Question

prove that two parallelograms on the same base and between same parallels are equal in area?

Answer 1

【 GIVEN】 :-

Two parallelograms ABCD and EFCD, on
the same base DC and between the same parallels
AF and DC are given

【TO PROVE THAT】 :-

We need to prove that ar (ABCD) = ar (EFCD).

【PROOF】 :-

In Δ ADE and Δ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)

Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)

Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now, ar (ABCD) = ar (ADE) + ar (EDCB)
= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.

Answer 2

【 GIVEN】 :-

Two parallelograms ABCD and EFCD, on

the same base DC and between the same parallels

AF and DC are given

【TO PROVE THAT】 :-

We need to prove that ar (ABCD) = ar (EFCD).

【PROOF】 :-

In Δ ADE and Δ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)

Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)

Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now, ar (ABCD) = ar (ADE) + ar (EDCB)

= ar (BCF) + ar (EDCB) [From(5)]

= ar (EFCD)

So, parallelograms ABCD and EFCD are equal in area.

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