Question 305638
<font face="Garamond" size="+2">


{{{drawing(
500, 250, -0, 30, -0, 15,
line(0,0,9,12),
line(9,12,25,0),
line(9,12,9,0),
line(0,0,25,0),
locate(9,6,12),
locate(4,8,b),
locate(16,8,a),
locate(4,1,x),
locate(14,1,25-x)
)}}}


And with a tip of the hat to Mr. Pythagoras we begin:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25^2\ =\ a^2\ +\ b^2]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ =\ 25^2\ -\ b^2]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ =\ 12^2\ +\ x^2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ =\ 12^2\ +\ (25\,-\,x)^2]


Which expands to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ =\ 12^2\ +\ 25^2\ -\ 50x\ +\ x^2]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25^2\ -\ b^2\ =\ 12^2\ +\ 25^2\ -\ 50x\ +\ x^2]


and simplifying


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\ b^2\ =\ 12^2\ -\ 50x\ +\ x^2]


but since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ =\ 12^2\ +\ x^2]


we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ =\ 25x]


But then we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25x\ =\ 12^2\ +\ x^2]


and in standard form:


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


which factors to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 9)(x\ -\ 16)\ =\ 0]


So 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 9]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 16]


which is the situation if the diagram was turned over.  Since the question asks for the longer of the two segments, the answer is 16.




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