(N/A) The farmer has a field in the form of a parallelogram $PQRS$,and a point $A$ is situated on $RS$.
Let us join $AP$ and $AQ$.
Obviously,the field is divided into three parts,i.e.,$\Delta APS$,$\Delta PAQ$,and $\Delta QAR$. These parts are triangular in shape.
Since $\Delta PAQ$ and parallelogram $PQRS$ are on the same base $PQ$ and between the same parallels $PQ$ and $RS$:
$\therefore \text{ar}(\Delta PAQ) = \frac{1}{2} \text{ar}(\text{parallelogram } PQRS) \dots(1)$
$\Rightarrow \text{ar}(\text{parallelogram } PQRS) - \text{ar}(\Delta PAQ) = \text{ar}(\text{parallelogram } PQRS) - \frac{1}{2} \text{ar}(\text{parallelogram } PQRS)$
$\Rightarrow [\text{ar}(\Delta APS) + \text{ar}(\text{QAR})] = \frac{1}{2} \text{ar}(\text{parallelogram } PQRS) \dots(2)$
From $(1)$ and $(2)$,we have:
$\text{ar}(\Delta PAQ) = \text{ar}(\Delta APS) + \text{ar}(\Delta QAR)$
Thus,the farmer can sow wheat in $\Delta PAQ$ and pulses in the combined area of $\Delta APS$ and $\Delta QAR$,or vice versa.