Question 257032
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The diagonal of a square is the hypotenuse of an isosceles right triangle.  So if *[tex \Large d] is the measure of the diagonal and *[tex \Large s] is the measure of the side of the square which is the same as the measure of the legs of the triangle, then Pythagoras says:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{s^2\ +\ s^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ s\sqrt{2}]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ \frac{d}{\sqrt{2}}\ =\ \frac{d\sqrt{2}}{2}]


Hence, if the diagonal measures 32 meters, then the side of the square measures *[tex \Large 16\sqrt{2}].


The area of a square is the measure of the side squared (hence the term 'squared'), so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ s^2\ =\ \left(16\sqrt{2}\right)^2\ =\ 256\,\cdot\,2\ =\ 512\text{ m^2}]



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