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Mathematics 22 Online
OpenStudy (anonymous):

I need help solving the length and width of a rectangle by using the Pythagorean Theorem to find it. The area is 108 square inches and it has a diagonal measure of 15 inches. I am having a lot of trouble understanding how to work this formula. I have come up with the answers of 14 for the length and 7.714 for the width. Does anyone have any idea on if this is correct?

OpenStudy (kirbykirby):

Area = length * width = \(\ell w=108\) And if you draw a diagram, you will see from the pythagorean theorem that \[\ell^2+w^2=15^2 \] Now you have 2 equations and 2 unknowns. If you solved that then you should have done it correctly. I didn't do it, but plugging in those numbers in both equations seems like it works :)!!

OpenStudy (anonymous):

I can't figure out how to correctly work the equation to come out with 15 equally. I can come close but not close enough to 15 exactly.

OpenStudy (kirbykirby):

\[\ell=\frac{108}{w} \] Plug this into the 2nd equation: \[\left( \frac{108}{w}\right)^2+w^2=15^2 \\ \frac{108^2}{w^2}+w^2=15^2,\text{multiply equation by }w^2\\ 108^2+w^4=15^2w^2\\ w^4-15^2w^2+108^2=0\] This is a quadratic formula, if you think of \(x=w^2\) \(x^2-15^2x+108^2=0\\ x^2-225x+11664=0\) \[x=\frac{-(-225)\pm\sqrt{(-225)^2-4(1)(11664)}}{2(1)}=81, 144 \] Since \(x=81\) or \(144\). then \(x=w^2=81,\implies w=\pm 9\), but only consider the positive root since we have distances. If \(w=9\), then \(\ell w=108 \implies \ell=12\) However, you might consider the other root 144: if \(x=144 = w^2 \implies w=12\) so, \(\ell w=108 \implies \ell = 9\). Since the width is shorter than the length, we assign \(w=9\) and \(\ell = 12\)

OpenStudy (kirbykirby):

I wonder how you got 14 and 7.714

OpenStudy (anonymous):

When I factored it down. 2(x-7.714)(x-14)

OpenStudy (kirbykirby):

did you get that by trial and error? Because it is only approximate (I initially thought before solving the problem that your 7.714 was some rounded solution to a square root)

OpenStudy (anonymous):

It was a rounded solution of a square root..

OpenStudy (anonymous):

Thank you for your help on this. I can better see the errors I was making.

OpenStudy (kirbykirby):

i still don't know how you get it. BUt those are the 2 equations are need and that's the solution to those equations

OpenStudy (kirbykirby):

yw :)

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