Question

In the given figure ABCD is a parallelogram . Prove that ar(∆APD)=ar(quadrilateral BPCD).
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Answer 1

Answer:

ABCD is a parallelogram (AB||DC & AD||BC)

Diagonal of a parallelogram divides it into two triangles of equal area.

Therefore, ar(triangle ABD)=ar(triangle BCD)----(1)

Triangle BPD & triangle BPC are on the same base BP and between the same parallel BP & DC

Therefore, ar(triangle BPD) = ar(triangleBPC)----(2)

Adding the corresponding sides of eqn. 1&2,

ar(triangle ABD) + ar(triangle BPD) = ar(triangle BCD) + ar(triangle BPC)

ar(triangle APD)= ar(quadrilateral BPCD)

Hence proved

Answer 2

Step-by-step explanation:

ABCD is parallelogram and e p divides into two equal area that is area of a triangle ATV and area of triangle BCD that is equation 1 and next step area of triangle BPT equal area of triangle BPC because they lie on the same base with parallel side is equation to adding corresponding sides that is equation 1 and 2 area of triangle ABC d plus area of triangle bpd equal area of triangle BCD plus area of triangle BPC nextstep area of triangle ABC d is equal area of triangle ABC the proved