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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Aicharya · Accepted Answer

In $	riangle 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}ight)=105^{\circ} \end{aligned} $ When $\quad C^{\prime}=135^{\circ}$, then $ \begin{aligned} & \quad A=180^{\circ}-\left(135^{\circ}+30^{\circ}ight)=15^{\circ} \ & 	ext { Area of } 	riangle A B C=\frac{1}{2} A B 	imes A C \sin A \ & =\frac{1}{2} 	imes 4 	imes 2 \sqrt{2} \sin \left(105^{\circ}ight) \ & =4 \sqrt{2} 	imes \frac{\sqrt{3}+1}{2 \sqrt{2}}=2(\sqrt{3}+1) \ & 	ext { Area of } 	riangle A B C^{\prime}=\frac{1}{2} A B 	imes A C \sin A \ & =\frac{1}{2} 	imes 4 	imes 2 \sqrt{2} \sin \left(15^{\circ}ight)=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 $	riangle A B C$ and $	riangle A B C^{\prime}$ $ \begin{aligned} & =	ext { Area of } 	riangle A C C^{\prime} \ & =\frac{1}{2} A D 	imes C C^{\prime}=\frac{1}{2} 	imes 2 	imes 4=4 \end{aligned} $

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…

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