In the given figure. ABCD is a rectangle. F is a point on AB and CE is drawn perpendicular to DF. If CE = 60 cm and DF = 40 cm. then what is the area $$(in cm^2)$$ of the rectangle ABCD? 

CE = 60 cm and DF = 40 cm

Let area of rectangle ABCD = $$(AB)\times(AD)=x$$ $$cm^2$$ -----------(i)

Area of rectangle ABCD = ar($$\triangle$$ CDF) + ar($$\triangle$$ ADF) + ar($$\triangle$$ BCF) 

=> $$x=(\frac{1}{2}\times60\times40)+(\frac{1}{2}\times AD\times AF)+(\frac{1}{2}\times BF\times BC)$$

=> $$x=(\frac{1}{2}\times60\times40)+(\frac{1}{2}\times AD\times AF)+(\frac{1}{2}\times BF\times AD)$$     [AD = BC in rectangle ABCD]

=> $$x=1200+\frac{1}{2}\times AD(AF+BF)$$     

=> $$x=1200+\frac{1}{2}\times AD\times AB$$

=> $$x=1200+\frac{x}{2}$$     [Using equation (i)]

=> $$x-\frac{x}{2}=1200$$

=> $$x=1200\times2=2400$$ $$cm^2$$

=> Ans - (C)