The possible range for x, in the terms, of BC is BC < x< 3BC.
Triangle Inequality Theorem
In mathematic, there is a rule for triangle side lengths. This rule says that the sum of the lengths of any two sides of a triangle is greater than the length of the one side (third side). Therefore,
[tex]L1+L2 > L3\; (1)\\ \\ L1+L3 > L2\; (2)\\ \\ L2+L3 > L1\; (3)[/tex]
The question gives three sides: AB (L1), BC (L2) and x (L3). Also, you know that AB=2 BC.
Applying the equation (1) of rule for triangle side lengths, you have:
L1+L2> L3
AB+BC > x
2BC+BC > x
3BC > x
x< 3BC
Applying the equation (2) of rule for triangle side lengths, you have:
L1+L3 > L2
AB+x >BC
2BC+x > BC
x > -BC
Applying the equation (3) of rule for triangle side lengths, you have:
L2+L3 > L1
BC + x > AB
BC + x > 2BC
x > BC
The intersection of the inequalities results is the solution of this problem. Then, the possible range for x, in the terms, of BC is BC < x< 3BC. See the attached image.
Read more about the triangle inequality theorem here:
brainly.com/question/2403556
