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 below:

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.

Triangle and Parallelogram Between the Same Parallel Lines

If a triangle and parallelogram are drawn on the same base and between the same parallel lines, then the area of the triangle is half of the area of the parallelogram. Consider the figure given below where △ABE and parallelogram ABCD are drawn on the same base AB and between the same parallels AB and CE.

Here, area (△ ABE) = 1/2 × area (ABCD). Let us prove the above theorem. Complete the parallelogram ABEF as shown below such that the triangle is exactly half of the parallelogram.

Here, area (ΔABE) = 1/2 × area (ABEF). But, as we know that two parallelograms on the same base and between the same parallel lines are equal in area. This implies, area (ABCD) = area (ABEF). So, we have,

area (ΔABE) = 1/2 × area (ABCD)

Hence it is proved that a triangle is half of the area of a parallelogram drawn on the same base and between the same parallels.

Solved Examples

Example 1: In the figure below, if triangle ABC and parallelogram ABCD lies in between the same parallel lines, then find the area of ΔABC.

Solution: In the given figure, ΔABC and parallelogram ABCD are on the same base and between same parallel lines, which means the area of the triangle is half of the area of the parallelogram.

⇒ area (Δ ABC) = 1/2 × area (ABCD)

Now, the area of the parallelogram formula is base × height. Here, base = 16 units, and height = 12 units. So, area (ABCD) = 16 × 12 square units.

⇒ area (ABCD) = 192 square units

⇒ area (ΔABC) = 1/2 × 192 square units

⇒ area (ΔABC) = 96 square units

Therefore, the area of triangle ABC is 96 square units.

Example 2: In the figure below in which MNOP and SNOR are two parallelograms on the same base ON, if the area (MNOR) = 100 square units, and area (SNOP) = 64 square units, find the area of ΔMNS.

Solution: In the above figure, MNOP and SNOR are two parallelograms on the same base ON and between the same parallel lines ON and MR. So, their areas are equal.

⇒ area (MNOP) = area (SNOR)

⇒ area (MNOP) - area (SNOP) = area (SNOR) - area (SNOP)

⇒ area (ΔMNS) = area (ΔPOR) ... (1)

Now, it is given that area (MNOR) = 100 square units, and area (SNOP) = 64 square units, so if we subtract the area of region SNOP from the total area of MNOR, we will get the area of two triangles.

⇒ area (ΔMNS) + area (ΔPOR) = area (MNOR) - area (SNOP)

⇒ area (ΔMNS) + area (ΔPOR) = 100 - 64

⇒ area (ΔMNS) + area (ΔPOR) = 36 square units

⇒ 2 × area (ΔMNS) = 36 square units (from 1)

⇒ area (ΔMNS) = 18 square units

Therefore, the area of triangle MNS is 18 square units.