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 $-$

• A

  2
• B

  $2 \sqrt{2}$
• Correct answer
  C

  $\sqrt{7}$
• D

  27

Solution

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.

Hence, the correct answer is option: 'C'.