Mysterious Triangle

A good thing to notice is that Heron's formula can be used to solve for one side of this triangle. In this case, the perimeter is $14$ , so the semi-perimeter is $7$ . You are given one side that is of length 3, and you are told that the area is $2\sqrt { 14 }$ . We can call one of the sides $x$ and the other $14-3-x$ , or more simply, $11-x$ . Using Heron's formula, $2\sqrt { 14 } =\sqrt { 7(7-3)(7-x)(7-(11-x)) }$ Now to simplify: ${ \left( 2\sqrt { 14 } \right) }^{ 2 }={ \sqrt { 7(7-3)(7-x)(7-(11-x)) } }^{ 2 }\\ 4(14)=7(4)(7-x)(-4+x)\\ 56=28(7-x)(-4+x)\\ 4=(7-x)(-4+x)\\ 4=-28+11x-{ x }^{ 2 }\\ { x }^{ 2 }-11x+30=0\\ (x-5)(x-6)=0\\ x=5 \quad or \quad x=6$ The fact that you are given two values for x does not change anything because, if you plug these values for x to find the triangle's sides, you will see that 11-x produces the other value of x. The positive difference of these values is 1.

If a , b , and c are the sides of a triangle such that a<b<c , then triangle can only be formed if , a+b>c

Given $a+b+c = 14$

And a = short side= 3

-----> $b+c = 11$

Possible combination of b and c are (1,10),(2,9),(3,8),(4,7) and (5,6) .

Among which, (b,c)=(5,6) satisfies condition $a+b>c$

Hence difference is 1 .