A farmer was havi

From the figure, it can be observed that point A divides the field into three parts. These parts are triangular in shape − ΔPSA, ΔPAQ, and ΔQRA

Area of ΔPSA + Area of ΔPAQ + Area of ΔQRA = Area of Parallelogram PQRS      ..............eq(i)

We know that if a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram

Therefore,

Area (ΔPAQ) = Area (PQRS)            .............eq(ii)

From equations (i) and (ii), we obtain

Area (ΔPSA) + Area (ΔQRA) + 1/2 Area (PQRS) = Area (PQRS)

Area (ΔPSA) + Area (ΔQRA) = 1/2 Area (PQRS) ....(iii)

Clearly, it can be observed that the farmer must sow wheat in triangular part PAQ

and pulses in other two triangular parts PSA and QRA or

wheat in triangular parts PSA and QRA and pulses in triangular parts PAQ.