Question

If the given figure shows parallelogram ABCD. Points P and Q trisect the side AB, then ____.

• A. $Area\left(△DPQ\right)=\frac{1}{3}\times Area\left(ABCD\right)$
• B. $Area\left(△DPQ\right)=\frac{2}{5}\times Area\left(ABCD\right)$
• C. $Area\left(△DPQ\right)=\frac{1}{6}\times Area\left(ABCD\right)$
• D. $Area\left(△DPQ\right)=\frac{4}{5}\times Area\left(ABCD\right)$

Answer 1

The correct option is C

$Area\left(△DPQ\right)=\frac{1}{6}\times Area\left(ABCD\right)$

Here DPQC is a trapezium,

We know that triangles between same parallels and on the same base have half the area of parallelogram.

So, $ar\left(△DQC\right)=\frac{1}{2}\times ar\left(ABCD\right)$

Let h be the height between the parallel lines

$PQ=\frac{1}{3}\times AB$

Hence, area of DPQC
$=\frac{1}{2}\times h\times (PQ+DC)$

$=\frac{h}{2}\times (DC+\frac{AB}{3})$
$=\frac{h}{2}\times (DC+\frac{DC}{3})$

$=\frac{h}{2}\times \left(\frac{4DC}{3}\right)$

$=\frac{2h\times DC}{3}$

Now, we know that $h\times DC=$ area of parallelogram ABCD.

Hence, $areaoftrapeziumDPQC=\frac{2}{3}\times areaofparallelogramABCD$

Now, $Area\left(DPQC\right)=Area\left(△DPQ\right)+\frac{1}{2}\times Area\left(ABCD\right)$

$\frac{2}{3}\times Area\left(ABCD\right)-\frac{1}{2}\times Area\left(ABCD\right)=Area\left(△DPQ\right)$

Or, $Area\left(△DPQ\right)=\frac{1}{6}\times Area\left(ABCD\right).$