Area of Parallelogram/Rectangle

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Theorem

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


Proof

Let $ABCD$ be a rectangle.

Cua1.PNG

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:

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

$\blacksquare$


Sources