The field is divided into three parts each in triangular shape.
Let us assume,
ΔPSA, ΔPAQ and ΔQAR be the triangles.
Area of (ΔPSA + ΔPAQ + ΔQAR) = Area of PQRS — (i)
Area of ΔPAQ = ½ area of PQRS — (ii)
From the diagram we observe that the triangle and parallelogram are on the same base and in-between the same parallel lines.
From (i) and (ii) we get,
Area of ΔPSA + Area of ΔQAR = ½ area of PQRS — (iii)
From (ii) and (iii), we can conclude that,
The farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.