Question 514051
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The 5 inch hypotenuse has to be on the smaller triangle, so we know two things assuming we use *[tex \Large a] and *[tex \Large b] to represent the legs of the smaller triangle:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ +\ b^2\ =\ 25]  (Thank you, Mr. Pythagoras)


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{ab}{2}\ =\ 6]


from which we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ \frac{12}{b}]


Substitute into the Pythagorean relationship


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{12}{b}\right)^2\ +\ b^2\ =\ 25]


A little Algebra music, Sammy:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^4\ -\ 25b^2\ +\ 144\ =\ 0]


Let *[tex \Large u\ =\ b^2] and solve the resulting quadratic for *[tex \Large u]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 25u\ +\ 144\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 16] or *[tex \LARGE u\ =\ -9]


Toss the negative root (we want positive real number measures of length).


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ =\ 16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ \pm4]


Toss the negative root again


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ 4]


If the hypotenuse is 5 and the long leg is 4, then the short leg has to be 3 (3-4-5 right triangle).


150 divided by 5 is 30.  30 times 4 is 120, 30 times 3 is 90, 120 plus 90 is 210 which is the desired sum.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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