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Homework Help: Triangle problem, finding unknown sides

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Alexander has a 6.0m long pole. He wants to use the pole to make a right triangle. One of the legs, meaning not the hypotenuse, is 2.0m long. Calculate the length of the tho other sides in the triangle.

    2. Relevant equations
    Phytagorah theorem asquared + b squared is equal to c squared.

    3. The attempt at a solution
    What i did was i tried to set up an equation. I wrote the other leg as x, and the hyoptenuse as 6-x because it was originally a 6m long pole. I then had 2squared+xsquared equal to 6-xsquared. I found that x is equal to the squareroot of 2 which is not correct and i do not know how to get further.
  2. jcsd
  3. Apr 15, 2015 #2


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    What about the other 2 m long leg?
  4. Apr 15, 2015 #3
    Instead of making the hypotenuse 6-x try 4-x because you already used 2 up in the other leg.
  5. Apr 16, 2015 #4
    Other 2m long leg? We only know that one side is 2m.
  6. Apr 16, 2015 #5
    think about what you have.

    The formula for adding up all the sides of a right triangle is
    a + b + c = Perimeter

    You know the total Perimeter is going to equal 6
    2 + b + c = 6

    You know the formula for a right triangle is
    (a^2) + (b^2) = (c^2)

    Is there something you can do with these two equations?
  7. Apr 16, 2015 #6
    2^2+b^2=6^2 ?
  8. Apr 16, 2015 #7
    Close but no cigar.

    Look at the perimeter formula again.
    2 + b + c = 6
    What does this really mean? A side that is 2 units long, plus another side, plus the largest the side, is all equal to 6. What happens if we try to isolate for something here and rearrange this formula a bit.

    We know these individual pieces are related to each other and they all add up to six right? What happens if we isolate one of the unknown pieces and try to get it by itself?
    2 + b + c = 6
    2 + b + c - 2 = 6 - 2
    b + c = 4
    c = 4 - b

    We have isolated c which is our hypotenuse variable. This statement reads "The hypotenuse is equal to 4 units minus some unknown number of units"

    Now back to our right triangle formula.
    (a^2) + (b^2) = (c^2)

    We know one of the sides is 2 meters right? How bought we assign our known "2 unit" long section to the "a" variable. Like in your post above.

    How does this look?
    (2^2) + (b^2) = (c^2)

    Do you notice something that we can do with our isolated " c" variable from earlier?
    c = 4 - b

    What can you do with these two formulas? Hint: the c here is the exact same as the squared c that is in the Pythagorean formula.
  9. Apr 16, 2015 #8
    I feel really stupid but i can't seem to figure it out. I guess you could do c^2=2^2-b^2 but then you have 2 variables in one equation. Are you 100% sure the prerimeter has to be 6 though? when you bend the sides won't the total length be changed somewhat?
  10. Apr 16, 2015 #9
    The sides are not bent. They are folded at angles.The sides of a triangle are constant and do not bend. Take a piece of string and arrange it in the shape of a triangle. Does the amount of string ever change?

    (2^2) + (b^2) = (c^2)
    c = 4 - b

    I'll give you another step.
    (2^2) + (b^2) = ((4 - b)^2)

    Do you see what I've done?
  11. Apr 16, 2015 #10
    a2 + b2 = c2

    Ok so here what you know the total length of the pole is 6m. You also know that at least one side but not the hypotenuse has a length of 2m you can set this to either a or b.


    because side a is taking up at least 2m then remaining is now 4m what ever length side b is we will have to subtract from the remaining length thus leaving us with side c so we can say c=4-b

    a2 + b2 = c2
    Plug in

    22 + b2 = (4-b)2

    4 + b2 = b2-8b+16
    then just solve for b
  12. Apr 17, 2015 #11
    Ahh i get it now, thanks for the help guys :)
  13. Apr 18, 2015 #12


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    So, what did you get for a result ?
  14. Apr 19, 2015 #13
    B is 1.5, hypotenuse then has to be 2.5
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