1. 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?
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By: anil on 05 May 2019 01.47 am
CE = 60 cm and DF = 40 cm Let area of rectangle ABCD = $$(AB) imes(AD)=x$$ $$cm^2$$ -----------(i) Area of rectangle ABCD = ar($$ riangle$$ CDF) + ar($$ riangle$$ ADF) + ar($$ riangle$$ BCF) => $$x=(frac{1}{2} imes60 imes40)+(frac{1}{2} imes AD imes AF)+(frac{1}{2} imes BF imes BC)$$ => $$x=(frac{1}{2} imes60 imes40)+(frac{1}{2} imes AD imes AF)+(frac{1}{2} imes BF imes AD)$$ [AD = BC in rectangle ABCD]
=> $$x=1200+frac{1}{2} imes AD(AF+BF)$$ => $$x=1200+frac{1}{2} imes AD imes AB$$ => $$x=1200+frac{x}{2}$$ [Using equation (i)] => $$x-frac{x}{2}=1200$$ => $$x=1200 imes2=2400$$ $$cm^2$$ => Ans - (C)
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=> $$x=1200+frac{1}{2} imes AD(AF+BF)$$ => $$x=1200+frac{1}{2} imes AD imes AB$$ => $$x=1200+frac{x}{2}$$ [Using equation (i)] => $$x-frac{x}{2}=1200$$ => $$x=1200 imes2=2400$$ $$cm^2$$ => Ans - (C)