Question 277181
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Let *[tex \Large x] represent the measure of one of the sides.  Then *[tex \Large x\ +\ 2] is the measure of the diagonal which is the hypotenuse of an isosceles right triangle.


There are two ways to go about this one.  The 'charge straight at it' way uses the Pythagorean Theorem.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 2)^2\ =\ x^2\ +\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ +\ 4\ =\ 2x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ -\ 4\ =\ 0]


Using the Quadratic Formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ = \frac{-(-4)\ \pm\ \sqrt{(-4)^2\ -\ 4(1)(-4)}}{2(1)}\ =\ \frac{4\ \pm\ \sqrt{32}}{2}\ =\ \frac{4\ \pm\ 4\sqrt{2}}{2}\ =\ 2\ \pm\ 2\sqrt{2}]


Exclude the negative root because we are looking for a positive measure of length and we are left with:


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


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A slightly simpler method requires that you recall that an isosceles right triangle has three sides in proportion:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\,:\,1\,:\,\sqrt{2}]


Hence, if *[tex \Large x] is the measure of one side, then *[tex \Large x\sqrt{2}] is the measure of the hypotenuse.  But we are given that the measure of the hypotenuse is *[tex \Large x\ +\ 2], therefore:


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


Solving for *[tex \Large x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\sqrt{2}\ -\ x\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(\sqrt{2}\ -\ 1\right)\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{2}{\sqrt{2}\ -\ 1}]


Finally, using the conjugate of the denominator to rationalize the denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \left(\frac{2}{\sqrt{2}\ -\ 1}\right)\left(\frac{\sqrt{2}\ +\ 1}{\sqrt{2}\ +\ 1}\right)\ =\ 2\ +\ 2\sqrt{2}]


(Verification of that last step is left as an exercise for the student)


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