The given figure shows parallelogram ABCD. Points P and Q trisects the side AB. To prove: Area of △DPQ = 16 × Area of (parallelogram ABCD)
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm2. Calculate : (i) area of triangle CDP, (ii) area of parallelogram ABCD.