In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$. Prove that $ \operatorname{ar}(\Delta \mathrm{RQC})=\frac{3}{8} \operatorname{ar}(\triangle \mathrm{ABC}) $.
Given:
In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$.
To do:
We have to prove that \( \operatorname{ar}(\Delta \mathrm{RQC})=\frac{3}{8} \operatorname{ar}(\triangle \mathrm{ABC}) \).
Solution:
Join $AQ$ and $PC$.
$CR$ is a median of $\Delta CAP$
This implies,
$\operatorname{ar}(\Delta \mathrm{ARC})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{CAP})$
$=\frac{1}{2}[\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABC})]$ ($\mathrm{CP}$ is the median of $\triangle \mathrm{ABC}$)
$=\frac{1}{4} \text { ar }(\triangle \mathrm{ABC})$.........(i)
$\mathrm{RQ}$ is a median of $\Delta \mathrm{RBC}$
This implies,
$\operatorname{ar}(\Delta \mathrm{RQC})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{RBC})$
$=\frac{1}{2}[\operatorname{ar}(\Delta \mathrm{ABC})-\operatorname{ar}(\Delta \mathrm{ARC})]$
$=\frac{1}{2}[\operatorname{ar}(\triangle \mathrm{ABC})-\frac{1}{4} \operatorname{ar}(\triangle \mathrm{ABC})$ [From (i)]
$=\frac{1}{2}[\frac{3}{4} \operatorname{ar}(\Delta \mathrm{ABC})]$
$=\frac{3}{8} \) ar \( (\triangle \mathrm{ABC})$
Hence proved.
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