Finding the range of third side of triangle

[Shogun]

I did this for a homework assignment and got it wrong. I do not see where I made my mistake. Can someone please correct me?

Given two sides of a triangle with the following measurements, (20 and 66) find the range of possible measures for a third side.

A^2 + B^2+C^2

20^2 + B^2 + 66^2

400 + B^2 + 4356
- 400 -400

B^2 =3956

square root of 3956 = 62.9

final answer is : 62.9 X < 86

 

[pka]

The Inequality for Triangles states that:
The sum of the lengths of any two sides must exceed the length of the third side.

You have two sides of 20 & 66. Thus, the third side must be less than 86 (20+66) and the third side must be greater than 46 (66-20).

Here is the rule.
If a & b are the lengths of two sides of a triangle then if x is the length of the third side we have: |a-b|<x<(a+b).


You seem to be using the Pythagorean Theorem; it has nothing to do with these sorts of problems. Use Inequality for Triangles stated above.