Answer my 7th grade math question?

Question

please help


the perimeter of any triangel is the sum of the lengths of its sides.the lengths of the sides of a certain triangle,in feet, are consecutive even numbers. the perimeter of this triangle is between 10 feet to 24 feet inclusive.list all possible lengths for the longest side of the triangle.

Answer 1

6 ft, 8ft, 10 ft

a. the sides would be 2 ft, 4 ft, and 6 ft....the perimeter would be 12 ft, and the longest side would be 6 ft.

b. the sides would be 4 ft, 6 ft, and 8 ft...the perimeter would be 18 ft and the longest side would be 8 ft

c. the sides would be 6 ft, 8 ft, and 10 ft...the perimeter would be 24 ft and the longest side would be 10 ft.

Hope this helps!

Answer 2

Lets assign variables for each side.

Consecutive even integers so that would be N, N+2, N+4

If you add the three sides you get the perimeter so that means the perimeter is equal to 3N+6

We know the perimeter is between 10 and 24
That can be rewriten as 10<3N+6<24

solve out each inequalitie seperately

10<3N+6
4<3N
4/3 so the smallest integer side that fits that inequalitie is 2

then the second inequalitie

3N+6<24
3N<18
N<6

The largest integer side that satisfies this inequalitie is 5

This makes the sides that are possible {2, 4, 6}; {4,6,8}

These are the only two possibilities

The two N lengths are 2 and 4 b/c they must be satisfy both inequalities and be even as the problem states. Then you just solve for the other sides.

So all the possible lengths for the longest side are the largest in each set which I gave. Those are 6 and 8. One problem however.
The first set is not probable. The law of triangles states that any two sides of a triangle when added together are larger than the third if the triangle is possible. So the first set is impossible because 2+4=6 and is not greater that 6. So the only real side possible side is 8.

I hope that helps and that I am right!

Answer 3

List all the possible perimeter values first, then figure out the sides for each situation, then take from each situation the longest side.

12: 2 + 4 + 6
18: 4 + 6 + 8
24: 6 + 8 + 10

So the longest side could be 6 feet, 8 feet, or 10 feet.

Answer 4

Oh wow, I haven't done probability in a while. Hmm... first, figure out the total number of possible combinations from the set. So that's 5+4+3+2+1 (because 6 has 5 possible pairings, then 5 already had 5*6 taken up, so it only has 4 possible pairings, etc... if this doesn't make sense, draw a picture. Just do 2 columns of numbers (6, 5, 4, 3, 2, and 1), then connect numbers on the left to possible pairings on the right, and count the lines as you draw them. Don't do right to left, though, since 4*5 is the same thing as 5*4. You also don't have to do 6*6, 5*5, etc, since the question says DIFFERENT numbers) Then find the number of combinations that multiply to 12 - 6*2 and 3*4 (I hope I didn't miss any?), which is 2 combos out of a total 15 So, the probability is 2/15.

Answer 5

2+4+6=12
4+6+8=18
6+8+10=24
The possible values of the longest side of triangle are 6, 8 and 10.

Answer 6

a. the perimeter is 2+4+6 with the longest side being 6
b. the perimeter is 4+6+8 with the longest side being 8
c. the perimeter is 6+8+20 with the longest side being 10

In essence, there are three choices being 6,8,or 10 feet.

Answer 7

6, 8, and 10

Answer 8

Hi. The question implies a right triangle. The first set (not even) is 3,4 5 triangle. Double these and you get 6,8,10, so 10 is one of them. Want to try the rest on your own?

Answer 9

so basically
2x4x6 is 12 - 6
4x6x8 is 18 - 8
6x8x10is 24 - 10

three possibilities

6,8, and 10

Answer 10

POST THE OTHER ONES B/C U GET BETTER ANSWERS.