MCQ
In figure, $\text{ABCD}$ and $\text{FECG}$ are parallelograms equal in area. If $\text{ar} (\triangle\text{AQE})=12\text{cm}^2,$ then $\text{ar}(||^{\text{gm}}\text{FGBQ}) =$
- A$12\ cm^2$
- B$20\ cm^2$
- ✓$24\ cm^2$
- D$36\ cm^2$
✓
Answer
Correct option: C.
$24\ cm^2$
$\text{Ar}(||^{\text{gm}}\text{ABCD})=\text{Ar}(||^{\text{gm}}\text{FECG})$
Ar of $(||^{\text{gm}}\text{AQED})$ is common in both,
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{AQED})=\text{Ar}(||^{\text{gm}}\text{FGBQ})\ ...(1)$
Now $AE$ is diagonal of $\text{AQED}.$
$\Rightarrow\text{Ar}(\triangle\text{AQE})=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{AQED})$
$\Rightarrow12\text{cm}^2=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{AQED})$
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{AQED})=2\times12\text{cm}=24\text{cm}^2$
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{FGBQ})=24\text{cm}^2 \ [$From $(1)]$
Hence, correct option is $(c)$.
Ar of $(||^{\text{gm}}\text{AQED})$ is common in both,
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{AQED})=\text{Ar}(||^{\text{gm}}\text{FGBQ})\ ...(1)$
Now $AE$ is diagonal of $\text{AQED}.$
$\Rightarrow\text{Ar}(\triangle\text{AQE})=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{AQED})$
$\Rightarrow12\text{cm}^2=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{AQED})$
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{AQED})=2\times12\text{cm}=24\text{cm}^2$
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{FGBQ})=24\text{cm}^2 \ [$From $(1)]$
Hence, correct option is $(c)$.
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