Question 1162374
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Let *[tex \Large x] represent the width of the shelf and then *[tex \Large 2x\ +\ 3] must represent the length of the brace.  Assuming that the shelf is exactly perpendicular to the wall, the line segment represented by *[tex \Large h], the brace, and the shelf form a right triangle where the brace is the hypotenuse.  Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (2x\,+\,3)^2\ =\ x^2\ +\ 12^2]


Expand


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4x^2\ +\ 12x\ + 9\ =\ x^2\ +\ 144]


Collect like terms in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  3x^2\ +\ 12x\ -\ 135\ =\ 0]


Divide by the lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ 4x\ -\ 45\ =\ 0]


Factor


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x\ -\ 5)(x\ +\ 9)\ =\ 0]


So either *[tex \Large x\ =\ 5] or *[tex \Large x\ =\ -9].  Since a negative number for the width of anything is absurd, discard the negative root.  That means *[tex \Large x\ =\ 5], and therefore the brace is *[tex \Large 2(5)\ +\ 3\ = 13]


Check the work.


*[tex \Large 13^2\ =\ 169], *[tex \Large 12^2\ =\ 144], *[tex \Large 5^2\ =\ 25], and *[tex \Large 144\ +\ 25\ =\ 169] -- Checks.


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