Difference Between Areas of Two Non Congruent Triangles

To find the absolute value of the difference between the areas of the two triangles $ABC$ and $ABC'$, we apply the Law of Cosines to $\triangle ABC$.

Let $c = AB = 4$, $b = AC = 2\sqrt{2}$, $\angle B = 30^\circ$, and let $a = BC$ be the length of the third side.

By the Law of Cosines:
$b^2 = a^2 + c^2 - 2ac \cos B$

Substituting the given values:
$(2\sqrt{2})^2 = a^2 + 4^2 - 2(a)(4)\cos 30^\circ$
$8 = a^2 + 16 - 8a \left(\frac{\sqrt{3}}{2}\right)$
$a^2 - 4\sqrt{3}a + 8 = 0$

Solving this quadratic equation for $a$:
$a = \frac{4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 - 4(1)(8)}}{2}$
$a = \frac{4\sqrt{3} \pm \sqrt{48 - 32}}{2}$
$a = \frac{4\sqrt{3} \pm \sqrt{16}}{2} = 2\sqrt{3} \pm 2$

Since both roots are positive, the two possible values for the side opposite to vertex $A$ are:
$a_1 = 2\sqrt{3} + 2 \quad \text{and} \quad a_2 = 2\sqrt{3} - 2$

These correspond to the two non-congruent triangles $ABC$ and $ABC'$, with base lengths $BC = a_1$ and $BC' = a_2$.

The area of a triangle with sides $a$, $c$ and included angle $B$ is given by:
$\Delta = \frac{1}{2}ac\sin B$

Thus, the absolute difference between the areas of the two triangles is:
$|\Delta_1 - \Delta_2| = \frac{1}{2} c \sin B \, |a_1 - a_2|$

Substituting $c = 4$, $B = 30^\circ$, and $|a_1 - a_2| = (2\sqrt{3} + 2) - (2\sqrt{3} - 2) = 4$:
$|\Delta_1 - \Delta_2| = \frac{1}{2} \times 4 \times \sin 30^\circ \times 4 = \frac{1}{2} \times 4 \times \frac{1}{2} \times 4 = 4$