Question

The given figure shows a parallelogram ABCD. Point P and Q trisect the side AB. Show that : area ($\Delta DPQ$) = area (BQC) = $\frac{1}{6}\times area\left(/{/}_{gm}ABCD\right)$

Answer 1

Given a parallelogram ABCD . Point P and Q trisect the side AB

Then AP=PQ=QB

Then area of $△DPQ$

Area of $△DPQ=\frac{1}{2}\times PQ\times height$....................(1)

Then area of $△BCQ$

And $△BCQ=\frac{1}{2}\times QB\times height$...........................(2)

But ABCD is a parallelogram

Where AB||DC

Then height of both triangle Is equal let is h

And PQ =QB

Then $Area△DPQ=areaof△BCQ=\frac{1}{2}PQ\times h$..............(3)

And area of parallelogram ABCD=$AB\times height\Rightarrow 3PQ\times h$................................................................(4)

Then (3) and (4) we get

Area of triangle DPQ =Area of triangle BQC =$\frac{1}{6}$ area of parallelogram ABCD