SOLUTION: Activity: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. What can you conclude about AC in ABC? (Hint: Use the above exampl

Question 1207717: Activity: The sum of the lengths of any two sides of a triangle is greater than the length of the third
side. What can you conclude about AC in ABC? (Hint: Use the above example as a guideline.)
2. BC = 7; AC = 2 + AB
3. AB + AC = 5; AC + BC = 4
4. BC + AC = 22; AB + BC = 12
These are the only questions I have. I don't want to make three different emails of practically the same thing.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

AB, BC, and AC are the three sides of triangle ABC.

The Triangle Inequality Theorem generates these 3 inequalities
AB + BC > AC
AB + AC > BC
AC + BC > AB
I'll refer to them as inequality (1) through (3).

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Problem 2

We're given that
AC = 2 + AB

Let's solve for AB
AC = 2 + AB
AB = AC - 2

Then apply this to inequality (1)
AB + BC > AC
AC - 2 + 7 > AC
AC + 5 > AC
5 > 0
The last inequality is always true, so the first inequality is always true.
We don't really generate anything interesting here, so let's move on.

Now move to inequality (2)
AB + AC > BC
AC-2 + AC > 7
2AC-2 > 7
2AC > 7+2
2AC > 9
AC > 9/2
AC > 4.5
Inequality (2) is only true when AC is larger than 4.5

Lastly inequality (3)
AC + BC > AB
AC + 7 > AC-2
7 > -2
This is always true.
Like with the first case shown above, nothing interesting is here.

To wrap up problem 2, if triangle ABC has sides BC = 7 and AC = 2+AB, then we conclude that AC > 4.5

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Problem 3

AB + AC = 5 solves to AB = 5 - AC
AC + BC = 4 solves to BC = 4 - AC

Plug them into inequality (1)
AB + BC > AC
5-AC + 4-AC > AC
9-2AC > AC
9 > 3AC
AC < 9/3
AC < 3

Also, plug them into inequality (2)
AB + AC > BC
5-AC + AC > 4-AC
5 > 4-AC
5-4 > -AC
1 > -AC
AC > -1
This will always be true since AC is a positive length.
This subcase isn't useful so we move on.

Lastly inequality (3)
AC + BC > AB
AC + 4-AC > 5-AC
4 > 5-AC
4+AC > 5
AC > 5-4
AC > 1
1 < AC

Combine 1 < AC with AC < 3 to conclude 1 < AC < 3 is the answer to problem 3.

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Problem 4

BC + AC = 22 solves to BC = 22 - AC

Plug this into the other equation so we get something in terms of AC.
AB + BC = 12
AB + 22-AC = 12
AB - AC = 12-22
AB - AC = -10
AB = AC-10

The key equations we'll be using are
BC = 22-AC and AB = AC-10
Like with the previous problem, they are useful for substitutions.

Let's revisit inequality (1)
AB + BC > AC
AC-10 + 22-AC > AC
12 > AC
AC < 12

Now focus on inequality (2)
AB + AC > BC
AC-10 + AC > 22-AC
2AC-10 > 22-AC
2AC+AC > 22+10
3AC > 32
AC > 32/3
AC > 10.6667 approximately

And now inequality (3)
AC + BC > AB
AC + 22-AC > AC-10
22 > AC-10
22+10 > AC
32 > AC
AC < 32

Combine or overlap AC < 12 with AC < 32, so we determine overall that AC < 12.
If some length of AC makes AC < 12 true, then it autometically makes AC < 32 true as well (but not always the other way around).

Combine 32/3 < AC with AC < 12 to determine 32/3 < AC < 12
In other words, 10.6667 < AC < 12 when going with the approximate form.

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Answers:
Problem 2) AC > 4.5
Problem 3) 1 < AC < 3
Problem 4) 32/3 < AC < 12 (where 32/3 = 10.6667 approximately)

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