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Area of a right-angled triangle

  1. Jan 27, 2004 #1
    In a given right-angled triangle, the length of the hypotenuse is 10 and the length of one of the heights is 6. What is the area of the triangle?

    a) 12
    b) 30
    c) 24
    d) 24 or 30

    Please explain your answer. :smile:
     
    Last edited: Jan 28, 2004
  2. jcsd
  3. Jan 27, 2004 #2

    chroot

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    The area of any triangle is

    [tex]\frac{1}{2} b h[/tex]

    where b is the length of the base of the triangle, and h is its height.

    - Warren
     
  4. Jan 27, 2004 #3

    HallsofIvy

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    I always get a kick out of people who just quote a problem. Do we get points for a correct answer?

    A (almost!*) always, Chroots answer is correct- the area of any triangle is (1/2)bh where b is the "base", one of the sides of the triangle and h is the "height", measured perpendicular to the base.

    In a right triangle, the two legs are perpendicular so you can use one as base and the other as height.
    Use the Pythagorean theorem to find the length of the other leg of the triangle, then apply "(1/2)bh".


    * I, myself, never make mistenk(s!
     
  5. Jan 28, 2004 #4
    Ahh, but I do make mistakes. The problem should read "the length of one of the heights is 6". What would be the correct answer then? (Bearing in mind that the given height could be the height to the hypotenuse.)
     
  6. Jan 28, 2004 #5

    chroot

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    The triangle has more than one height?

    Listen, we've told you how to do this problem, but we're not going to do it for you. Try it, it's not hard.

    - Warren
     
  7. Jan 28, 2004 #6
    I do know how to solve this problem and already have, years ago. I am just trying to see if you can, but clearly I overestimated you.

    And yes, every triangle - the thing with three sides, you know - has three heights. Not just one. But three. (And it's this little factoid that makes the problem more than just a 1st grade exercise.)
     
    Last edited: Jan 28, 2004
  8. Jan 28, 2004 #7

    chroot

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    No, if you first select one side as a base, a triangle has only one height. Sorry.

    - Warren
     
  9. Jan 28, 2004 #8
    Where have I selected a base? All the problem says is that the length of one of (the three) heights in this triangle is 6.
     
  10. Jan 28, 2004 #9

    chroot

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    We don't really have time here for your little riddles. Sorry. At the very least, if you're "challenging" us with a riddle, label it as such. We thought you were a schoolchild who didn't understand how to find the area of triangles.

    - Warren
     
  11. Jan 28, 2004 #10

    Integral

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    Your initial statement completely specifies the triangle, where is the difficulty?

    The triangle you describe is a 3-4-5 triangle, one of the most useful combinations of all right triangles. Now tell us why finding this area should be difficult?
     
  12. Jan 28, 2004 #11
    Is it not possible that the height whose length is 6, is perpendicular to the hypotenuse of this triangle?
     
  13. Jan 28, 2004 #12
    I did not realize that the answer to a question depends on the inquirer. And if you really thought I were a schoolchild you could at least pretend to be nice about it, because if I really were a 10 year old your replies would have been insulting to me. Luckily I have passed that age and I am just saddened by the fact that it is a mentor - who should set an example for others - that speaks like this.
     
    Last edited: Jan 28, 2004
  14. Jan 28, 2004 #13

    matt grime

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    So you've got one those infinitely large triangles lying around have you?
     
  15. Jan 28, 2004 #14

    chroot

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    Your question, as originally posted, was "what is the area of a right triangle with one leg 6 feet and hypotenuse 10 feet?", which is a schoolchild's question.
    Oh no.

    - Warren
     
  16. Jan 28, 2004 #15

    Integral

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    Perhaps a link to a drawing to this "triangle" would clear up the questions. It is said a drawing is worth a 1000 words. Perhaps english is not your native tounge and you do not realize what you are saying.
     
  17. Jan 29, 2004 #16
    Lets calm down everyone. I think he means he knows the hypotenuse and one leg. If so, it's [tex]10^2=6^2+A^2[/tex]. Therefore, [tex]100=36-A^2[/tex] and 64=A^2 and A=8. 8 is your other leg. 8*6=48, so your area is 24. It's pretty late here, so I hope I didnt skip anything, its been so long since I even touched a problem like this.

    Please stop fighting, people. There's no reason to act negative towards others here.

    That's a good theory, Integral. His profile says he is from Israel, and if it is true, it makes sense to me to say some things incorrectly.
     
    Last edited: Jan 29, 2004
  18. Jan 29, 2004 #17

    Integral

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    I don't think he is talking about side length, but rather perpendicular bisectors.
     
  19. Jan 29, 2004 #18
    like the triangle is sitting on the hypotonuse???? is this a 45-45-90 triangle or can we not make that assumption?
     
  20. Jan 29, 2004 #19

    chroot

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    Either 24 or 30. If the triangle is sitting on either of its legs, it's 24. If the triangle is sitting on its hypotenuse, it's 30.

    - Warren
     
  21. Feb 1, 2004 #20
    i guess, everybody had his/her contribution. i hope that chen would have understood it in anyway.

    However, if chen haven't pictured the situation, then there is still a need for us to draw the figure to help him/her out.
     

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