# Thread: Please help me with this simliarity sum..

1. ## Please help me with this simliarity sum..

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!

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

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

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

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

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

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

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

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

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

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

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

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

### ABCD is a parallelogram. P is a point on BC such that

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