In the given figure, ABCD is a parallelogram P is a point on BC, such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given area of triangle CPQ = 20 m², calculate the area of triangle DCP.
 

Given: ABCD is a parallelogram P is a point on BC, such that BP : PC = 1 : 2. DP produced meets AB produced at Q.
Area of triangle CPQ = 20 m².

To Find: Area of triangle DCP. Construction: Join DB.

∠ BPQ = ∠ DPC  (Vertically opposite angles)
∠ BQP = ∠ PDC (alternate angles, BQ || DC, DQ meets them)
∴ Δ BPQ ∼ Δ CPD (AA similarity)
BP/CP = ½ (Given)

∴ Area D CPD = Area D BPQ
and area ΔCPQ = 20 cm² (Given)
Area ΔBPQ = 10 cm²
Area ΔCPD = 40 cm²
 (Proportional to bases BC and PC)
∴ Area Δ DBC = 40 × 3/2 = 60 cm².