A triangle having angles 30, 60 \& 90 , the corresponding opposite sides are $\(\mathrm{a}, \mathrm{a} \sqrt{3} \& 2 \mathrm{a}\)$ respectively. Similarly the right triangle having angles $\(90,45 \& 45\)$, the corresponding opposite sides are $\(\mathrm{a} \sqrt{2}\)$, a \& a respectively.
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GRE triangle triplete.png [ 41.93 KiB | Viewed 54 times ]
Using the above two right triangle rules we can find the remaining sides in the given figure.
In triangle ADC , the side opposite to 45 degree is $\(\mathrm{x}=\mathrm{AD}\)$, so the side AC would also be x Next in triangle ABC , angles are $90,60 \& 30$, where the side opposite to 30 is $\(\mathrm{AC}=\mathrm{x}\)$, so the remaining sides opposite to $\(60 \& 90\)$ would be $\(\mathrm{x} \sqrt{3}(=\mathrm{BA}) \& 2 \mathrm{x}(=\mathrm{BC})\)$ respectively.
The sides when mentioned in the figure we get
Attachment:
GRE triangle sides.png [ 78.73 KiB | Viewed 54 times ]
Hence the perimeter of the triangle $\(A B C\)$ is $\(x+x \sqrt{3}+2 x=3 x+x \sqrt{3}\)$, which for $\(A D=3(=x)\)$ comes out to be $\(9+\sqrt{27}\)$, so the answer is (D).