question_answer

ABCD is a rectangle with O as any point in its interior. If \[ar(\Delta AOD)=3c{{m}^{2}}\]  and \[ar\,(\Delta BOC)\]\[=6\,c{{m}^{2}},\]then area of rectangle ABCD is

A) \[~9\,c{{m}^{2}}\]                             

B)        \[~12\text{ }c{{m}^{2}}\]      

C)        \[~15\,c{{m}^{2}}\]                           

D)        \[~18\,c{{m}^{2}}\]

Correct Answer: D

Solution :

Draw a parallel line EF to AD and BC Then, area of\[\Delta AOD\] \[=\frac{1}{2}\](area of rectangle AFED)              ?(i) Also \[ar(\Delta BOC)=\frac{1}{2}\](rectangle FBCE)         ?(ii) Adding (i) & (ii), we get area \[\Delta \Alpha {\mathrm O}D+\]area of\[\Delta \Beta {\mathrm O}C\] \[=\frac{1}{2}(area\,of\,rectangle\,AFED\,+area\,of\,FBCE)\] \[\Rightarrow \]\[(3+6)c{{m}^{2}}=\frac{1}{2}\](area of rectangle ABCD) \[\Rightarrow \]area of rectangle\[ABCD=18\,c{{m}^{2}}\]