# How to know if the lengths may form a triangle? For example. I have 3 cm, 10 cm, 15 cm sticks, will it form a triangle? How to guess/predict just with the info's about the lengths? not by trying it in actual.

2
by JC10

2016-01-19T21:09:14+08:00
The sum of the two sides must be greater than the other side. For example, 3 + 10 = 13, 13 < 15, therefore, you can't form a trigon with a measurement of 3, 10, and 15 units.

Maybe....
a + b > c

where a, b, and c are the lengths.

For example..
10 + 18 > 25
28 > 25
Therefore, you may form a trigon with sides having lengths of 10, 18, and 25.

Actually, I'm not sure about my answer/explanation. Why? because it's my assignment either ._. but don't worry, with the use of examples given to us, I can tell that it's sometimes true ☺. But still, I hope it help. Oh, and also, I'm not sure if the sum of the two sides is just the same as the other side, not sure if it will form a triangle.

I hope it help. ☺

sorry if I've answered it here even if, I'm not sure because it's my assignment either ._.
it's okay :) Wow, what a coincidence!
you sure it's a coincidence? ._. It seems like you know and just pretending that you don't.
lol
• Brainly User
2016-01-20T04:55:14+08:00
Since triangle or trigon has three sides, let:

a = side 1
b = side 2
c = side 3

If a and b are the two sides, then, to form a triangle the third side must satisfy the condition:
(a - b ) < c < a + b

Side c must be greater than the difference of sides a and b, but less than the sum of a  and b.

Given:
Option 1:
Side a = 3 cm
Side b = 10 cm
Side c = 15 cm

(10-3) < 15 < (3+ 10)
7 < 15 < 13       ⇒    False
⇒   The third side must be
{8, 9, 10, 11, 12},
not {15}

Option 2:
Side a = 10 cm
Side b = 15 cm
Side c = 3 cm

(15-10) < 3 < (10 + 15)
5     < 3 <   25   ⇒   False
⇒   The third side must be
{6, 7, 8, 9, 10, ..., 23, 24, 25},
not {3}

Option 3:
Side a = 15 cm
Side b = 3 cm
Side c = 10 cm

(15-3) < 10 < (15 + 3)
12   < 10 < 18       ⇒   False
⇒   The third side must be
{13, 14, 15, 16, 17}
not {10}

Thanks!
You're welcome:-)  In Option 2, the solution set must be {6,7,...,23,24); sorry typo error :-)  Further, given any two sides in Options 1 to 3, the solution sets also provide for the possible number of triangles that can be formed:-)