Question 256212
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I'm not sure why you introduced the idea of the area of the rectangle into this.  You are dealing with a right triangle which has sides of 50 and *[tex \Large x] and a hypotenuse of *[tex \Large x\,+\,10]


At that, you almost had it right.  Using Pythagoras:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 50^2\ =\ \left(x\ +\ 10\right)^2]


I'm assuming that the first relationship you wrote just had a typo in it and that you meant what I just wrote above.


Now here is where you really got off track.  You tried to say that *[tex \Large \left(x\,+\,10\right)^2\ =\ x^2\ +\ 10^2].  No! No! No! 1000 times No!


In general, *[tex \LARGE \left(a\,+\,b\right)^2\ =\ \left(a\,+\,b\right)\left(a\,+\,b\right)\ =\ a^2\ +\ 2ab\ +\ b^2]  (Remember FOIL?)


So, your next step SHOULD have been:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2500\ =\ x^2\ +\ 20x\ +\ 100]


Which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 20x\ =\ 2400]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 120]


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