If the tangents at the point  P and  Q on the circle $x^2+y^2-2x+y=5$ meet at the point $R\left(\frac{9}{4},2\right)$ then area of the triangle $PQR$ is

$s=x^2+y^2-2x+y-5=0$

$r=\sqrt{\left(s_{11}\right)}=\sqrt{\left(1+\frac{1}{4}-2-\frac{1}{2}-5\right)}$$=\frac{5}{2}$

chos d of contact of $p\left(\frac{9}{4},2\right)$ to 

$s=os,=o$

$x\frac{9}{4}+y\left(2\right)-1\left(x+\frac{9}{4}\right)+\frac{1}{2}\left(y+2\right)-5=0$

$x+2y-5=0$

$PQ=2\sqrt{r^2-p^2}=\sqrt{5}$

$h=\frac{\left|\frac{9}{4}+4-5\right|}{\sqrt{5}}=\frac{\sqrt{5}}{4}$

$\begin{array}{l}Areaof\\ {△}^{le}=\frac{1}{2}b\times b=\frac{5}{8}\end{array}$