Question 187398
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Your thought about the Pythagorean Theorem was a very good thought indeed because that is the only way to solve this problem.  Instead of <i>a</i> and <i>b</i> for the measures of the legs of your triangle, let's use <i>l</i> and <i>w</i> because we are talking about the sides of a rectangle after all.


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  l^2 + w^2 = (6.5)^2]


shows the relationship between the lengths of the sides of the rectangle with the rectangle's diagonal.  By itself, that doesn't do much good because there are an infinite number of pairs of <i>l</i> and <i>w</i> values that would make this true.


Fortunately, we are also given another relationship between the length and the width of the rectangle, namely that the length is 3.5 inches longer than the width, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l = w + 3.5]


Knowing that, we can substitute <i>w</i> + 3.5 anywhere we see an <i>l</i>, so let's put it into the Pythagorean equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (w + 3.5)^2 + w^2 = (6.5)^2]


Expand using FOIL:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2 + 7w + (3.5)^2 + w^2 = (6.5)^2]


Collect terms and add *[tex \Large -(3.5)^2] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2 + 7w  = (6.5)^2-(3.5)^2]


Notice that you have the difference of two squares on the right, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2 + 7w  = (6.5+3.5)(6.5-3.5) = 10\,\cdot\,3 = 30]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2 + 7w  = 30]


Put into standard form by adding -30 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2 + 7w  -30= 0]


And finally, solve the quadratic which very conveniently factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (w + 6)(2w - 5)= 0]


The Zero Product Rule gives us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w + 6 = 0 \ \ \Rightarrow\ \ w = -6]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w - 5 = 0 \ \ \Rightarrow\ \ w = 2.5]


Notice that one of the roots of the quadratic is negative.  Since it is absurd to have a negative value for the measure of length of something, we can exclude that root as being extraneous -- having been introduced by the act of squaring the variable in the process of solving the problem.  The other root, the positive one, is the desired answer for the measure of the width.  And the value of the measure of the length comes from simply adding 3.5 inches to the calculated width, or 2.5 + 3.5 = 6.


Check the answer:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2.5)^2 = 6.25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^2 = 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6.25 + 36 = 42.25 = (6.5)^2]


Answer checks.


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