Question 235902
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If you have a right triangle, then the measure of the legs of the triangle are also the measure of the base and altitude (doesn't matter which you call which).


Since the area of any triangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{bh}{2}]


And if we let *[tex \Large x] represent the length of the shorter leg so that *[tex \Large x\ +\ 5] represents the length of the longer leg, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 250\ =\ \frac{x\left(x\ +\ 5\right)}{2}]


Or equivalently (Verification of equivalence left as an exercise for the student)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 5x\ -\ 500\ =\ 0]


Leaving you with a factorable quadratic equation.  Since you are looking for a positive measure of length, exclude the negative root.  The positive root will be the measure of the shorter leg.  Add 5 to get the measure of the longer leg.


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