MCQ
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
- A$2$
- B$9$
- ✓$4$
- D$5$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{a^2+8}{8 a}$
$\Rightarrow a^2-4 \sqrt{3} a+8=0$
$\Rightarrow a_1+a_2=4 \sqrt{3}, a_1 a_2=8$
$\Rightarrow\left|a_1-a_2\right|^2=\left(a_1+a_2\right)^2-4 a_1 a_2=48-32=16$
$\Rightarrow\left|a_1-a_2\right|=4$
$\therefore\left|\Delta_1-\Delta_2\right|=\frac{1}{2}\left|a_1-a_2\right| \cdot c \sin$ $B=\frac{1}{2} \times 4 \sin 30^{\circ} \times 4=4$
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