Question 188306
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Sorry, but there is something wrong with your problem statement.  Either you have a rectangle rather than a square as you said, or the legs of the triangle are both the same.


Since the legs of the triangle are different, I'll presume a rectangle.  Since it is a rectangle, the included angle of the two side lengths given must be a right angle, but definition of a rectangle.  So the triangle must be a right triangle.


Just use the Pythagorean Theorem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a^2\ +\ b^2\ =\ c^2 \ \ \Rightarrow\ \ c = sqrt{a^2\ +\ b^2}]


So, for this problem <i>c</i> = <i>x</i>, and <i>a</i> and <i>b</i> are 3 and 5, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = sqrt{3^2\ +\ 5^2}]


You can use a calculator as well as I can, although if your instructor expects the exact representation of the answer, you need to leave it in radical form.


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