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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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Verified Answer
The correct answer is:
4
In $\triangle A B C$, by sine rule,
$$
\begin{aligned}
& \frac{a}{\sin A}=\frac{2 \sqrt{2}}{\sin 30^{\circ}}=\frac{4}{\sin C} \\
\Rightarrow \quad & C=45^{\circ}, C^{\prime}=135^{\circ} \\
\Rightarrow & A=180^{\circ}-\left(45^{\circ}+30^{\circ}\right)=105^{\circ}
\end{aligned}
$$
When $\quad C^{\prime}=135^{\circ}$, then
$$
\begin{aligned}
& \quad A=180^{\circ}-\left(135^{\circ}+30^{\circ}\right)=15^{\circ} \\
& \text { Area of } \triangle A B C=\frac{1}{2} A B \times A C \sin A \\
& =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(105^{\circ}\right) \\
& =4 \sqrt{2} \times \frac{\sqrt{3}+1}{2 \sqrt{2}}=2(\sqrt{3}+1) \\
& \text { Area of } \triangle A B C^{\prime}=\frac{1}{2} A B \times A C \sin A \\
& =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(15^{\circ}\right)=2(\sqrt{3}-1)
\end{aligned}
$$
Difference of areas of triangles
$$
=|2(\sqrt{3}+1)-2(\sqrt{3}-1)|=4
$$
$$
A D=2, D C=2
$$
Difference of areas of $\triangle A B C$ and $\triangle A B C^{\prime}$
$$
\begin{aligned}
& =\text { Area of } \triangle A C C^{\prime} \\
& =\frac{1}{2} A D \times C C^{\prime}=\frac{1}{2} \times 2 \times 4=4
\end{aligned}
$$
$$
\begin{aligned}
& \frac{a}{\sin A}=\frac{2 \sqrt{2}}{\sin 30^{\circ}}=\frac{4}{\sin C} \\
\Rightarrow \quad & C=45^{\circ}, C^{\prime}=135^{\circ} \\
\Rightarrow & A=180^{\circ}-\left(45^{\circ}+30^{\circ}\right)=105^{\circ}
\end{aligned}
$$
When $\quad C^{\prime}=135^{\circ}$, then
$$
\begin{aligned}
& \quad A=180^{\circ}-\left(135^{\circ}+30^{\circ}\right)=15^{\circ} \\
& \text { Area of } \triangle A B C=\frac{1}{2} A B \times A C \sin A \\
& =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(105^{\circ}\right) \\
& =4 \sqrt{2} \times \frac{\sqrt{3}+1}{2 \sqrt{2}}=2(\sqrt{3}+1) \\
& \text { Area of } \triangle A B C^{\prime}=\frac{1}{2} A B \times A C \sin A \\
& =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(15^{\circ}\right)=2(\sqrt{3}-1)
\end{aligned}
$$
Difference of areas of triangles
$$
=|2(\sqrt{3}+1)-2(\sqrt{3}-1)|=4
$$
$$
A D=2, D C=2
$$
Difference of areas of $\triangle A B C$ and $\triangle A B C^{\prime}$
$$
\begin{aligned}
& =\text { Area of } \triangle A C C^{\prime} \\
& =\frac{1}{2} A D \times C C^{\prime}=\frac{1}{2} \times 2 \times 4=4
\end{aligned}
$$
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