Question

In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP$:$PC$=1:2$. DP produced meets AB produced at Q. Given ar($\Delta$CPQ)$=20c{m}^2$. Calculate ar($\Delta$CDP).

Answer 1

For $△BPQ$ and $△DCP$

$\angle BPQ=\angle DPC\because ofverticallyoppositeangles$

$\angle BQP=\angle PDC\because ofalternateangles$

$\therefore △BPQ\sim △DCP$

$\therefore \frac{BQ}{DC}=\frac{QP}{PD}=\frac{BP}{PC}$

Given:$BP:PC=1:2$

$\Rightarrow \frac{BP}{PC}=\frac{1}{2}$

$\therefore \frac{QP}{PD}=\frac{1}{2}$

For $△CPQ$ and $△CDQ$ are on the same base at vertex $C$

$\frac{Areaof△CPQ}{Areaof△CDP}=\frac{Qp}{PD}=\frac{1}{2}$

$\therefore \frac{20}{Areaof△CDP}=\frac{1}{2}$

$\therefore Areaof△CDP=40sq.cm$