$SD =2-\frac{1}{3}=\frac{5}{3}$

Area of $OABS$ is $A _{1}$

Area of $SCD$ is $A _{2}$

Distance $=\left| A _{1}\right|+\left| A _{2}\right|$

$A _{1}=\frac{1}{2}\left[\frac{13}{3}+1\right] 4=\frac{32}{3}$

$A _{2}=\frac{1}{2} \times \frac{5}{3} \times 2=\frac{5}{3}$

Distance $=\left|A_{1}\right|+\left|A_{2}\right|$

$=\frac{32}{3}+\frac{5}{3}$

$=\frac{37}{3}$