Express the length of the hypothenuse as a function of the perimeter.by ialink Tags: express, function, hypothenuse, length, perimeter 

#1
Aug111, 02:34 AM

P: 24

1. The problem statement, all variables and given/known data
The altitude perpendicular to the hypothenuse of a right triagle is 12 cm. Express the length of the hypothenuse as a function of the perimeter. I found a very elegant solution for the problem and even mathematia agrees on that asuming the equations i start with are right. The problem as i will show is checking the solution. When i check it i get an imaginary solution. 2. Relevant equations After drawing this image I used three equations for start:
p stands for perimeter. a,b,h correspond to the image 3. The attempt at a solution From eq 1 i derive (ph)^2 = (a+b)^2 from which the right part can be written as a^2 + b^2 + 2ab a^2 + b^2 = h^2 (eq 2) and a = 12h/b (eq 3). This results in (ph)^2 = h^2 + 24h expanding this results in p^2 + h^2  2ph = h^2 + 24h which equals p^2 = 2ph +24h which can be written as h(p) = p^2/(2p+24) Seems very elegant but i trie to check this solution as following Assume p = 20 h(p) = 25/4 so ph = a+b = 2025/4 = 55/4 ab = 12h = 75 So we have a+b=55/4 and ab=75. 2 variables and two equations. Piece of cake so it seems but the resulting equation is b^2 (55/4)b+75 = 0 wich has only imaginary solutions I don't know the correct answer, only my work. It's an even problem from review chapter 1 from calculus stewart 5th edition. What am I doing wrong? I spend hours trying to find my mistake..... I'd very much appreciate help on this one grt Ivar 



#2
Aug111, 02:48 AM

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Standard proof by contradiction  make an assumption, derive a result that is false, conclude the initial assumption was false. Assuming all of your work is correct, of course. But I claim once you accept the idea that your initial assumption could be invalid, there's an easy argument that p cannot be 20. (In fact, the easy argument I have in mind shows that p must be bigger than 36) 



#3
Aug111, 03:24 AM

P: 24

Thanks for your help 



#4
Aug111, 03:36 AM

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Express the length of the hypothenuse as a function of the perimeter.I knew that the minimum was somewhat larger than 36 (in fact, I knew it was larger than 40.97), but I didn't go further than that. The easy argument I had in mind is to make observations a>12, b>12, and h>a. I thought about suggesting finding the minimum as an exercise  and I agree with your expectation that the symmetric case is probably where the minimum occurs (especially given the numerical evidence). I bet you can prove it to be the minimum by repeating the argument in your first post, but instead finding b as a function of p... then showing that b is imaginary whenever p is less than your conjectured minimum. 



#5
Aug111, 05:01 AM

P: 24

I hope you forgive me my lazyness. Expressing b as a function of p is quitte difficult. Though i was insterested and i used mathematica. It gave me the function b(p). In it is a root. The root of a division to be exact which offcourse can't be less then zero. Solving this gives the proof that p > 24*(1+2^0.5) if we look at the positive domain.
I guess this was what you were meaning, correct? 



#6
Aug111, 09:19 AM

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(p.s. I have nothing against using Mathematica, as long as you aren't using it to skip over things you should be learning. ) (p.p.s. the FullSimplify would do a better job if you gave it the second parameter Assumptions > p>0) 



#7
Aug111, 10:15 AM

P: 24

Well what I should be learning is quitte relative since i'm currently not studying. I have a bachelors degree on civil engeneering and 2 years of university (civil engineering). At this point im preparing for the fact that over a year or so i want to study physics. So i grabbed my calculus book from university and i'm doing selfstudy on it now so it's 100% pure interest i'm doing these excersises. Your point still applies but i only use mathematica to check my answers and as you see i check by hand first but this seemed to me to much of a challenge.
Thanks for your help! grt Ivar PS you were right about the assumption but it still gives a quadratic one 



#8
Aug111, 11:19 AM

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P: 16,101

Oh, since we're playing with Mathematica... I'm sure there's a way to use Reduce to straight out tell you what the range on p is.
I bet if you tried adding inequalities to say that a, b, and h were >= 0, told it to reduce over the reals... and I forget what you can pass in as the other options, but I think one of them would let you ask for what it can tell you in terms of p. 



#9
Aug111, 02:20 PM

P: 24

That's something for tomorrow because it's Xfiles time!!!!
As the famous filosofer said:"I'll be back" grtz ivar 



#10
Aug111, 03:38 PM

P: 5

Have you ever heard of the 306090 triangle? If I put two of those together, I end up with your triangle:
Where every side is proportional to each other by those constants. Having this information, finding the length of the hypotenuse as a function of the perimeter is a walk in the park: [itex]h = \frac{2P}{3+\sqrt{3}}[/itex] Where h is the hypotenuse and P is the perimeter. There is no need to make things more complicated than they really are! Ask if you have any questions. 



#11
Aug111, 04:11 PM

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#12
Aug211, 01:31 AM

P: 24

The simpeler the solution the better but this is too simple in my opnion. 



#13
Aug211, 02:03 AM

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P: 5,674





#14
Aug211, 02:15 AM

P: 24

With too simple i meant the same as you. It doesn't cover the question. I think we can agree on that
grtz Ivar 



#15
Aug211, 10:31 AM

P: 5

With what I said above, I simply wanted to point out the important fact that, regardless of their setup, the triangles will remain proportional to each other by a certain constant which depends on one of the angles.
But since you apparently want an answer to this problem, I will have to post it: [itex]h = \frac{\sqrt{tan^{2}\alpha + 1}}{\sqrt{tan^{2}\alpha + 1} + tan\alpha + 1}P[/itex] It is still a simple solution, in my opinion. 



#16
Aug211, 03:50 PM

P: 24

It's not that i don't appreciate help but i'd more appreciated it if you had read the postings before grtz Ivar ps for P = 24+24*2^2 both formulas return 24 (with angle Pi/4 and x = 12, P = 24+24*2^2) but your answer still is wrong cause alpha should not be in the formula. h as a function of p was asked. Fun though that two completely different approaches give the same result 



#17
Aug211, 04:25 PM

P: 5

You are right. I apologize.




#18
Aug311, 12:17 PM

P: 24

that's oke



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