Question

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

• A. Area of ΔDPQ = (1/3) area of (ABCD)
• B. Area of ΔDPQ = (2/5) area of (ABCD)
• C. Area of ΔDPQ = (1/6) area of (ABCD)
• D. Area of ΔDPQ = (4/5) area of (ABCD)

Answer 1

The correct option is C

Area of ΔDPQ = (1/6) area of (ABCD)

Here DPQC is a trapezium,

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

So, Area of $△$DQC = $\frac{1}{2}$ (ABCD)

Let h be the height between the parallel lines

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

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

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

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

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

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

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

Hence, area of trapezium DPQC = $\frac{2}{3}\times$ area of parallelogram ABCD

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

$\frac{2}{3}$ Area of ABCD - $\frac{1}{2}$ Area of ABCD = Area of $△$DPQ

Or, Area of $△$DPQ = $\frac{1}{6}$ Area of ABCD