MCQ
$ABCD$ is a trapezium in which $AB\ ||\ DC$. If $\text{ar}(\triangle\text{ABD})=24\text{cm}^2$ and $AB = 8\ cm$, then height of $\triangle\text{ABC}$ is:
- A$3\ cm$
- B$4\ cm$
- ✓$6\ cm$
- D$8\ cm$
✓
Answer
Correct option: C.
$6\ cm$
$\triangle\text{ABD}\ \&\ \triangle\text{ABC}$ are on same base $AB$ and are between same parallels.
$\Rightarrow\text{Ar}(\triangle\text{ABD})=\text{Ar}(\triangle\text{ABC})$
$\text{Ar}(\triangle\text{ABD})=\frac{1}{2}\times8\times\text{h}=24\text{cm}^2$
$\Rightarrow\text{h}=6\text{cm}$
Now, $\text{Ar}(\triangle\text{ABC})=\frac{1}{2}\times8\times\text{h}=24\text{cm}^2$
$\Rightarrow\text{h}=6\text{cm}$
Hence, correct option is $(c)$.
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