Question 192391
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Exactly the same way you would find the area or perimeter of a square with any other number as a side length.  I would simplify *[tex \LARGE  5\sqrt{12} = 5\sqrt{4\,\cdot\,3} = 10\sqrt{3}] first. Then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A = s^2 = (10\sqrt{3})^2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P = 4s = 4(10\sqrt{3})]


You can do your own arithmetic.



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