Question 253712
1. 5 and 9
<pre><font size = 4 color = "indigo"><b>
Let side a = 5.  Let side b = 9.  Then the triangle has
sides

a=5, b=9, and c=?.  Any side of a triangle must be less 
than the other two sides added together, so

a < b+c, b < a+c, and c < a+b

Substituting:

 5 < 9+c, 9 < 5+c, and c < 5+9

Solving all three for c:

-4 < c, 4 < c, c < 14

The first one is of course true, so ignore it.
Put the second and third together and get 

4 < c < 14

That's it. c must be more than 4 and less than 14.

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2. 7 and 14 
<pre><font size = 4 color = "indigo"><b>
Let side a = 7.  Let side b = 14.  Then the triangle 
has sides

a=7, b=14, and c=?.  Any side of a triangle must be less 
than the other two sides added together, so

a < b+c, b < a+c, and c < a+b

Substituting:

7 < 14+c, 14 < 7+c, and c < 7+14

Solving all three for c:

-7 < c, 7 < c, c < 21

The first one is of course true, so ignore it.
Put the second and third together and get 

7 < c < 21

That's it. c must be more than 7 and less than 21.

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3. 8 and 13 
<pre><font size = 4 color = "indigo"><b>
Let side a = 8.  Let side b = 13.  Then the triangle 
has sides

a=8, b=13, and c=?.  Any side of a triangle must be less 
than the other two sides added together, so

a < b+c, b < a+c, and c < a+b

Substituting:

8 < 13+c, 13 < 8+c, and c < 8+13

Solving all three for c:

-5 < c, 5 < c, c < 21

The first one is of course true, so ignore it.
Put the second and third together and get 

5 < c < 21

That's it. c must be more than 5 and less than 21.

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4. 10 and 12 
<pre><font size = 4 color = "indigo"><b>
Let side a = 10.  Let side b = 12.  Then the triangle 
has sides

a=10, b=12, and c=?.  Any side of a triangle must be less 
than the other two sides added together, so

a < b+c, b < a+c, and c < a+b

Substituting:

10 < 12+c, 12 < 10+c, and c < 10+12

Solving all three for c:

-2 < c, 2 < c, c < 22

The first one is of course true, so ignore it.
Put the second and third together and get 

2 < c < 22

That's it. c must be more than 2 and less than 22.

They are all done the same way. You can do the other half
by yourself.

Edwin</pre>