Question

In $\Delta PQR$, base QR is divided at X such that $QX=\frac{1}{2}XR$. Prove that $ar\left(\Delta PQX\right)=\frac{1}{3}ar\left(\Delta PQR\right)$

Answer 1

$QX=\frac{1}{2}XR$
$LetQX=x$
$XR=2X$
$QR=QX+XR=X+2X=3X$
Let 'h' be the height of the traingles between the two lines.
$\frac{Areaof\Delta PQX}{Areaof\Delta PQR}=\frac{\frac{1}{2}\times QX\times h}{\frac{1}{2}\times QR\times h}=\frac{x}{3x}=\frac{1}{3}$
$Areaof\Delta PQX=\frac{1}{3}Areaof\Delta PQR$