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Hi, its me again...

OK can someone check my answer to see if it is complete?

This time the problem goes like this:

The perimeter of a triangle is 28. What are the possible lengths of the sides of the triangle?

My answer:

9,11,8
10,10,8
11,9,8
12,6,19
7,10,11
6,12,10

Is my list complete? If not, can you tell me what to add?

Bonus: Is there a rule to figuring out a problem like this?*


Thanks so much :)

2006-09-20 08:04:46 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

No, your list is not complete. Actually there are not only infinitely many possibilities, but uncountably many, because the sides are real numbers.

If a, b and c are the triangle sides, then, since each side is less than the sum of the other 2, a, b, c must satify the following conditions:

a + b + c = 28 => a = 28 - b - c
a < b + c => 28 - b - c < b + c => b + c > 14
b < a + c => b < 28 - b => b < 14
c < a + b => c < 28 - c => c < 14

and, of course, a, b , c> 0. There are uncountably many pairs (b, c) of positive numbers satisfying b < 14, c < 14 and b + c > 14. To each of such pais there corresponds a unique value for a = 28 - b - c

2006-09-20 08:32:24 · answer #1 · answered by Steiner 7 · 0 0

But you didn't specify that the sides had to be whole
numbers! If you allow fractions, then there are
infinitely many possibilities.
The rule is that x + y + z = 28 AND
the sum of any two of these numbers must be
greater than the third.
For example, one of the triples in your
list, 12, 6, 19, would not make a triangle
because 12 + 6 = 18 < 19.
This is a typo anyway. I think you meant 12, 6, 10 here.
This is the same triangle as 6, 12, 10.
In your list, you are missing 5, 10,13; 8, 8, 12;
7, 9,12; 7,8,13; 6,9,13; 4,11,13; 3,12,13; 2,13,13;
6,11,11; 4,12,12; 5,11,12.
Do you spot a pattern in what I'm doing?
Let's list the shortest side first.
It's not hard to see that these range from
2 to 10.
Now start with 2 and find all the possibilities
with one side 2, listing the shorter one second.
Then decrease the middle one by 1 and increase
the third one by 1 till the sum of 2 of these
becomes equal to the third.
So, 2,13, 13 is the only triangle with 2.
Now let's do 3.
We get 3,12,13 and that's the only one with 3.
Now let's do 4.
The possibilities are
4,12,12
4,11,13
One more: Let's do 5.
The possibilities are
5,11,12
5,10,13.
I'll let you carry on from here. Hope this makes
some sense!

2006-09-20 09:02:13 · answer #2 · answered by steiner1745 7 · 0 0

12+6+19=37 this isn't an answer

Assuming whole numbers.

From the rule that two sides of a triangle must sum to be a greater number then any third side. I took your perimeter, 28, divided it by two, 14. But it divides evenly so I took the two numbers 15 and 13. Then I broke the number 15 into two other numbers and started a list of possible matches as follows:
3, 12, 13
4,11,13
5,10,13
6,9,13
7,8,13

Then I went to 16 and 12:
5,11,12
6, 10, 12 you already have this one
7,9,12
8,8,12

Then 17 and 11
7, 10, 11 you already have this one
8, 9, 11 you already have this one

Then 18 and 10
8, 10, 10 you have this one
9, 9, 10

Then 19 and 9 and this combination won't work.

2006-09-20 08:24:46 · answer #3 · answered by Kaedence 2 · 0 0

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2016-12-12 11:46:55 · answer #4 · answered by ? 4 · 0 0

Your list is not complete, and can never be complete. Consider: 10.34, 8.65, 9.01 is such a triangle. It is obvious that the list is not only infinite, but uncountably infinite, as the sides can be any real numbers. (Unless there was a restriction to integers, which you did not state.) There is a typographic error in the problem as stated -- 12,6,19 should be 12,6,10.

2006-09-20 08:13:17 · answer #5 · answered by Anonymous · 0 0

9,11,8 yes
10,10,8 yes
11,9,8 yes
12,6,19 no
7,10,11 yes
6,12,10 yes

2006-09-20 08:13:17 · answer #6 · answered by raj 7 · 0 0

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