SOLUTION: use interger values for the lengths of the sides or use 12 matchsticks to investigate how to make up your triangles

Question 947866: use interger values for the lengths of the sides or use 12 matchsticks to investigate how to make up your triangles
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
What is the question? I w3ill make up a question myself.

How many different triangles can you make with a perimeter of 12, and integer side lengths?

The length of one side of a triangle,
let's say side AB,
is the distance from A to B along a straight line.
Of course, it is less than the total distance covered
first going from A to C, and then
going from C to B.
In other words,
the length of one side of a triangle,
is less than the sum of the lengths of the other two sides.

So if the longest side's length must be less than the sum of the lengths of the other two sides,
in a triangle where all side lengths add up to 12,
the longest side's length must be less than 6.

If the longest side's length is 5,
the other two side lengths must add up to 12-5=7.
Their length could be 4 and 3 (because 4+3=7),
or 5 and 2 (because 5+2=7),.
(In that last case, two sides have a length of 6,
and are tied for longest side).
Counting all the options above, we have 2 triangles so far.

If the longest side's length is 4,
the other two side lengths must add up to 12-4=8,
and since 4+4=8,
our only option is an equilateral triangle where all sides have a length of 4.

NOTE 1: Take into account that I consider the two mirror-image triangles in the drawing below to be the same triangle, just flipped over like a pancake.
, so I count them as 1 triangle.

NOTE 2: If that was not your question, ask again.
We are good guessers, but we cannot read minds clearly over long distances.
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