`A B C D` is a rectangle with `O` as any point in its interior. If `a r\ ( A O D)=3\ c m^2a n d\ a r\ ( B O C)=6\ c m^2,` then area of rectangle `A B C D` is: `9c m^2` (b) `12\ c m^2` (c) `15 c m^2` (d) `18\ c m^2`,

Explanation

`ABCD` is a rectangle, and O is any point in its interior.
Area (`/_\AOD)=3sqcm` and Area `/_\BOC=6sq cm.`
From `O` draw a line `EF || AB` intersecting `AD \at\ E` and `BC \at\ F.`
Therefore,`OE_|_AD \and\ OF_|_BC`
Let `OE =x \and\ OF =y`
`OE +OF=EF =AB`
Therefore `x+y=AB` ....(1)
`ar /_\AOD+ Area /_\BOC=3+6=9sq. cm.`
`1/2xxADxxOE+1/2xxOFxxBC=9sq.cm`
`1/2(ADxxOE+OFxxBC)=9sq.cm`
`(ADxxOE+OFxxBC)=9xx2=18sqcm`
Since `BC =AD` (opposite sides of a rectangle)
`=>AD(OE+OF)=18sq.cm`
`ADxxEF=18sq.cm`

So, `EF =AB`
`=>ADxxAB=18sq.c
m`
Area of the rectangle `ABCD =ABxxAD=18sq.cm`