Triangle or not triangle, that is the question

Since 56+72=128, triangle ABC has no height. Therefore, its area is 0

As the sum of AB and BC (56+72) is equal to AC therefore the triangle is not possible

Solution using trigonometry.

By the law of cosines.

$(BC)^2=(AC)^2+(AB)^2-2(AC)(AB)(\cos A)$

$72^2=128^2+56^2-2(128)(56)(\cos A)$

$A=0^\circ$

So the area is also $\boxed{0}$ .

coinciding points

there is not a triangle as 128=72+56

The sides doesn't fit the triangle inequality theorem, therefore it's a triangle with 0 height.