Question 194818
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Let <b><i>x</i></b> represent the width, then <b><i>x</i> + 2</b> represents the length, in other words the two legs of the right triangle formed by the diagonal.


So:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + x^2 + 4x + 4 = 100]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2 + 4x - 96 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 +  2x - 48 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x + 8)(x - 6) = 0]


Discard the negative root because we are looking for a positive measure of length.  Width is 6, Length is then 6 + 2 = 8.  You can calculate the perimeter for yourself.




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