The perimeter of a triangle is equal to 12 inches. The length of the longest side is equal to the sum of the other two sides minus 2. If you multiply the shortest side by 3 and add one, it will equal twice the longest side. Find the side lengths.
@Elysse2012 I have done some work on this problem but I need you to check it. If you have not already solved it, let me know.
yeah?
Let L represent the longest side of the triangle. Let M represent the second longest side of the triangle. Let S represent the smallest side of the triangle. Now, to write the equations.
The perimeter of a triangle is equal to 12 inches. ------------------------------------------ L + M + S = 12
length of the longest side is equal to the sum of the other two sides minus 2. --------------------------------- L = M + S - 2
multiply the shortest side by 3 and add one, it equals twice the longest side -------------------------------- 3S + 1 = 2L
Agree with those three equations as accurate expressions of the relationships of the sides of this triangle? @Elysse2012
Waiting for a reply before I proceed.
@Elysse2012 Agree or disagree with the three equations?
agree
1) L + M + S = 12 2) L = M + S - 2 --> L - M - S = -2 3) 3S + 1 = 2L
Add the terms of equations (1) and (2) 1) L + M + S = 12 2) L - M - S = -2 --------------------- 2L + 0 + 0 = 10 2L = 10 L = ? @Elysse2012
5
Take the third equation: 3) 3S + 1 = 2L 3S + 1 = 2*5 where L =5 from above work. Solve the equation for S (no decimals at this point, please) @Elysse2012
3S+1=32 3S=31 S= 10 1/3
OH OOPP..
nevermind
wait.... is that right?
I am not following how this: 3S + 1 = 2L 3S + 1 = 2*5 3S + 1 = 10 became this: 3S+1=32 3S=31 I may have messed you up talking about decimals. I had a typo of 17 as the perimeter which is not right and sent me to fraction-ville.
>wait.... is that right? NO
Solve this for S: 3S + 1 = 10 We're almost finished.
@Elysse2012 Question?
sorry I didn't look carefully at the 2*5.. thought it was a 2^5
3S = 9 S=3
and so M is 4
Yes. That's correct.
ok :) thank you!
Glad to help.
Join our real-time social learning platform and learn together with your friends!