Let A B C and A B C^{prime} be two non-congru | vedclass

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$	ext { Using cosine rule } \cos B=\frac{a^2+c^2-b^2}{2 a c}=\frac{a^2+16-8}{2 	imes a 	imes 4}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{a^2+8}{8

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$4$

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$	ext { Using cosine rule } \cos B=\frac{a^2+c^2-b^2}{2 a c}=\frac{a^2+16-8}{2 	imes a 	imes 4}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{a^2+8}{8 a}$

$\Rightarrow a^2-4 \sqrt{3} a+8=0$

$\Rightarrow a_1+a_2=4 \sqrt{3}, a_1 a_2=8$

$\Rightarrow\left|a_1-a_2ight|^2=\left(a_1+a_2ight)^2-4 a_1 a_2=48-32=16$

$\Rightarrow\left|a_1-a_2ight|=4$

$	herefore\left|\Delta_1-\Delta_2ight|=\frac{1}{2}\left|a_1-a_2ight| \cdot c \sin$ $B=\frac{1}{2} 	imes 4 \sin 30^{\circ} 	imes 4=4$

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

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