ABCD is a rectangle with O as any point in its interior. If ar( )=3cm^2 and ar( )=6cm^2, then area of rectangle ABCD is:

ABCD is a rectangle with O as any point in its interior. If ar( )…

MCQ

$ABCD$ is a rectangle with $O$ as any point in its interior. If $\text{ar}(\triangle\text{AOD})=3\text{cm}^2$ and $\text{ar}(\triangle\text{ABOC})=6\text{cm}^2,$ then area of rectangle $ABCD$ is:

A
$9 \mathrm{~cm}^2$
B
$12 \mathrm{~cm}^2$
C
$15 \mathrm{~cm}^2$
✓
$18 \mathrm{~cm}^2$

✓

Answer

Correct option: D.

$18 \mathrm{~cm}^2$

A line $PQ$ is drawn fromn $AB$ parallel to $AD$ & $BC$.
Now, $\triangle\text{AOD}$ has height $= AP$
And, $\triangle\text{BOC}$ has height $= BP$
Area of $\triangle\text{AOD}=\frac{1}{2}\times\text{AD}\times\text{AP}=3\text{cm}^2$
$\Rightarrow\text{AD}\times\text{AP}=6\text{cm}^2\ ...(1)$
$\text{Ar}(\triangle\text{BOC})=\frac{1}{2}\times\text{BC}\times\text{BP}=6\text{cm}^2$
$\Rightarrow\text{BC}\times\text{BP}=12\text{cm}^2\ ...(2)$
Adding equation $(1)$ and $(2)$, we get
$\text{AD}\times\text{AP}+\text{BC}\times\text{BP}=18\text{cm}^2$
$\Rightarrow\text{AD}\times\text{AP}+\text{AD}\times\text{BP}=18\text{cm}^2$ $(AD = BC)$.
$\Rightarrow\text{AD}(\text{AP}+\text{BP})=18\text{cm}^2$
$\Rightarrow\text{AD}\times\text{AB}=18\text{cm}^2$ = Area of rectangle $ABCD$
Hence, correct option is $(d)$.