Question

in a trapezium abcd ab parallel cd the diagonal ac and bd intersect at p if ab:cd =2:1 then area of triangle cpd:ar of triangle apb​

Answer 1

Answer:

Step-by-step explanation:

Let ABCD is a quadrilateral having diagonal AC and BD. These diagonals intersect at point P.

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)