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In a triangle $\mathrm{ABC}$, it is known that $\mathrm{AB}=\mathrm{AC}$. Suppose $\mathrm{D}$ is the mid- point of $\mathrm{AC}$ and $\mathrm{BD}=\mathrm{BC}=2$. Then the area of the triangle $\mathrm{ABC}$ is $-$
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Verified Answer
The correct answer is:
$\sqrt{7}$
We know
$\begin{array}{l}
\mathrm{AB}^{2}+\mathrm{BC}^{2}=2\left(\mathrm{CD}^{2}+\mathrm{BD}^{2}\right) \\
\mathrm{AB}^{2}+4=2\left(\frac{\mathrm{AB}^{2}}{4}+4\right) \\
\mathrm{AB}^{2}+4=\frac{\mathrm{AB}^{2}}{2}+8 \\
\frac{\mathrm{AB}^{2}}{2}=4 \\
\mathrm{AB}^{2}=8 \\
\mathrm{AB}=2 \sqrt{2}
\end{array}$
Now
Area $=\frac{1}{2} \times 2 \times \cdot \sqrt{7}=\cdot \sqrt{7}$ square unit.
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