A triangle PQR, with PQ = PR = 5 and QR = 6 is inscribed in a circle C. The radius of the circle with centre on  $\overline{QR}$ which is tangent to both ‘C’ and side  $\overline{PQ}$  is R, the value of  $\frac{9R}{4}$  is equal to _______________

PD is perpendicular bisector of side QR.
O is centre of circumcircle of  $\Delta \mathcal{l}e$ PQR.
M is centre of the circle which touches the side PQ and circumcircle of  $\Delta \mathcal{l}e$  PQR
S is point of contact of the two circles 
 
 $QD=DR=\frac{QR}{2}=\frac{6}{2}=3$
 $OM=OS-MS$
= R – r
MT = r
 $\sin \theta =\frac{4}{5}=\frac{r}{QM}\Rightarrow QM=\frac{5r}{4}$
 $MD=QD-QM=3-\frac{5r}{4}$
 
 $\frac{PR}{\sin \theta }=2R\Rightarrow \frac{5}{\left(\frac{4}{5}\right)}=2R\Rightarrow R=\frac{25}{8}$
 $OD=PD-OP=4-R=4-\frac{25}{8}=\frac{7}{8}$
In  $\Delta \mathcal{l}e$ OMD,  $O{D}^2+M{D}^2=O{M}^2$
 $\Rightarrow {\left(\frac{7}{8}\right)}^2+{(3-\frac{5r}{4})}^2={(\frac{25}{8}-r)}^2$
 $\Rightarrow r=\frac{20}{9}$
 $\frac{9}{4}r=\frac{9}{4}\left(\frac{20}{9}\right)=5$