ABCD is a //gm, P is a point on BC such that BP:PC=1:2 and DP produced meets AB produced at Q. If area of △CPQ = 20 sq.cm, find:
1. area of △BPQ
2. area of △CPD
3. area of //gm ABCD.

pls. open the attached word document to view the figure...

2. Hello, snigdha!

$ABCD$ is a parallelogram.
Point $P$ is on $BC$ such that: . $BP:PC\,=\,1:2$
$DP$ produced meets $AB$ produced at $Q.$
If area of $\Delta CPQ = 20 \text{ cm}^2$, find:

1. area of $\Delta BPQ$
2. area of $\Delta CPD$
3. area of parallelogram $ABCD$
Code:
            D                   R    C
o - - - - - - - - - + - -o
/    *               :   /
/         *           :  /
/              *       : /
/                   *   :/
/                        o P
/                        /:   *
/                        / :       *
o - - - - - - - - - - - -o- + - - - - - o
A                        B  S           Q

Draw line segment $CQ.$
. . Area of $\Delta CPQ\,=\,20$
Draw altitude $RS$ through point $P\!:\;\;h = RS.$

Since $BP:PC = 1:2$, then: . $SP:PR = 1:2$
. . That is: . $SP = \tfrac{1}{3}h,\;PR = \tfrac{2}{3}h$

Area of $\Delta BCQ \,=\,\tfrac{1}{2}(BQ)(h)$
Area of $\Delta BPQ \,=\,\tfrac{1}{2}(BQ)\left(\tfrac{1}{3}h\right)\,= \,\tfrac{1}{6}(BQ)(h)$
. . Hence: . $\Delta BPQ \:=\:\tfrac{1}{3}\Delta BCQ$

Then: . $\Delta CPQ \,=\,\tfrac{2}{3}\Delta BCQ \quad\Rightarrow\quad 20 \,=\,\tfrac{2}{3}\Delta BCQ$

Hence: . $\Delta BCQ\,=\,30 \quad\Rightarrow\quad \Delta BPQ \,=\,10$ .(1)

We see that: . $\Delta CPD \sim \Delta BPQ$
. . and $\Delta CPD$ has twice the base and twice the height of $\Delta BPQ$
Hence: . $\Delta CPD \:=\:4\times\Delta BPQ \:=\:4(10) \quad\Rightarrow\quad \Delta CPD\:=\:40$ .(2)

The area of $\Delta CPD$ is: . $\tfrac{1}{2}(CD)\left(\tfrac{2}{3}h\right) \:=\:40 \quad\Rightarrow\quad \tfrac{1}{3}(CD)(h) \:=\:40$

Hence:. . $(CD)(h) \:=\:120$

Note that: . $(CD)(h)$ = area of parallelgram $ABCD.$

Therefore: .area of $ABCD \:=\:120$ .(3)