# The question is below?

## Let $A B C$ and $A B {C}^{'}$ be two non congruent triangles with sides $A B = 4 , A C = A C ' = 2 \sqrt{2}$ and angle $\angle B = {30}^{\circ}$. The absolute value of the difference between the areas of these triangles is?

Jun 6, 2018

The two non congruent triangles $\Delta A B C \mathmr{and} \Delta A B C '$ differ in sides BC and BC'.Their possible lengths can be found out by solving the following quadratic equation of $x$ obtained by using cosine law of triangle. Here two roots of $x$ represents the lengths of $B C \mathmr{and} B C '$. Let $B C ' > B C$

Now

$\cos {30}^{\circ} = \frac{{x}^{2} + {4}^{2} - {\left(2 \sqrt{2}\right)}^{2}}{2 \cdot 4 \cdot x}$

$\implies \frac{\sqrt{3}}{2} = \frac{{x}^{2} + 16 - 8}{8 x}$

$\implies {x}^{2} - 4 \sqrt{3} x + 8 = 0$

$\implies x = \frac{4 \sqrt{3} \pm \sqrt{{\left(4 \sqrt{3}\right)}^{2} - 4 \cdot 8 \cdot 1}}{2}$

$\implies x = 2 \sqrt{3} \pm 2$

So $B C ' = 2 \sqrt{3} + 2 \mathmr{and} B C = 2 \sqrt{3} - 2$

So $B C ' - B C = 4$

Now $\Delta A B C ' = \frac{1}{2} \cdot A B \cdot B C ' \cdot \sin {30}^{\circ}$
$= \frac{1}{2} \cdot 4 \cdot B C ' \cdot \frac{1}{2} = B C '$

Similarly

$\Delta A B C = \frac{1}{2} \cdot A B \cdot B C \cdot \sin {30}^{\circ}$
$= \frac{1}{2} \cdot 4 \cdot B C \cdot \frac{1}{2} = B C$

So $\Delta A B C ' - \Delta A B C = B C ' - B C = 4$ squnit