See If I answered these right?

Question

AC + BC > AB 1st Triangle Inequality condition
AB + BC > AC 2nd Triangle Inequality condition
AB + AC > BC 3rd Triangle Inequality condition

 

: 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?

 

1. BC = 7; AC = 2 + AB 

2.AB + AC = 5; AC + BC = 4 
3. BC + AC = 22; AB + BC = 1

 

 

1. I solved for AB using AC = 2+AB and got -2 + AC = AB

Using the conditions above, I plugged each value in and solved to get these answers: 

 

condition 1)  2 + AB + 7 > -2 + AC 

                         AB + 7 > -2 + 2 + AC    ----- >  AC < AB + 7  (my answer) 

 

condition 2) -2 + AC + 7 > 2 + AB

                         AC + 7 > 2 + (-2) + AB    -----> AC > AB - 7 (my answer)  

 

condition 3) -2 + AC + (-2 +AB) > 7

                         AC  + AB > 7 + 2 +2   -----> AC > 11 - AB (my answer)

 

 

Please see if I answered these right, if not give an explanation, thank you for your help. 

 

 

Answer 1

You need the answer to be a compound inequality (with 2 inequality signs)

 

AC must be greater than the difference of the other two sides and less than the sum of the other two sides

 

For example, using a different triangle, if side #1 = 8 and side #2 = 10, then side #3 is defined as:

 

(10-8) < side #3 < (10 + 8)

 

1) BC - AB < AC < AB + BC

 

We know BC = 7 and AB = AC - 2

 

7 - (AC - 2) < AC < (AC - 2) + 7

 

split these two inequalties up into

 

7 - (AC - 2) < AC   and AC < (AC - 2) + 7

 

9 < 2AC                   and    on this side, you get AC < AC + 5

 

4.5 < AC                  and    the right side will always be true

 

so What can you conclude about AC  in (1) ? 

 

4.5< AC

 

I hope this help with the first part