Question

In $△PQR$, $A$ and $B$ are the mid point of the sides $PQ$ and $PR$ respectively, then the ratio of area of $(△GAQ+△GBR+△GQR)$ to the area of $△PQR$, where $G$ is the centriod is

Answer 1

Let the vertex $P$ be the origin and position vector of $Q$ and $R$ are $\overrightarrow{q}$ and $\overrightarrow{r}$ respectively.
So we can write,
$\overrightarrow{PG}=\frac{\overrightarrow{q}+\overrightarrow{r}}{3}$
$\overrightarrow{PA}=\frac{\overrightarrow{q}}{2}$ and $\overrightarrow{PB}=\frac{\overrightarrow{r}}{2}$
Required area = area of $△PQR-$ area of quadrilateral $PAGB$
Area of $△PQR$
$=\frac{1}{2}|\overrightarrow{P}Q\times \overrightarrow{P}R|=\frac{1}{2}|\overrightarrow{q}\times \overrightarrow{r}|$

Area of quadrilateral $PAGB=$ area of $△PAG+$ area of $△PGB$
Area of $△PAG$
$=\frac{1}{2}|\overrightarrow{PG}\times \overrightarrow{PA}|$
$=\frac{1}{2}|\frac{\overrightarrow{q}+\overrightarrow{r}}{3}\times \frac{\overrightarrow{q}}{2}|=\frac{1}{12}|\overrightarrow{q}\times \overrightarrow{r}|$

Area of $△PBG$
$=\frac{1}{2}|\overrightarrow{PG}\times \overrightarrow{PB}|$
$=\frac{1}{2}|\frac{\overrightarrow{q}+\overrightarrow{r}}{3}\times \frac{\overrightarrow{r}}{2}|=\frac{1}{12}|\overrightarrow{q}\times \overrightarrow{r}|$

Area of quadrilateral
$PAGB=\frac{1}{6}|\overrightarrow{q}\times \overrightarrow{r}|=\frac{1}{3}\left(△PQR\right)$
Required area$=△PQR-\frac{1}{3}△PQR=\frac{2}{3}\left(△PQR\right)$
Hence the ratio will be,
$\frac{\frac{2}{3}△PQR}{△PQR}=\frac{2}{3}=0.667$