The figure shows the  p-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse Then, 

The process  $A\to B$ lies on a circle having centre  (2, 2)  and radius 1 unit. The  equation of the circle is given by   ${(p-2)}^2+{(V-2)}^2=1$. 
 The ideal gas equation is  $pV=nRT.$ The product  pV not constant in the process $A\to B$  and   hence the process is not isothermal .  
 In the process  $B\to C\to D$ the product  pV decreases from 6 units to a value less than 2 units  Thus, the temperature of the gas reduces in this process.  Hence, internal energy of the gas decreases i.e.,  $\Delta U<0.$  Also, work done by the gas is negative, $\Delta W<0$ (since  volume decreases ) . Substitute in the first law of thermodynamics,  $\Delta Q=\Delta U+\Delta W,$ to get  $\Delta Q<0$ The  negative sign of  $\Delta Q$ indicates  heat flows  out of the gas. The  Work done by the gas is given by area under the  p-V diagram, It is positive for the process  $A\to B$ (volume increases) and negative for the process $B\to C$  (volume decreases). The magnitude of area under AB is more than area under BC. Note that,
   $W_{AB}=2+\pi /2$ 
 $W_{BC}=-(1+\pi /2)$ 
  $W_{ABC}=W_{AB}+W_{BC}=1$
Thus, the work is done by the gas in the process $A\to B\to C$  as  $W_{ABC}>0.$
Using the same argument for the cycle ABCDA, positive work is done by the gas in this cycle. This work  is equal to the area enclosed by the curve ABCDA .