Math problem?

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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!

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  • Angie_1991
    Angie_1991 Posts: 447 Member
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    11 and 22 wasn't right? BLAST
  • OH_matt
    OH_matt Posts: 228 Member
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    i drew it in my CAD program at work and got that they are 11 and 33.
  • ewl6850
    ewl6850 Posts: 158 Member
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    Thanks. The trouble now is showing my work.
  • ewl6850
    ewl6850 Posts: 158 Member
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    11 and 33 makes perfectly logical sense, proving it is the hard part.
  • OH_matt
    OH_matt Posts: 228 Member
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    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
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    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
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    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
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    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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