The sum of any two sides must be greater than the third side for a triangle to exist
let the third side be x
x+11>17 AND x+17>11 AND 11+17> x
x > 6 AND x>-6 AND x < 28
so 6 < x < 28
So how many positive integers can you count between 6 and 28 exclusive ?
Two sides of a triangle are 11 and 17. How many possible lengths are there for the third side, if it is a positive integer?
I tried listing them all but I get confused...Please help thank you!
5 answers
PLEASE DO NOT TRY TO CHEAT AND USE THE INTERNET FOR YOUR ANSWERS TO YOUR HOMEWORK THANKS!
Let the third side be $n$. Then by the triangle inequality,
\begin{align*} n + 11 &> 17, \\ n + 17 &> 11, \\ 11 + 17 &> n, \end{align*}
which gives us the inequalities $n > 6$, $n > -6$, and $n < 28$. Therefore, the possible values of $n$ are 7, 8, $\dots$, 27, for a total of $27 - 7 + 1 = \boxed{21}$ possible values.
Let the third side be $n$. Then by the triangle inequality,
\begin{align*} n + 11 &> 17, \\ n + 17 &> 11, \\ 11 + 17 &> n, \end{align*}
which gives us the inequalities $n > 6$, $n > -6$, and $n < 28$. Therefore, the possible values of $n$ are 7, 8, $\dots$, 27, for a total of $27 - 7 + 1 = \boxed{21}$ possible values.
Please don't cheat or we will track your IP address and ban you from AoPS. These problems are meant to challenge you and asking for help won't get you hard. Thank you!
Scroll down all the way to the bottom of this message
GET TROLLED HAHA...
GET TROLLED HAHA...
PLEASE DONT CHEAT OR I WILL EAT YOUR BOOTY OFF YOUR OG AKA MOMMA
1 answer