In the given figure, PB is one-third of AB and BQ is one-third of BC. If the area of BPDQ is 20 $$cm^{2}$$, then what is the area (in$$\ cm^{2}$$) of ABCD?

Given PB is one-third of AB and  BQ is one-third of BC

Area of $$\triangle$$ ADP : Area of $$\triangle$$ PDB = 2 : 1 and 

Area of $$\triangle$$ BDQ : Area of $$\triangle$$ QDC = 2 : 1 

The area of BPDQ is 20 $$cm^{2}$$ (given) then area of $$\triangle$$ PDB = 10 $$cm^{2}$$ and area of $$\triangle$$ BDQ = 10 $$cm^{2}$$

Total area of ABCD is given by,

Area of ($$\triangle$$ ADP + $$\triangle$$ PDB + $$\triangle$$ BDQ + $$\triangle$$ QDC) = 20 + 10 + 10 + 20 = 60 $$cm^{2}$$

Hence, option D is the correct answer.