Question

in a trapezium ABCD AB//CD, the diagonals AB, BD intersect at 'p'. IfAB:CD=2:1,then area of triangle CPD:area of triangle APD​

Answer 1

Answer:Given: In quadrilateral ABCD ,AD||BC and AP=1/3AC

To prove: DP=1/2BP

Proof:

since,AP=1/3AC

,AP/AC=1/3

Let AP=x and AC=3

w

Where x is an undefined number

ACP=AC - AP =3x-x=2x

Now , in quadrilateral ABCD

AD||BC

By Alternate angle theorem

angle PAD=angle PCB

angle PDA=angle PBC

By AA test of similarity

∆APD~∆CPB

therefore By the property of similar triangles

AP/PD=CP/PB

x/PD=2x/PB

PB/PD=2/1

PD/PB=1/2

Therefore PD=1/2PB

Step-by-step explanation:

Answer 2

Answer:

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