Math Is Fun Forum

Winifred

Guest

triangles. please help?

I don't know how I am supposed to solve these problems!

Tell whether or not they can be the lengths of the sides of a triangle:

1. 2 ft, 7 ft, and 9 ft.
2. 12 cm, 7 cm, 15 cm
3. 27 yd, 4 yd, 30 yd

and: Find all the sets of four numbers with a sum of 16.

Zach

Member

Registered: 2005-03-23

Posts: 2,075

Re: triangles. please help?

2, 7 and 9
2² = 4
7² = 49
9² = 81

A = 2
B = 7
C = 9

A² + B² ≠ C²
A² + C² ≠ B²
C² + B² ≠ A²

Then just shuffled them around. Remember the almighty formula for Triangles;
A² + B² = H² (or C²)

Which is the sum of the squares of each side is equal to the square of the hypotenuse.

Of course, this is presuming you're doing right-angled triangles.

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I bet you didn't know it, but I'm a fiddle player too.
And if you'd care to take a dare, I'll make a bet with you.

Winifred

Guest

Re: triangles. please help?

No, we are not doing right-angled triangles.

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: triangles. please help?

Find all the sets of four numbers with a sum of 16.

I *hope* they mean positive whole numbers, or the answer is going to be *very* long.

A "set" normally implies no duplicates, too.

So (making all these assumptions) you could approach it this way: find all combinations of *three* numbers, then calculate the 4th like this:

1+2+3 = 6, 16-6=10: result: 1,2,3,10
1+2+4 = 7, 16-7=9: result: 1,2,4,9
...

The trick is to get all sensible combinations.

You could use my Combinations and Permutations Calculator to help

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: triangles. please help?

On the trianle issue:   The sum of the length of any 2 sides must be greater than the 3rd side.    So sides of 2, 7 and 9 is not possible because 2 + 7 is not greater than 9.  The next set (7,12, 15) is valid because 7 + 12 > 15, 12 + 15 > 7 and 7 + 15 > 12.   Really all you need to check is to make sure the sum of the 2 shorter sides is greater than the longest side.  So the third set is also valid since 4 + 27 > 30.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: triangles. please help?

pi man's right about the triangle thing. If you've only been given side lengths, then the only thing you need to check to make sure a triangle can be drawn is the triangle inequality. There are some other checks involving angles as well, but as you don't have any angles you don't need to worry about them.

With the numbers adding to 16, it actually doesn't take too long to write out all the combinations. Just make sure you follow the rule that each number you write is more than or equal to the number before it, so that you don't repeat combinations in a different order.

Here are all the sets that contain 1: (1,1,1,13), (1,1,2,12), (1,1,3,11), (1,1,4,10), (1,1,5,9), (1,1,6,8), (1,1,7,7,), (1,2,2,11), (1,2,3,10), (1,2,4,9), (1,2,5,8), (1,2,6,7), (1,3,3,9), (1,3,4,8), (1,3,5,7), (1,3,6,6), (1,4,4,7), (1,4,5,6), (1,5,5,5).

That's more than half of them already. Just keep going methodically and you should manage to find all the others fairly easily. I got that there are [thirty-four] different possible sets. See if you agree.

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John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: triangles. please help?

mathsy seems to be putting the digits in increasing order of size; they get bigger to the right, so I'll continue...
(1,5,5,5)...(2,2,2,10),(2,2,3,9),(2,2,4,8),(2,2,5,7),(2,2,6,6),(2,3,3,8),(2,3,4,7),(2,3,5,6),(2,4,4,6),(2,4,5,5),(2,5,5,nope),
(2,5,6,nope),(2,6,6,nope),(3,3,3,7),(3,3,4,6),(3,3,5,5),(3,4,4,5),(3,5,nope),(4,4,4,4).  All done!!
Thirty-four of them.  Is that right??

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igloo myrtilles fourmis

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: triangles. please help?

Should be 64 .

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John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: triangles. please help?

For triangles, any two sides you pick are beside one another and these two sides have to total longer than the other side inorder to reach both ends of the other side.  So for sides A,B,C:  A < B+C and  B < A+C and C < A + B.
So for 2, 7, and 9,   9 = 2 + 7, which is not bigger, so it makes a straight line doubled back on itself, not a triangle.

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igloo myrtilles fourmis

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: triangles. please help?

For the 2nd question, if you allow zeroes, there are 64 combinations:

16,0,0,0

15,1,0,0

14,2,0,0    14,1,1,0

13,3,0,0    13,2,1,0    13,1,1,1

12,4,0,0    12,3,1,0    12,2,2,0    12,2,1,1

11,5,0,0    11,4,1,0    11,3,2,0    11,3,1,1    11,2,2,1

10,6,0,0    10,5,1,0    10,4,2,0    10,4,1,1    10,3,3,0    10,3,2,1    10,2,2,2

9,7,0,0    9,6,1,0    9,5,2,0    9,5,1,1    9,4,3,0    9,4,2,1    9,3,3,1    9,3,2,2

8,8,0,0    8,7,1,0    8,6,2,0    8,6,1,1    8,5,3,0    8,5,2,1    8,4,4,0    8,4,3,1    8,4,2,2    8,3,3,2

7,7,2,0    7,7,1,1    7,6,3,0    7,6,2,1    7,5,4,0    7,5,3,1    7,5,2,2    7,4,4,1    7,4,3,2    7,3,3,3

6,6,4,0    6,6,3,1    6,6,2,2    6,5,5,0    6,5,4,1    6,5,3,2    6,4,4,2    6,4,3,3

5,5,5,1    5,5,4,2    5,5,3,3    5,4,4,3

4,4,4,4

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: triangles. please help?

I told you

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Hooyoo

Member

Registered: 2006-09-08

Posts: 2

Re: triangles. please help?

It's very easy to solve the 2nd question if you can program! ^_^

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: triangles. please help?

and if u allow negative numbers, there are some more combinations