Question

the perimeter of a triangle is 57 in. The Lengths of the sides are represented by three consecutive odd integers. Find the sides of the triangle.

Answer 1

A+B+C = 57
A = B+2 = C+4.
C+4+C+2+C = 57. Find C, then A, B.

Answer 2

wouldnt it be something like 1+(x+1)+(x+2)+(x+3)=57????

Answer 3

No. Odd numbers increase by 2.

Answer 4

so it would be x+x+2+x+4

Answer 5

Yes. Remember to use x to find the remaining two sides.

Answer 6

Like we have 3 numbers
The 3 numbers are consecutive odd numbers
For example the following are consecutive odd numbers:
1,3,5,7,9,11
Notice that this can be written as 1,1+2,1+2(2),1+2(3),1+2(4),1+2(5)
Do you see a pattern:
This pattern can be simplified to 1,1+2,1+4,1+6,1+8,1+10
What if we let n be 1, then we can write this as n,n+2,n+4,n+6,n+8,n+10

But we have only 3 consecutive numbers in your problem so we will only need the first 3 from the above pattern,
Your first odd number will be n
Your second odd number will be n+2
Your third odd number will be n+4


So we have the perimeters is just us adding the sides
So we have n+(n+2)+(n+4)=57
Or you drop the parentheses and just say n+n+2+n+4=57

Answer 7

so that would be 3n+6=57 then 57-6=51.. so then we have 3n=51 51/3= 17... so would that be the answer??

Answer 8

That is one side.
The other sides are n+2 and n+4.

Answer 9

There are three answers.

Answer 10

17, 19, and 21???

Answer 11

Check by adding them up.

Answer 12

so it would be 17+2 and 17 +4???

Answer 13

for the other 2 sides???

Answer 14

thank yuo guys! :)

Answer 15

Maybe. Check by adding the results. If you get 57, you're correct.
You're welcome.