Question

Am I correct?
If not, please explain why.
Thanks! ^.^

Answer 1

The lengths of two sides of a triangle are 15 and 26. The length of the third side must be greater than ___ but less than ____.

I have 41 and 11.

Answer 2

I agree with your answer ^_^

Answer 3

I would change the order.

Answer 4

Oh, yep @SolomonZelman is right

Answer 5

greater than 41, and less than 11 ?

Answer 6

you meant it the other way most likely, didn't you ?

Answer 7

I don't remember how I got the numbers, but that's the order I had them in.. 41 for the first blank, 11 for the second.

Answer 8

2 sides of a triangle must add to be greater than the 3rd side.

Answer 9

\(\large\color{slate}{\displaystyle a+b>c}\)
you have sides of `15` and `26`

If 26 is the largest side, then
\(\large\color{slate}{\displaystyle a+15>26}\)
(then a>11)

If the largest side, is unknown then
\(\large\color{slate}{\displaystyle 26+15>c}\)
(then 41>c)

Answer 10

So if I took 26+15, I'd get 41.. If I subtract 15 from 26, I'd get 11..
Is this idea included in any simple way to find the numbers?

Answer 11

yes, you are simply doing a logical thinking.
You have two sides \(\large\color{slate}{15}\) and \(\large\color{slate}{26}\).

(obviously, 15 can't be the largest side)

Answer 12

So either \(\large\color{slate}{26}\) is the largest side, and then we will need to find the smallest size for the smallest side.
\(\large\color{slate}{a+15>26}\)
Any number that is greater than 11 makes it true., so a>11.

OR, 26 is one of the smaller sides, and the largest side is unknown.
Then you need to find the maximum size of the largest side.
\(\large\color{slate}{26+15>c}\)
\(\large\color{slate}{41>c}\)

Answer 13

So, whatever the unknown side is, whether it is the largest side or the smallest one, it''s range is:

\(\large\color{slate}{11>x>41}\)

Answer 14

Okay, I think I'm getting it.. I'm still a bit confused, though..

Answer 15

why are you confused?

Answer 16

I don't know. Maybe I need another example.

Answer 17

So the other side is 30?

Answer 18

(where x, is the 3rd unknown side)

Answer 19

you keep putting the inequality backwards :p

Answer 20

I did,

Answer 21

both times even after correcting him on his direction of inequality :p

Answer 22

Wait, you are confusing me. I don't I made any mistake in this reply. (even though I make them fairly often).

Answer 23

I don't think* I made... (there you go, lol)

Answer 24

\[\large\color{slate}{11>x>41} \\ \large\color{slate}{30>x>70} \]
both of these are backwards

Answer 25

11 is not greater than 41
30 is not greater than 70

Answer 26

lol, I am thinking with my butt, sorry

Answer 27

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You have two sides, \(\large\color{slate}{20}\) and \(\large\color{slate}{30}\).
(you need a third side, don't you ? )

Your form is: a+b>c


If the unknown side is the smaller,
THEN, a+20>50
(solving for a, by subtracting 20 from both sides)
a>30


If the unknown side is the largest,
THEN, 50+20>c
(solving for c, by adding everything on the left side)
70>c

(or, same as, c<70)


So, whatever the unknown side is (be it smaller side, or the largest side)
it must be greater than 30 (if it is a smaller side), and less than 70 if it is (the largest side).

And thus, 30<x<70

Answer 28

tnx for correcting me.

Answer 29

meow

Answer 30

me 0 ?

Answer 31

that means np in kitty talk

Answer 32

sure:)

Answer 33

I confused you Whitemonster, sorry.

Answer 34

I think I'm getting it now haha, and thanks. :3

Answer 35

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Basically for any two sides C and D .
(where C < D )

The third side can either be:
x < C + D
(if the third side is the largest)

Or the third side can be:
x+C<D
x<D-C

So, for any 2 sides C and D in such a case, a third side, is

D-C < x < C + D

Answer 36

Oh, I did it correctly this time. COngrats, lol

Answer 37

Math hasn't even been my thing, if I helped you anyhow, you are welcome.

Answer 38

Yes, this one makes much more sense, now..
\(\Large\bf\color{magenta}{\bigstar~Thanks!~\bigstar}\)