Question 340347
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As soon as I see a hypotenuse that measures some multiple of 5, I always suspect that I am dealing with a 3-4-5 right triangle.  Sure enough, 30 times 40 is 1200.  And there's the answer without doing any work.  But I suspect you need to see the method, so here goes:


Let *[tex \Large x] represent one leg of the triangle (which is to say one dimension of the rectangle) and let *[tex \Large y] represent the other leg.


Since we know the area to be 1200 square inches, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xy\ =\ 1200]


From which we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{1200}{x}]


From the Pythagorean Theorem we know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ =\ 50^2]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \left(\frac{1200}{x}\right)^2\ =\ 50^2]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{1440000}{x^2}\ =\ 2500]


Multiply both sides by *[tex \Large x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4\ +\ 1440000\ =\ 2500x^2]


Ewww -- that's ugly!  Not to worry though.


Let *[tex \Large u\ =\ x^2], substitute, and gather everything on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 2500u\ +\ 1440000\ =\ 0]


Hmmm, 16 times 9 is 144 and 16 plus 9 is 25...


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ -\ 1600)(u\ -\ 900)\ =\ 0]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 1600\ \Rightarrow\ \ x\ =\ \pm40]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 900\ \Rightarrow\ \ x\ =\ \pm30]


Toss the negative roots (we are looking for positive measures of length) and if *[tex \Large x\ =\ 40], then *[tex \Large y\ =\ \frac{1200}{40}\ =\ 30] or if *[tex \Large x\ =\ 30], then *[tex \Large y\ =\ \frac{1200}{30}\ =\ 40] 


And there you have it.


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