Question

In the figure, $ABCD$ is a parallelogram. $P$ is the point on $BC$ such that $\frac{BP}{PC}=\frac{1}{4}$ and $DP$ is produced to meet $AB$ produced at $Q$. The area of triangle BPQ = k (area of triangle $CPD$). Find the value of $k$.

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{1}{32}$

Answer 1

The correct option is C $\frac{1}{16}$

Given that:

ABCD is a parallelogram.

$\frac{BP}{PC}=\frac{1}{4}$

$ar\left(△BPQ\right)=k\left(ar\right(△CPD\left)\right)$

To Find: $k=?$

Solution:

In $△BPQ$ and $△CPD$

$\angle BPQ=\angle CPD$ (Vertically opposite angles)

$\angle BQP=\angle CDP$ (Alternate interior angles)

Therefore,$△BPQ\sim △CPD$

Thus $\frac{ar\left(△BPQ\right)}{ar\left(△CPD\right)}={\left(\frac{BP}{CP}\right)}^2={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$

$\Rightarrow ar\left(△BPQ\right)=\frac{1}{16}\left(ar\right(△CPD\left)\right)$

So, value of $k=\frac{1}{16}$

Hence, C is the correct option.