Hello Aspirant..

trapezium ABCD AB//CD,the diagonals AC and BD intersect at p.if AB:CD-2:1 then area of CPD:area of APB

Step-by-step instructions:

Let ABCD be a quadrilateral with AC and BD as diagonals. At point P, these diagonals cross.

Now area of a triangle  = 1/2 * base * height

=> Area(ΔAPB) * Area(ΔCPD)  = {1/2 * BP*AM}*{1/2 * PD*CN}

=> Area(ΔAPB) * Area(ΔCPD)  = 1/4 * BP*AM* PD*CN .........1

Again

Area(ΔAPD) * Area(ΔBPC) = {1/2 * PD*AM}*{1/2 *CN*BP}

=> Area(ΔAPD) * Area(ΔBPC)  = 1/4 * BP*AM* PD*CN ..........2

From equation 1 and 2, we get..

Area(ΔAPB) * Area(ΔCPD)  = Area(ΔAPD) * Area(ΔBPC)

I hope this information helps.

Regards.