Question

Prove that area of the parallelogram are on the same base between the same parallel are equal​

Answer 1

Parallelograms on the Same Base and Between Same Parallels

Parallelograms on the same base and between the same parallel lines are equal in area. It is a very important theorem in mathematics that is used to find and compare the areas of parallelograms.

Prove that Parallelogram on the Same Base and Between the Same Parallels are Equal in Area

Consider two parallelograms ABCE and ABDF on the same base AB, and between the same parallels AB and CF, as shown above:

What will be the relation between the areas of these two parallelograms? The theorem on parallelograms on the same base and between the same parallels states that such parallelograms are equal in area. Let us prove this theorem in the next section.

Proof:

The parallelograms on the same base and between the same parallel lines are equal in area. Consider the figure presented above. Can you see that ΔBCD and ΔAEF might be congruent? This is easy to show. We have:

BC = AE (opposite sides of a parallelogram are equal)

∠BCD = ∠AEF (corresponding angles when BC || AE and CE is the transversal)

∠BDC = ∠AFE (corresponding angles when BD || AF and FD is the transversal)

By the ASA criterion, the two triangles are congruent, which means that their areas are equal. It can be written as area (BCD) = area (AEF). Now, if we add the area of the region ABDE, we will get,

area (BCD) + area (ABDE) = area (AEF) + area (ABDE) [by area addition postulate]

area(ABCE) = area(ABDF)

Hence, it is proved that the two parallelograms drawn on the same base and between the same parallels are equal in area.

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