SOLUTION: The lengths of the sides of a triangle are consecutive odd numbers. What is the length of the longest side if the perimeter is 45 centimeters?

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Question 1181833: The lengths of the sides of a triangle are consecutive odd numbers. What is the length of the longest side if the perimeter is 45 centimeters?
Found 3 solutions by MathLover1, MathTherapy, greenestamps:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

a pair of any consecutive odd integers are also units apart
let consecutive odd numbers be , ,
if the perimeter is , we have





then


the length of the longest side is


Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
The lengths of the sides of a triangle are consecutive odd numbers. What is the length of the longest side if the perimeter is 45 centimeters?
Let longest side be L
Then other 2 are: L - 2, and L - 4
We then get: L + L - 2 + L - 4 = 45
3L - 6 = 45
3L = 51
Longest side, or
Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


One tutor used x, x+2, and x+4 for the three consecutive odd integers; the other used x, x-2, and x-4. Both choices are valid and lead to the correct answer if used properly. Perhaps the latter is a slightly better choice, since the problem asks for the length of the longest side; letting x be the longest side saves an extra step at the end.

Often the work to solve a problem like this is easier if we let x be the length of the middle number. The sides are then x-2, x, and x+2; the sum is 3x=45 so x is 15; then the longest side is x+2=17.

I personally prefer this last method, even though the solution to my equation is not the answer to the problem; overall, the work getting to the answer is a bit easier.

The three methods for setting up the problem in the three responses you have received are an example of the fact that, when you are learning to solve problems like this, you should try different methods for setting up the problem to see what methods "work" for you.


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