Question

prove that parallelograms on the same base and between the same parallels are equal​

Answer 1

Refer to the attachment attached to the answer.

In the diagram, we are given

• Two parallelograms on the same base and b/w the same parallel lines.

Construction:

• Drop a perpendicular on the line m, from the point E such that it is the height of both the parallelograms.
• ||gm is short for Parallelogram.

You may think why the height of both the parallelograms is the same. that's because they are between the same parallel lines which means the perpendicular distance b/w the two lines is the same everywhere, So the height of the parallelograms b/w these parallel lines..

Now, In ||gm ABCD,

• Height = EG
• Base = AB

Area of ||gm is given by,

⇒ Area = Base × Height

⇒ Area = AB × EG ...(i)

Similarly, In ||gm ABFE,

• Height = EG
• Base = AB

Area of ||gm is given by,

⇒ Area = Base × Height

⇒ Area = AB × EG ...(ii)

From eq.(i) and eq.(ii) , We observe that the area of both the parallelograms is the same which means they will cover each other completely when put on another, so they are equal.

Hence, Proved!

Answer 2

RTP :

• Parallelograms on the same base and between the same parallels are equal​.

PROOF :

→ Let ABCD and EFCD be two parallelograms lying on the same base CD and between the same parallels AF and CD.

✮ We shall prove that area of the parallelograms ABCD and EFCD are equal.

➛ In Δ ADE and Δ BCF, we have :

• ∠ DAE = ∠ CBF
• ∠ AED = ∠ BFC

( ∵ corresponding angles are equal )

➛ AD = BC

( ∵ Opposite sides of the parallelogram are equal )

So ,

• Δ ADE ≅ Δ BCF

( ∵ ASA rule )

Also,

• Congruent figures have the same area.

Hence,

• ar ABCD = ar (BCF) - (I)

Now,

• ar ABCD = ar (ADE) + ar(EDCB)
• ar (BCF) + ar(EDCB) ( ∵ I )
• ar (EDCF)

Thus, the areas of the two parallelograms ABCD and EFCD are the same.