Question

In the adjoining figure, $DP$ is parallel to $AC$, then the ratio of the area of triangle $PCB$ and quadrilateral $ABCD$ is:

• A. $1:1$
• B. $1:2$
• C. $1:4$
• D. $2:3$

Answer 1

The correct option is A $1:1$

Area of $\Delta PCB=\frac{1}{2}\times PB\times CE$

Area of quadrilateral $ABCD=\left(\text{area of}\Delta ABC\right)+\left(\text{area of}\Delta DCA\right)$

Area of quadrilateral $ABCD=(\frac{1}{2}\times CE\times AB)+(\frac{1}{2}\times AF\times DC)$

$\because DC\parallel PA$, so triangles are congruent.

Also, the quadrilateral $PACD$ is a parallelogram.

So, $DC\parallel AB$ and

$DC=PA$, $DP=AC$ and $DC=AB$

Area of quadrilateral $ABCD=(\frac{1}{2}\times CE\times AB)+(\frac{1}{2}\times CE\times AB)$

$=2(\frac{1}{2}\times CE\times AB)$

$\because PA+AB=PB$

$\therefore PA=AB,AB+AB=PB\Rightarrow 2AB=PB$

$\frac{\text{area of}\Delta PCB}{\text{area of quadrilateral}ABCD}=\frac{\frac{1}{2}\times 2AB\times CE}{2\times \frac{1}{2}\times CE\times AB}=\frac{1}{1}$

The ratio of $\Delta PCB$ and quadrilateral $ABCD$ is $1:1$.