A thermodynamic system under goes cyclic process ABCDA as shown in figure. The work done by the system

$$\begin{gathered} W_{BCOB} = - Area of triangle BCO = -P_0{V}_0/ 2=negative \sin ce cycle is anti clock wise \\ {W}_{AODA} = + Area of triangle AOD = + P_0{V}_0 / 2=positive \sin ce cycle is clock wise \\ (here each triangle area is half of rec\tan gle area) \\ work done by the system = sum of the areas under the curve=\frac{-P_0{V}_0}{2}+\frac{P_0{V}_0}{2}=0 \end{gathered}$$