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.

Area-of-Rectangle.png

Then construct the square with side length:

$\map \Area {AB + BI}$

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:

$\map \Area {ABCD} = \map \Area {CHGF}$ (where $\map \Area {FXYZ}$ denotes the area of the plane figure $FXYZ$).

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

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

\(\ds \paren {a + b}^2\) \(=\) \(\ds a^2 + 2 \map \Area {ABCD} + b^2\)
\(\ds \paren {a^2 + 2 a b + b^2}\) \(=\) \(\ds a^2 + 2 \map \Area {ABCD} + b^2\)
\(\ds a b\) \(=\) \(\ds \map \Area {ABCD}\)

$\blacksquare$


Sources

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