In the figure, $5\mathit{\Omega }$ and $4\mathit{\Omega }$ are in series, So their resultant will be $5+4$ i.e. $9\mathit{\Omega }$.
Now, $9\mathit{\Omega }$ and $3\mathit{\Omega }$ are in parallel, so their resultant ${\mathit{R}}_{\mathit{P}}$ is calculated as :
$\frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{9}+\frac{1}{3}$
$\frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1+3}{9}$
$\frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{4}{9}$
$\therefore {\mathrm{R}}_{\mathrm{P}}=\frac{9}{4}=2.25\mathit{\Omega }$
Now, $8\mathit{\Omega }$ and $2.25\mathit{\Omega }$ are in series,
$\therefore {\mathrm{R}}_{\mathrm{AB}}=8+2.25$
$=10.25\mathit{\Omega }$