### Theory:

In this topic, let us learn the parallelograms on the same base and between the same parallels are equal in the area using an activity, and we shall prove it as a theorem.

Activity:

Consider drawing parallelograms $$PQRS$$ and $$EFRS$$ in a graph sheet with the same base and between the same parallels. The above figure shows that the parallelograms are on the same base $$SR$$ and between the same parallels $$SR$$ and $$EQ$$.

In grade $$6$$, we have learnt how to find the area by counting the number of squares in the graph. Click here to recall them.

Now, applying this concept, let us find the area of the parallelograms $$PQRS$$ and $$EFRS$$.

Consider the parallelogram $$PQRS$$. Area of fully-filled squares $$= 8$$ $$= 8 \times 1 = 8 \ sq. \ cm$$

Area of half-filled squares $$= 0$$

Area of more than half-filled squares $$= 4$$ $$= 4 \times 1 = 4 \ sq. \ cm$$

Area of less than half-filled squares $$= 4$$ [Which can be neglected]

Therefore, the area of the parallelogram $$PQRS$$ is $$8 + 4 = 12 \ sq. \ cm$$.

Similarly, consider the parallelogram $$EFRS$$. Area of fully-filled squares $$= 8$$ $$= 8 \times 1 = 8 \ sq. \ cm$$

Area of half-filled squares $$= 0$$

Area of more than half-filled squares $$= 4$$ $$= 4 \times 1 = 4 \ sq. \ cm$$

Area of less than half-filled squares $$= 4$$ [Which can be neglected]

Therefore, the area of the parallelogram $$EFRS$$ is $$8 + 4 = 12 \ sq. \ cm$$.

Since the areas of both the parallelograms $$PQRS$$ and $$EFRS$$ are the same, we arrive at the result that "parallelograms on the same base and between the same parallels are equal in the area".
Theorem

Statement: Parallelograms on the same base and between the same parallels are equal in area.

Proof: Consider $$2$$ parallelograms $$PQRS$$ and $$QBRS$$ on the same base $$RS$$ and between the same parallels $$RS$$ and $$PB$$. To prove: Area of parallelorgram $$PQRS =$$ Area of parallelogram $$ABRS$$

Construction: Draw $$AX \perp SR$$ and $$QY \perp SR$$ Area of parallelogram $$PQRS = \text{base} \times \text{height}$$

Area of parallelogram $$PQRS = SR \times AX$$ ---- ($$1$$)

Similarly, Area of parallelogram $$ABRS = \text{base} \times \text{height}$$

Area of parallelogram $$ABRS = SR \times QY$$ ---- ($$2$$)

We know that "the perpendicular distance between parallel lines are equal".

Therefore, $$AX = QY$$ ---- ($$3$$)

Substituting equation ($$3$$) in ($$1$$), we get:

Area of parallelogram $$PQRS = SR \times QY$$

Area of parallelogram $$PQRS =$$ Area of parallelogram $$ABRS$$ (Using ($$2$$))

Therefore, parallelograms on the same base and between the same parallels are equal in area.

Hence proved.