Question

In the fig. (3) Given below, ABCD is a parallelogram. P is a point on DC such that area of $\Delta DAP=25{cm}^2$ and area of $\Delta BCP=15{cm}^2.$ Find
(i) area of $\parallel$ gm $ABCD$
(ii) DP: PC.

Answer 1

(c) Given:$ABCD$ is a parallelogram.
$P$ is a point on $DC$ such that area of $\Delta DAP=25{cm}^2$ and area of $\Delta BCP=15{cm}^2$
To Find:
(i) area of $\parallel gmABCD$
(ii) DP: PC Now let us find, From fig ( 3 )
(i) we know that, ar $\left(\Delta APB\right)=1/2$ ar $(\parallel gmABCD)$
Then,
$\begin{array}{cc}1/2ar(\parallel gmABCD)& =\mathrm{ar}\left(\Delta DAP\right)+\mathrm{ar}\left(\Delta BCP\right)\\ & =25+15\end{array}$
$=40{cm}^2$
So, ar $(\parallel gmABCD)=2\times 40=80c{m}^2$
(ii) we know that, $△ADP$ and $\Delta BCP$ are on the same base $CD$ and between same parallel lines $CD$ and $AB$.
$\mathrm{ar}\left(\Delta DAP\right)/\mathrm{ar}\left(\Delta BCP\right)=DP/PC$
$25/15=DP/PC$
$5/3=DP/PC$
So, DP: $PC=5:3$