Question 772415
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The diagonal of a square is the hypotenuse of an isosceles right triangle with two adjacent sides of the square forming the legs of the triangle.  Let *[tex \LARGE x] represent the measure of the given diagonal.  Let *[tex \LARGE s] represent the measure of the side of the square.  Mr. Pythagoras said, if you recall, that the square of the hypotenuse, *[tex \LARGE x^2] in this case, is equal to the sum of the squares of the other two sides, that is *[tex \LARGE s^2\ +\ s^2] in this case, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 2s^2]


Multiply by *[tex \LARGE \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{2}\ =\ s^2]


But *[tex \LARGE s^2] is the area of the square, so to find the area of a square given the measure of a diagonal, square the diagonal and divide the result by 2.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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