If the angle between the pair of lines  $x^2+2\sqrt{2}xy+k{y}^2=0,(k>0)$ is ${45}^o$, then the area of the triangle formed by the pair of bisectors of angles between the given lines and the line x + 2y + 1 = 0 is ...… (in sq units)

$\begin{array}{l}{x}^2+2\sqrt{2}xy+k{y}^2=\mathrm{0.....}\left(1\right)\\ x+2y+1=0\mathrm{...}\left(2\right)\end{array}$
The angle b/w the lines (1) is 45^o
$\begin{array}{l}\cos {45}^o=\left|\frac{1+k}{\sqrt{(1-k)+8}}\right|=\frac{1}{\sqrt{2}}\\ \Rightarrow k^2+6k-7=0\Rightarrow (k+7)(k-1)=0\\ \therefore k=1(\because k>0)\end{array}$
The equation of angular bisectors of (1) is  
$$\begin{gathered} h(x^2-y^2)-(a-b)xy=0 \\ \begin{array}{l}\Rightarrow \sqrt{2}(x^2-y^2)-(1-k)=0\\ \therefore x^2-y^2=\mathrm{0....}\left(3\right)\end{array} \end{gathered}$$
The req area formed by the lines (2), (3) is 
$\frac{n^2\sqrt{h^2-ab}}{\left|a{m}^2-2hlm+b{l}^2\right|}=\frac{1}{3}$