# 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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Updated on: 10-Oct-2022

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