in a trapezium ABCD AB//CD,the diagonals AC and BD intersect at p.if AB:CD-2:1 then area of ∆CPD:area of ∆APB
Answer (1)
16th Aug, 2021
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.
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