Math problem?

ewl6850
ewl6850 Posts: 158 Member
in Chit-Chat
I will befriend, and maybe include another small prize, per your request to the first person who can tell me how to solve the problem in my profile picture. Ready? GO!

Replies

  • Angie_1991
    Angie_1991 Posts: 447 Member
    11 and 22 wasn't right? BLAST
  • OH_matt
    OH_matt Posts: 228 Member
    i drew it in my CAD program at work and got that they are 11 and 33.
  • ewl6850
    ewl6850 Posts: 158 Member
    Thanks. The trouble now is showing my work.
  • ewl6850
    ewl6850 Posts: 158 Member
    11 and 33 makes perfectly logical sense, proving it is the hard part.
  • OH_matt
    OH_matt Posts: 228 Member
    Those three parallel lines make 3 right triangles. Since all three have the same angles, the lengths of the other sides don't matter. Just look at the relationship.
    Smaller triangle has side A and side B with hyp C.
    Medium triangle has side 2A and side 2B with hyp D.
    Large triangle has side 3A with side 3B with hyp E.
    so...

    Small triangle
    A^2 + B^2 = C^2
    Medium Triangle
    (2A)^2 + (2B)^2 = D^2
    Large Triangle
    (3A)^2 + (3B)^2 = E^2

    Square root all three equations,
    A + B = C
    2A + 2B = D OR 2A + 2B = 22 OR 2*(A + B) = 22
    3A + 3B = E OR 3*(A + B) = E

    Medium triangle hyp is double the small triangle's hyp and large triangle hyp is 3x small triangle.

    So C=11, D=22, and E=33
  • ehs5mw
    ehs5mw Posts: 65
    Well, I don't really want to solve it all, but if I had to, I would look at the following:

    You know BAC is a right triangle.

    AD^2 + AE^2 = DE^2
    (2AD)^2 + (2AE)^2 = FG^2 = 22^2 = 484
    (3AD)^2 + (3AE)^2 = BC^2

    I think that should get you there, because you can just plug in a variable to represent the line segments AD and AE, which will make it look easier for you.
  • ehs5mw
    ehs5mw Posts: 65
    I got annoyed that I had written the equations and hadn't solved, and it is pretty easy to simplify my equations and get what you see above.

    Okay, now I made myself confused, though.

    If you said it was a 3,4,5 triangle, then you would have 3^2 + 4^2 = 5^2, but if you took the square roots the way it looks in your equations, you would have 3 + 4 = 5, which is clearly not right. What am I missing now?
  • krowanvil
    krowanvil Posts: 49 Member
    Use Pothagorean Theorem (and keep using it) since these are a right triangles. a*a+b*b=c*c.
    c=22 for the mid size triangle and a=b since the parrallel lines cut AB and AC into congruent sections (equal lengths).
    (AD=DF=FB=AE=EG=GC); (AB=3*AD, AF=2*AD)
    In the end you will end up with approximately 11 and 33.

    As a former teaching assistant and tutor, I wil leave the details up to you so that you actually learn to do this.:smile:
    But I have given you more than enough clues to solve it.

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