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Since a^2 + b^2 = 50 and a, b are (nonnegative) integers, you know that neither a nor b can be more than 8 (since 7^2 = 49 is the biggest square that "fits" within the 50-area constraint.

That immediately should show you that a=7, b=1 works. 49 + 1 = 50.


The other one is equally easy: a=5, b=5. 25+25=50.


So (7,1) and (5,5) are the two combinations of side lengths.

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Q: You are told to draw two squares so that the sum of their areas is 50 square inches and the side length of each square is a whole number of inches There are two combinations of side lengths for the t?
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