Question

The given figure shows parallelogram ABCD. Points P and Q trisects the side AB.
To prove: Area of $△$DPQ = $\frac{1}{6}$ × Area of (parallelogram ABCD)

Answer 1

We know that triangles between same parallels and on the same base have half the area of the parallelogram.
So, $Areaof△DQC=\frac{1}{2}$ (area of parallelogram ABCD )
Let 'h' be the height between the parallel lines.
$PQ=\frac{1}{3}AB$ (P and Q trisects AB)
Hence, $Areaof△DPQ=\frac{1}{2}\times \frac{1}{3}\times AB\times h$
$Areaof△DPQ=\frac{1}{6}\text{(area of}\text{parallelogram ABCD)}$