# Math Help - geometry problem

1. ## 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

2. 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

3. #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.

4. 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??

5. Originally Posted by unify34
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'

6. 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

7. Originally Posted by unify34
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 ...

8. 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

9. Originally Posted by unify34
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 ...