108]

Pretty cool

[@dogeidc8698]

I didn’t understand shit

[@person42069]

label overlaps a, b, and c. then area of grey is (area of square with side length 11 + area of square with side length 7) - a (from square with side 11) - b - c (from square with side 11). then to calculate black we need (area of square with side length 9 + area of square with side length 5) - a - b (from square with side 9) - c (from square with side 5). that is (11^2 + 7^2 - a - b - c) - (9^2 + 5^2 - a - b - c), and simplifying, we get 170 - a - b - c - 106 + a + b + c. the a, b and c cancel out, and we are left with 170 - 106 = 64