Question

Determine the possible lengths for the third side of a triangle whose first two sides measure a and b.

Answer 1

A + b > x
x + a > b
x + v > a

Answer 2

@Mimi_x3, @paki, @chosenmatt,

Answer 3

@mathmath333

Answer 4

would it be b - a < x < a + b

Answer 5

wait

Answer 6

ok

Answer 7

tirangle inequality theorm states that

if the triangle having three sides \(a\) , \(b\) and \(x\)

then it must satisfies the condition

three conditions
\(\large\tt \begin{align} \color{black}{
a+b>x \hspace{.33em}\\~\\
x+b>a \hspace{.33em}\\~\\
x+a>b \hspace{.33em}\\~\\
}\end{align}\)

Answer 8

I know so, is my inequality correct then?

Answer 9

b - a < x < a + b

Answer 10

well can u rewrite
the three inequalities

Answer 11

If b > a, then b - a < x < a + b right?

Answer 12

what this,
a + b > x
x + a > b
x + v > a

Answer 13

where does \(\huge v\) come from

Answer 14

o sorry I meant
a + b > x
x + a > b
x + b > a

Answer 15

yes that's correct

Answer 16

so is the inequality b - a < x < a + b correct? Because they want me to write an inequality.

Answer 17

oh so u must write that specifically in ur question

Answer 18

yes because it asks for the possible length. So, wouldn't that mean for an inequality?

Answer 19

this is what they gave me. I almost forgot

Answer 20

well can u tell what was the exact question
do they want only one inequality or all

Answer 21

they what the possible lengths for x so I guess they want an inequality

Answer 22

is in the diagram given \(x+ ?\) like that or u did by sellf

Answer 23

no sorry it gives x = ?

Answer 24

well is it has options ?

Answer 25

no

Answer 26

then u should write as

\(a+b\geq x\)

Answer 27

thats why I thought they wanted an inequality

Answer 28

what about the a - b part?

Answer 29

this one b - a < x < a + b ?

Answer 30

yeah, is it wrong?

Answer 31

where did u get this from ?

Answer 32

idk
I tried working them out first

a + b > x
a + b - b > x - b

x + a > b
x + a - a > b - a
x > b - a

x + b > a
x + b - b > a - b

Answer 33

i dont think thats needed

Answer 34

o on the first one i mean
a + b > x
a + b - b > x - b
a > x - b

Answer 35

so what do I do to find the possible lengths for x?

Answer 36

lol just write
\(a+b\geq x\)

Answer 37

and on the last one I meant
x + b > a
x + b -b > a - b
x > a -b

Answer 38

but I need to show my work

Answer 39

u can give the reason as by " triangle inequality theorm" thats it

Answer 40

but how did you find the inequality?

Answer 41

u can find the length here as it only given a and b and not numbers like 7 ,8 9 etc

Answer 42

inequalities look like this
> x >
so I did, a + b > x > b - a
because you already know that a + b > x in the beginning
and I you work out x + a > b
it turns into x > b - a
so, a + b > x > b - a

Answer 43

thank you for the help

Answer 44

then by that way it is also possible that

a+b>x

and x+b>a
x>a-b
so

a+b>a-b

and by the way by ur reasoning which is

a + b > x > b - a

b gets cancelled
and a shifts to left hand side
and remains

2a>b

Answer 45

and it is not necessary for inequalities to look like these

\(<x<\)