Question

In the figure given above, $ABCD$ is a parallelogram P is a point on $BC$ such that $PB:PC=1:2.DP$ produced meets $AB$ produced at Q. If the area of the triangle $BPQ$ is $20$ square units, what is the area of the triangle $DCP$?

• A. $20$ square units
• B. $30$ square units
• C. $40$ square units
• D. None of the above

Answer 1

The correct option is D None of the above

Since $BP:PC=1:2$, so the triangles $BPQ$ and $DCP$ are similar whose sides are in the ratio $1:2$.
$\Rightarrow$ The area of these triangles is in the ratio of $1:4$.
$\Rightarrow$ The area of the $△DCP$ is $4\times 20$, i.e., $80$ square units.