Question

In the above figure, what is the perimeter of $\Delta$ $ACD$?

• A. $2\sqrt{6}-2\sqrt{2}$
• B. $4\sqrt{3}$
• C. $4+2\sqrt{6}+2\sqrt{2}$
• D. $2\sqrt{6}+6\sqrt{2}$

Answer 1

The correct option is C $4+2\sqrt{6}+2\sqrt{2}$

In $\Delta ABD,\angle ABD+\angle ADB+\angle BAD={180}^o$

$\Rightarrow \angle BAD={60}^o$

$\therefore \angle BAC=60-15={45}^o$

$\therefore \angle ACB={45}^o$ and $\Delta ABC$ is an isosceles triangle.

$AB=BC=2\sqrt{2}$

$\tan \left({30}^o\right)=\frac{AB}{BD}$

So, $BD=2\sqrt{6}$

$CD=BD-BC=2\sqrt{6}-2\sqrt{2}$ ...(1)

$\sin \left({30}^o\right)=\frac{AB}{AD}$

So, $AD=4\sqrt{2}$ ...(2)

Now, using equations (1) and (2), we get the required perimeter to be

$AC+CD+AD=4+2\sqrt{6}-2\sqrt{2}+4\sqrt{2}$

$=4+2\sqrt{6}+2\sqrt{2}$