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Math Help - geometry problem

  1. #1
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    geometry problem

    i have the slightest idea on what to do can someone please explain? step by step and conclude on how the answer was determined?



    all help is highly appreciated
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  2. #2
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    for the first one i know AB is parrallel to CD right?

    and if agh makes an angle that equals to 2x and ghc = x-30

    does that mean i have to 2x + x-30 = 180?

    this is confusing... some basic rules i guess if any one can give out im really lost guys i feel llike i cant get the right anwers
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  3. #3
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    #1 hint ...

    same side interior angles are supplementary.


    #2 hints ...

    alternate interior angles are equal

    \angle LQP and \angle PQV are supplementary


    #3 hints ...

    \angle V = 2\angle B - 40

    three angles of any triangle sum to 180.


    #4 hint ...

    alternate interior angles are equal


    #5 hints ...

    three angles of any triangle sum to 180.

    \angle ACB and \angle BCD are supplementary


    #6 hints ...

    exterior angle of a triangle = sum of the two remote interior angles

    shortest side of a triangle is opposite the smallest angle


    #7 hints ...

    angle-side-angle

    CPCTC


    last one ...

    x^2 - 2x - 80 = 0

    left side will factor.
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  4. #4
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    Excellent... mind checkin over my answers??? im stuck on like two problems though...

    for #1 i have x = 70 , agh = 140, and ghc = 40

    #2 i have... M<2 = 62 degrees and m<3 = 118

    #3 ... 2(110)-40=180 and x = 110

    #4 i have m<x = 50

    #5 m<d = 80
    m<cbd = 30
    m<bcd = 70
    m<abc=30

    #6 i am stuck in..... but i have x = 35 , R=80 and angle 1 = 65 but i dont know which side is the shortest? how can i determine?

    #7 im working on...

    and #8 how do i begin factoring? do i do (x+1)(x=1) + 2x=80??
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  5. #5
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    Quote Originally Posted by unify34 View Post
    Excellent... mind checkin over my answers??? im stuck on like two problems though...

    for #1 i have x = 70 , agh = 140, and ghc = 40

    correct

    #2 i have... M<2 = 62 degrees and m<3 = 118

    correct

    #3 ... 2(110)-40=180 and x = 110

    what is x?

    #4 i have m<x = 50

    correct

    #5 m<d = 80 no
    m<cbd = 30
    m<bcd = 70 no
    m<abc=30



    #6 i am stuck in..... but i have x = 35 (ok), R=80 (ok) and angle 1 = 65 (ok) but i dont know which side is the shortest? how can i determine? look at the hint I gave before.


    #7 im working on...

    and #8 how do i begin factoring? do i do (x+1)(x=1) + 2x=80??

    look at the hint again ...
    keep chuggin'
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  6. #6
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    for #3 what do you mean "what is x?"

    X is 110


    180 = 2(X) - 40

    riiight?

    and i mean how can i determine the side? do i write RT? i think its the bottom side

    and for #8 i got
    x= -10

    can you help me setup#7 with the prooffs and given? i dont know where to begin
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  7. #7
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    Quote Originally Posted by unify34 View Post
    for #3 what do you mean "what is x?"

    X is 110


    180 = 2(X) - 40

    how is angle V = 180 ?

    riiight?

    and i mean how can i determine the side? do i write RT? i think its the bottom side

    they don't want the length of the side, only which side is shortest.

    and for #8 i got

    x= -10

    that's not the only solution ... show what you did to get x = -10.

    can you help me setup#7 with the prooffs and given? i dont know where to begin

    triangle DAB and CBA are both right triangles; AD = BC and AB is congruent to itself ...
    keep going ...
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  8. #8
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    hmmm i really dont know how angle v is 180... you wrote any three sides of an angle = 180 so i just subsituded angle v for 180

    guess that was wrong huh? i dont know... i feel stupid


    the bottom side is shortest i think.

    and i got x2 - 2x - 80=0
    (x+10)(x-8)=0
    x+10=0 i subbtracted 10 from zero and got x-10

    and then i got x=+8 but when i check +8 it didnt add up so used -10

    and hmmmmm so far i wrote

    AD=BC is given and then AD is parral to ab and BC is parral to ab is given and then i wrote what DAB and CBA = right triangles, ad and bc congruent to itsself therefore DAB and CBA = a.s.a
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  9. #9
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    Quote Originally Posted by unify34 View Post
    hmmm i really dont know how angle v is 180... you wrote any three sides of an angle = 180 so i just subsituded angle v for 180

    guess that was wrong huh? i dont know... i feel stupid

    V = the vertex angle
    B = one base angle

    V + 2B = 180
    V = 2B - 40


    the bottom side is shortest i think.

    correct ... it's opposite angle x, the smallest angle. your given picture is not drawn to scale.

    and i got x2 - 2x - 80=0
    (x+10)(x-8)=0
    x+10=0 i subbtracted 10 from zero and got x-10

    and then i got x=+8 but when i check +8 it didnt add up so used -10

    my mistake ... original equation should be x^2 + 2x - 80 = 0


    and hmmmmm so far i wrote

    AD=BC is given and then AD is parral to ab and BC is parral to ab is given and then i wrote what DAB and CBA = right triangles, ad and bc congruent to itsself therefore DAB and CBA = a.s.a

    there you go ... clean it up like your teacher wants.
    almost there ...
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