Question

The given figure shows parallelogram ABCD. Points P and Q trisect the side AB, then which of the following is correct?

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

Answer 1

The correct option is C Area of $△$DPQ = $\frac{1}{6}$ Area of (parallelogram ABCD)

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, $Areaof△DPQ=\frac{1}{2}\times \frac{1}{3}\times AB\times h$
$Areaof△DPQ=\frac{1}{6}\text{(area of}\text{parallelogram ABCD)}$