Consider the cuves on the Argand plane as 

$C_1:\arg \left(z\right)=\frac{\pi }{4}$

$C_2:\arg \left(z\right)=\frac{3\pi }{4}$

and $C_3:\arg (z-5-5i)=\pi ,$ where $i=\sqrt{-1}.$

Statement- 1: Area of the region bounded by the curves $C_1,C_{2}and{C}_3$ is $\frac{25}{2}$ 
Statement-2 : The boundaries of $C_1,C_{2}and{C}_3$ constitute a right isosceless triangle.
 

We have $C_1:\arg \left(z\right)=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)=\frac{\pi }{4}[letz=x+iy]\Rightarrow \frac{y}{x}=\tan \frac{\pi }{4}=1\Rightarrow y=x$

$\therefore C_1:y=x\mathrm{.....}\left(i\right)$

$C_2:\arg \left(z\right)=\frac{3\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)=\frac{3\pi }{4}[Letz=x+iy]\Rightarrow \frac{y}{x}=\tan \frac{3\pi }{4}=-1\Rightarrow y=-x$

$\therefore C_2:y=-x\mathrm{.....}\left(ii\right)$

And $C_3:\arg (z-5-5i)=\pi \Rightarrow ta{n}^{-1}\left(\frac{y-5}{x-5}\right)=\pi [letz=x+iy]$

$\Rightarrow \frac{y-5}{x-5}=\tan \pi =0\Rightarrow y=5$

$\therefore C_3:y=5\mathrm{....}\left(iii\right)$

We get the following figure.

$\therefore$Area of the region bounded by $C_1,C_{2}and{C}_3=\frac{1}{2}\Vert \begin{array}{l}5-05-0\\ -5-05-0\end{array}\Vert =25$

$\therefore$Statement-1 is false

Now, $OA=5\sqrt{2},OB=5\sqrt{2}andAB=10$

$\because {\left(OA\right)}^2+{\left(OB\right)}^2={\left(AB\right)}^{2}andOA=OB$

Therefore, the boundary of $C_1,C_{2}and{C}_3$ constitutes right isosceless triangle. 
Hence, Statement-2 is true.