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Each of the overlapping areas contributes to the area of exactly two squares. So the total area of the three squares is equal to the area of the non-overlapping parts of the squares plus twice the total of the three overlapping areas, i.e. $(117 + 2(2 + 5 + 8))\;\mathrm{cm}^2 = (117 + 30)\;\mathrm{cm}^2 = 147\;\mathrm{cm}^2$.
So the area of each square is $(147 \div 3)\;\mathrm{cm}^2 = 49\;\mathrm{cm}^2$. Therefore the length of the side of each square is $7\;\mathrm{cm}$.
Intersecting Squares
Age 11 to 14
ShortChallenge Level 
Each of the overlapping areas contributes to the area of exactly two squares. So the total area of the three squares is equal to the area of the non-overlapping parts of the squares plus twice the total of the three overlapping areas, i.e. $(117 + 2(2 + 5 + 8))\;\mathrm{cm}^2 = (117 + 30)\;\mathrm{cm}^2 = 147\;\mathrm{cm}^2$.
So the area of each square is $(147 \div 3)\;\mathrm{cm}^2 = 49\;\mathrm{cm}^2$. Therefore the length of the side of each square is $7\;\mathrm{cm}$.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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