A trapezium has vertices marked as $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ $\text{(in that order anticlockwise).}$ The side PQ is parallel to side SR. Further, it is given that, $\mathrm{PQ}=11 \mathrm{~cm}, \mathrm{QR}=4 \mathrm{~cm}, \mathrm{RS}=6 \mathrm{~cm}$ and $\mathrm{SP}=3 \mathrm{~cm}$.
What is the shortest distance between PQ and SR (in cm )?
- $1.80$
- $2.40$
- $4.20$
- $5.76$
