Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

two sides of the triangle have the same lengths of 6cm and 6cm what is the possible length of the third side A)0

OpenStudy (anonymous):

@wolf1728

jimthompson5910 (jim_thompson5910):

hint: if two sides are x and y, then the third side z must satisfy this condition x-y < z < x + y note: x > y (ie x is the larger side compared to y or at least as long as y)

OpenStudy (anonymous):

I don't need a hint that would confuse me I just need an explanation on how to solve it

OpenStudy (wolf1728):

I notice A and B are the exact same answer. Shouldn't 1 of those be changed? I think the correct answer is the third side can be 0 < x < 12

jimthompson5910 (jim_thompson5910):

example: if two sides are known to be 3 and 5, then the third side z is represented by this expression 5-3 < z < 5 + 3 2 < z < 8

jimthompson5910 (jim_thompson5910):

ie if two sides are 3 and 5, then the third side must be between 2 and 8 (excluding both 2 and 8)

OpenStudy (anonymous):

oops b is actualy 0<x<12

OpenStudy (wolf1728):

Hey it looks like I spotted the error. Also I have attached a graphic that should make it easy to visualize.

OpenStudy (anonymous):

ok so my book logged out and I have a diff problem... two side lengths of a triangle are 8ft and 12ft what is the 3rd side a)4<x<8 b)4<x<20 c)8<x<20 d)12<x<20

OpenStudy (anonymous):

ido nt understand how you got the answser

OpenStudy (wolf1728):

I'd say B for that also. Look at the graphic I posted and substitute 8 and 12 for the 2 sixes.

jimthompson5910 (jim_thompson5910):

Here's another way to put it: Say you have a triangle with side lengths of 3, 7, and x The missing side must be larger than 7 - 3 = 4 AND the missing side must be smaller than 7 + 3 = 10 So this means 4 < x < 10

jimthompson5910 (jim_thompson5910):

Hopefully that makes sense

OpenStudy (anonymous):

I does but I still cant get the answr right

jimthompson5910 (jim_thompson5910):

if two sides are 8 and 12, then the third side must be larger than 12 - 8 = 4 if the third side is x, then what must it be less than?

OpenStudy (wolf1728):

GParsley - in the graphic (for the 6 & 6 triangle) if you have two sides very close to each other which leaves very little room for the third side

OpenStudy (anonymous):

but the onei n the other pic they were fare apart how can u tell how close they are that's confusingme

OpenStudy (wolf1728):

The "far apart" triangle can only be so far apart. (Since each side has a length of 6 you can't have anything greter than 6 & 6 (or 12) for the third side.

OpenStudy (anonymous):

ok so what about my toher problem

OpenStudy (wolf1728):

I could draw another graphic if you want

OpenStudy (anonymous):

no il just keep trying

OpenStudy (wolf1728):

Okay I made a graphic for the 12 8 triangle with some explanation

OpenStudy (wolf1728):

Here is the second graphic foe the 8 12 triangle

OpenStudy (anonymous):

soit has to be between 4 and 20?

jimthompson5910 (jim_thompson5910):

correct, if x is the third missing side then 4 < x < 20 this is if the two other sides are 8 and 12

OpenStudy (anonymous):

ok so what about 18 and 23 the answer would be 5<x<23???

jimthompson5910 (jim_thompson5910):

23 - 18 = 5 23 + 18 = 41 so you're half right and it should be 5 < x < 41

OpenStudy (anonymous):

pooo!

jimthompson5910 (jim_thompson5910):

but it looks like you're (sorta) getting the hang of things

OpenStudy (anonymous):

ok so know i have 20 and 35 15<x<55

OpenStudy (anonymous):

i think '

jimthompson5910 (jim_thompson5910):

good on both

OpenStudy (anonymous):

yay

jimthompson5910 (jim_thompson5910):

nice work

OpenStudy (anonymous):

5 and 16 11<x<21

jimthompson5910 (jim_thompson5910):

again good on both

jimthompson5910 (jim_thompson5910):

you got this down it looks like, so that's good

OpenStudy (anonymous):

thanks bunches!!

jimthompson5910 (jim_thompson5910):

you're welcome

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!