Question 270960
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#1:  Let *[tex \Large x] represent the measure of one leg of your triangle.  Let *[tex \Large y] represent the measure of the other leg.  Since the perimeter is 3 meters, which is to say 300 cm, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 145\ +\ x\ +\ y\ = 300]


It will be convenient to solve for *[tex \Large y] at this point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ = 155\ -\ x]


Since this is a right triangle, we can also write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ =\ 145^2]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \left(155\ -\ x\right)^2\ =\ 145^2]


Just solve for *[tex \Large x].  You will get two positive roots, one will be one leg of the triangle and the other root will be the other leg of the triangle.


#2:   Let *[tex \Large x] represent the width of the mat.  If the picture is 10 by 8, the mat has to be *[tex \Large 10\ +\ 2x] by *[tex \Large 8\ +\ 2x].


Now the only question is whether the area of the mat mentioned in the problem is the overall area of the mat before you cut the hole for the picture, or the area that remains after you cut the hole for the picture.


If it is the first way:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(10\ +\ 2x\right)\left(8\ +\ 2x\right)\ =\  2\cdot 10\cdot 8\ =\ 160]


which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 36x\ -\ 80\ =\ 0]


if it is the second way (and this is probably the correct interpretation):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(10\ +\ 2x\right)\left(8\ +\ 2x\right)\ -\ 80\ =\ 160]


Which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2\ +\ 36x\ -\ 320\ =\ 0]


Either way, solve for *[tex \Large x].  Either way you get a positive and a negative root.  Discard the negative root because you are looking for a positive measure of length.


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