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Cannot find the height of this triangle given only one leg

  1. Aug 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi, I have inserted a picture of the problem:

    17JUUIV.jpg

    I have to find x, the altitude, or height of the triangle.

    x is the height of the larger triangle, and is perpendicular to the base of 13. It divides the base into segments of length 4 and 9.

    This was a question on my math assessment test for community college. But I couldn't think of an answer, so I left it blank. It's impossible to find x without more information, such as the length of another leg of the triangle.

    If the other leg were y, then:

    x = (y + 4)(y - 4)

    but there is no y variable given.
     
    Last edited: Aug 12, 2013
  2. jcsd
  3. Aug 12, 2013 #2

    SteamKing

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    I think Pythagoras and a lot of algebra could have solved your problem.
     
  4. Aug 12, 2013 #3
    With the larger triangle, I have only 1 side to work with. And with the two smaller ones, I have only 2 sides each. In order to apply the Pythagorean Theorem, I need 3 sides.
     
  5. Aug 12, 2013 #4

    SteamKing

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    Take the two smaller triangles individually. You can use Pythagoras to write an expression for the length of the unknown side in each case. Once you have expressions for the unknown sides, then you know, also by Pythagoras, that the sum of the squares of the unknown lengths must equal 13^2. Now you apply a sufficient quantity of algebra and hard work, and you should be able to calculate the unknown distance 'x'.
     
  6. Aug 12, 2013 #5
    Ok, I got it! Thank you SteamKing.

    x = 6

    I'm very rusty when it comes to this, but it's slowly coming back.
     
  7. Aug 13, 2013 #6
    You can also use the fact that all of the triangles involved are similar, and so the ratios of the corresponding sides are all equal; i.e. x/9=4/x.
     
  8. Aug 13, 2013 #7
    It is not hard to get to:

    height² = projection a × projection b
    In this case:

    x² = 9 * 4
    x² = 36
    x = 6
     
  9. Aug 13, 2013 #8
    Thanks gopher_p. How could you tell the triangles were similar without finding the other legs first?
     
  10. Aug 13, 2013 #9
    I appreciate the info. I never heard of that projection formula. I can't kind even find it on the web, but what you wrote works!
     
  11. Aug 13, 2013 #10
    Some textbooks treat it as a formula and some say that the formula is [itex]\frac{m}{x}=\frac{x}{n}[/itex], from where you can get to [itex]m n = x^{2}.[/itex]
     
  12. Aug 13, 2013 #11
    Thanks! Are m and n the lengths of the line segments of the leg of a triangle that x, the height of a triangle, divides? And does x have to divide the right angle of the larger triangle, or can it be any angle?
     
  13. Aug 13, 2013 #12
    They have the same angles.
     
  14. Aug 13, 2013 #13
    Answering both questions:
    [itex]m[/itex] and [itex]n[/itex] are projections, that means that [itex]m\ +\ n[/itex] needs to equal to the side of the triangle that the height segment intercepts.

    The height intercepts a side perpendicularly. If [itex]x[/itex] divided another angle in a right triangle it would be superposed to one of the sides of the triangle.

    Legend:
    [itex]a[/itex] = cathetus (or side)
    [itex]b[/itex] = cathetus (or side)
    [itex]c[/itex] = hypotenuse
    [itex]m[/itex] = projection relative to a
    [itex]n[/itex] = projection relative to b
    [itex]x[/itex] = height

    Pythagorean theorem:
    [itex]a^{2}\ +\ b^{2}=\ c^{2}[/itex]

    Derived from it:
    [itex]a^{2}\ =\ m c[/itex]
    [itex]b^{2}\ =\ n c[/itex]
    [itex]x^{2}\ =\ m n[/itex]
    [itex]a b\ =\ x c[/itex]

    Using equation I is the easiest way to find a cathethus as you don't need the 'crude' theorem:
    If you needed, [itex]a[/itex], for instance, you would solve:
    [itex]a = \sqrt{mc}[/itex]
    [itex]a = \sqrt{13\cdot 9}[/itex]
    [itex]a = \sqrt{117}[/itex]
    [itex]a = 3\sqrt{13}[/itex]
     
  15. Aug 14, 2013 #14
    You're right! I had to really think about all three triangles for a while, but yes, they are similar, even though I didn't realize this at first. I was able to find that each of the two smaller triangles are similar to the large triangle containing them, by A.A.A (angle-angle-angle). Then, if the two smaller triangles are both similar to the larger triangle, they must be similar to each other, even though I couldn't find a direct A.A.A. relation between the two smaller triangles. Thank you!
     
  16. Aug 14, 2013 #15
    Wow. This is way above my head, but I am going to take some more time to study what you have written. They definitely don't teach this in high school geometry. I do appreciate this though!
     
  17. Aug 14, 2013 #16
    I do not study in the U.S., but people that do confirmed that this IS taught in HS. And if it is not, you should derive those expressions from the "original formula".

    The same way you can use the Pythagorean Theorem to derive:

    Diagonal of a square of side [itex]l[/itex]: [itex]l\sqrt{2}[/itex]
    Height of an equilateral triangle of side [itex]l[/itex]: [itex]...[/itex]

    If you don't know the last "formula" that I mentioned, for real, use the Pythagorean Theorem to get to discover it.
    Use paper if you can't do it by mind. Post your answer (or ask uncle Google if it is correct).

    Obs.: These two formulas are absurdly easy and useful, so if you need math (as a student, you probably do), try to memorize them!
     
  18. Aug 14, 2013 #17
    You're welcome. I needed to do a lot of problems involving triangles the "hard" way before I realized that recognizing/using similar triangle arguments made some of them much easier.
     
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