Question

In the given figure, $AD$ is a median of $△ABC$ and $P$ is a point on $AC$ such that:
$ar\left(△ADP\right)$ : $ar\left(△ABD\right)$ $=2:3.$
Find : (i) AP : PC (ii) $ar\left(△PDC\right)$ : $ar\left(△ABC\right).$

Answer 1

Let $ar\left(△ABD\right)=x$

since D is mid point of $BC$.

$ar\left(△ADC\right)=x.$

Given,

$ar\left(△ADP\right)=\frac{2}{3}ar\left(△ABD\right)$

$\Rightarrow ar\left(△ADP\right)=\frac{2}{3}x.$

$ar\left(△DPC\right)=ar\left(△ADC\right)-ar\left(△ADP\right)$

$=x-\frac{2x}{3}$

$=\frac{x}{3}$

i) $AP:PC=ar\left(△ADP\right):ar\left(△PDC\right)$ [since both triangles have height]

$=\frac{2x}{3}:\frac{x}{3}$

$=2:1$

ii) $ar\left(△PDC\right):ar\left(△ABC\right)=\frac{x}{3}:2x$

$=1:6$