Q.Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

The area of a parallelogram spanned by adjacent sides $\vec{u}$ and $\vec{v}$ is $|\vec{u}\times\vec{v}|$. Use this with the fact that "same base, same parallels" fixes the height.

Let the common base be the vector $\vec{b}$. Let the first parallelogram have the other side $\vec{p}$ and the second have the other side $\vec{q}$. Their opposite sides lie on the same line parallel to the base, so $\vec{q}$ is reached from $\vec{p}$ by sliding along the base direction: $\vec{q}=\vec{p}+\lambda\vec{b}$ for some scalar $\lambda$.

Then …

What the statement means

Two parallelograms are "on the same base and between the same parallels" when they share one side (the base) and their opposite sides both lie on a single line parallel to that base. The base length is common and the height (the gap between the two parallel lines) is common, so area $=$ base $\times$ height must match. The task is to prove this cleanly with vectors.

Setting up

Put the shared base along a vector $\vec{b}$.

• First parallelogram: adjacent sides $\vec{b}$ and $\vec{p}$.
• Second parallelogram: adjacent sides $\vec{b}$ and $\vec{q}$.

The area of a parallelogram spanned by two vectors is the magnitude of their cross product:

Using "between the same parallels"

The tips of $\vec{p}$ and $\vec{q}$ lie on the same line parallel to the base, so $\vec{q}$ is obtained from $\vec{p}$ by sliding along that line, i.e. by adding a multiple of the base direction:

In words, $\vec{p}$ and $\vec{q}$ have exactly the same perpendicular (height) component relative to $\vec{b}$.

The key cancellation …

Method: Proving an area equality by writing area as a cross product

Use this reasoning pattern to prove that two parallelograms on the same base and between the same parallels have equal area.

Steps

Step 1: Express each area as a cross product on the shared base.

Let the common base be $\vec{b}$ and the two "other" sides be $\vec{p}$ and $\vec{q}$. Then

Step 2: Encode "between the same parallels" algebraically.

Both far sides end on one line parallel to the base, so $\vec{q}$ differs from $\vec{p}$ only by a slide along the base:

$\vec{q}=\vec{p}+\lambda\vec{b}\quad\text{for some scalar }\lambda.$ …

Common Mistakes

Mistake 1: Thinking the two parallelograms must be congruent.

Why it's wrong: "same base, between the same parallels" fixes only the base length and the height, not the shape; the slanted sides can differ yet the areas match. Correct approach: prove equality of area via $|\vec{b}\times\vec{p}|=|\vec{b}\times\vec{q}|$, not congruence.

Mistake 2: Not encoding "between the same parallels" as $\vec{q}=\vec{p}+\lambda\vec{b}$. …