The given figure shows a trapezium in which AB || DC and diagonal AC and BD intersect at point P. If AP : CP = 3:5 and area of △APB is 27 m2, then the area of △CPD is ___.
75 m2
In ΔAPB and ΔCPD,
∠APB=∠CPD (vertically opposite angles)
∠PDC=∠PBA (alternate angles)
∴ΔAPB ∼ ΔCPD (by AA similarity criterion)
AP2CP2=ar(ΔAPB)ar(ΔCPD)
(Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.)
(APCP)2=ar(ΔAPB)ar(ΔCPD)
⇒(35)2=27 m2ar(ΔCPD)
⇒ar(ΔCPD)=27×259=75 m2