Question 289744
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If


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ =\ (51)^2\ +\ (140)^2]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ \sqrt{(51)^2\ +\ (140)^2}]


Just do the arithmetic. Use your calculator.  By the way, that is a perfect square, so correctly done, *[tex \Large c] will be an integer.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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