Question

$P$ and $Q$ are respectively the mid-points of sides $AB$ and $BC$ of a triangle $ABC$ and $R$ is the mid-point of $AP$, show that:

$ar\left(RQC\right)=\frac{3}{8}ar\left(ABC\right)$

Answer 1

Median and traingle concept :

As, $CR$ is a median of $∆CAP$

$$\begin{gathered} ar\left(ARC\right)=\frac{1}{2}ar\left(CAP\right) \\ =\frac{1}{2}\left[\frac{1}{2}ar\left(ABC\right)\right] \\ =\frac{1}{4}ar\left(ABC\right) \end{gathered}$$.

Since, $RQ$ is a median of $∆RBC$

$$\begin{gathered} ar\left(RQC\right)=\frac{1}{2}ar\left(RBC\right) \\ =\frac{1}{2}\left[ar\left(ABC\right)-ar\left(ARC\right)\right] \\ =\frac{1}{2}\left[ar\left(ABC\right)-\frac{1}{4}ar\left(ABC\right)\right] \\ =\frac{1}{2}\left[\frac{3}{4}ar\left(ABC\right)\right] \\ =\frac{3}{8}ar\left(ABC\right) \end{gathered}$$

Hence, it is proved that $\mathbf{ar}\mathbf{}\mathbf{\left(}\mathbf{RQC}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{8}}\mathbf{}\mathbf{ar}\mathbf{}\mathbf{\left(}\mathbf{ABC}\mathbf{\right)}$ .