Triangle area

I've just seen the following image on spikedmath.com:

The second answers seem to be obviously the correct ones, right? Wrong.

According to Heron's formula you can calculate a triangles area like this:

Let $a, b, c$ be the side lengths of the triangle. $s := \frac{a+b+c}{2}$ $T = \sqrt{s \cdot (s-a) \cdot (s-b) \cdot (s-c)}$

So the area of the first triangle is $s_1 := \frac{16}{2} = 8$ $T_1 := \sqrt{8 \cdot (8-5) \cdot (8-5) \cdot (8-6)} = \sqrt{8 \cdot 3 \cdot 3 \cdot 2} = 3 \cdot 4 = 12$

The area of the second one is $s_1 := \frac{18}{2} = 9$ $T_1 := \sqrt{9 \cdot (9-5) \cdot (9-5) \cdot (9-8)} = \sqrt{9 \cdot 4 \cdot 4 \cdot 1} = 3 \cdot 4 = 12$

Both triangles have the same area!

When you draw it, it looks like this: