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How to solve this triangle problem.

  1. Aug 22, 2012 #1
    1. The problem statement, all variables and given/known data

    The altitude perpendicular to the hypotenuse of a right triangle is 12 cm. Express the length of the hypotenuse as a function of the perimeter.

    2. Relevant equations
    3. The attempt at a solution


    So I have to express the hypotenuse (h) in terms of the perimeter. I tried to find the area of the triangle using two different equations

    1) A = (1/2)*12*h = 6h
    2) A = (1/2)*xy, let x and y be the other sides of the triangle

    I can set them to each other, but I get 6h = 0.5xy

    The perimeter (P) is x+y+h.

    I have no idea how to express h in terms of P since now I have too many variables.

    I tried substitution, and the best I got (without using x and y) was some complicated polynomial with P and h that I couldn't just simply solve for h.

    Any tips are appreciated.
     
  2. jcsd
  3. Aug 22, 2012 #2

    ehild

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    Think of Pythagoras' Theorem.

    ehild
     
  4. Aug 22, 2012 #3
    I certainly did. But I still have to assign a variable to one of the sides, ie x, and then the other side would be [itex]\sqrt{h^{2} - x^{2}}[/itex]

    It doesn't help me in putting h in terms of P however, because now

    P = [itex]x + h + \sqrt{h^{2} - x^{2}}[/itex]

    How can I express [itex] h [/itex] only in terms of [itex] P [/itex]?
     
  5. Aug 23, 2012 #4

    Curious3141

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    You basically have to solve for x in terms of h.

    Use [itex]x\sqrt{h^2 - x^2} = 12h[/itex], which you've already figured out.

    Square both sides to get a quartic in x that can be reduced to a quadratic by a sub like [itex]z = x^2[/itex], then solve for x in terms of h.

    The algebra can be a little messy, though.
     
  6. Aug 23, 2012 #5
    Ok. If I only have to express it in terms of x then I can do it. Glad to know that it's physically impossible to express it in P. Thanks!
     
  7. Aug 23, 2012 #6

    Curious3141

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    NO! You *solve* for x in terms of h, then replace all the x terms in [itex]P = x + h + \sqrt{h^2 - x^2}[/itex] with this expression. Finally, rearrange/solve to get h in terms of P.

    As I said, it's messy. But completely possible.
     
  8. Aug 23, 2012 #7
    LOL. I like your emphatic NO. But yea, that sounds very messy indeed.... I see some [itex]x^{2}h^{2} - x^{4} = 144h^{2}[/itex]... Just this alone makes me not want to do it...
     
  9. Aug 23, 2012 #8

    Curious3141

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    I just did it. Yes, it's tedious. But once you work through the algebra, you get a nice simple expression for h in terms of P.

    Some things to note. You'll get two solutions for the quadratic in [itex]x^2[/itex]. Keep both of them. You'll find a nice symmetry in the expression for [itex]\sqrt{h^2 - x^2}[/itex] when you use each of the values of [itex]x[/itex] that'll simplify your work (so you don't have to do everything twice!). Also keep the identity [itex](a + b)(a - b) = a^2 - b^2[/itex] in mind because you'll find that a great timesaver.

    Other than that, you just have to roll up your sleeves and get your hands dirty. I can't give you any more help without giving you the complete solution. :biggrin:

    (I was hoping there'd be a simpler way with similar triangles, but no such luck).
     
  10. Aug 23, 2012 #9

    ehild

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    From Pythagoras' Theorem, x2+y2=h2, and you know that xy=12h.
    If you add 2xy to both sides to Pythagoras' equation you get the square of x+y on he left side:

    (x+y)2=h2+2xy=h2+24h.
    x+y=P-h. Substitute for x+y, expand, simplify. The solution is two more lines (h2 cancels).

    ehild
     
    Last edited: Aug 23, 2012
  11. Aug 23, 2012 #10

    ehild

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    Also keep the identity (a+b)2=a2+b2+2ab
    in mind because you'll find that a great timesaver. :biggrin: in problems where sum and product of two variables are involved.

    ehild
     
  12. Aug 23, 2012 #11

    Curious3141

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    Well, I thought that one went without saying. :biggrin:
     
  13. Aug 23, 2012 #12

    Curious3141

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    Wow, this is really, really neat!:approve:

    Mine was like :eek: then :yuck:, finally ending up :smile:.

    But I'm :redface: I didn't see your way from the start. I blame the fever I'm having! :tongue:
     
  14. Aug 23, 2012 #13
    Thanks guys, that was good!
     
  15. Aug 23, 2012 #14

    ehild

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    Get well soon! :smile:

    ehild
     
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