# Area of Parallelogram/Rectangle

Jump to: navigation, search

## Theorem

The area of a rectangle equals the product of one of its bases and the associated altitude.

## Proof

Let $ABCD$ be a rectangle.

Then construct the square with side length $\left({AB + BI}\right)$ where $BI = BC$, as shown in the figure above.

Note that $\square CDEF$ and $\square BCHI$ are squares.

Thus $\square ABCD \cong \square CHGF$.

Since congruent shapes have the same area, $\left({ABCD}\right) = \left({CHGF}\right)$ (where $\left({FXYZ}\right)$ is the area of the plane figure $FXYZ$).

Let $AB = a$ and $BI = b$.

Then the area of the square $AIGE$ is equal to:

 $\displaystyle \left({a + b}\right)^2$ $=$ $\displaystyle a^2 + 2 \left({ABCD}\right) + b^2$ $\quad$ $\quad$ $\displaystyle \left({a^2 + 2ab + b^2}\right)$ $=$ $\displaystyle a^2 + 2 \left({ABCD}\right) + b^2$ $\quad$ $\quad$ $\displaystyle a b$ $=$ $\displaystyle \left({ABCD}\right)$ $\quad$ $\quad$

$\blacksquare$