Area of Parallelogram

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Theorem

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


Proof

There are three cases to be analysed: the square, the rectangle and the general parallelogram.


Square

AreaOfParallelogram-Square.png

From Area of Square:

$\paren {ABCD} = a^2$

where $a$ is the length of one of the sides of the square.

The altitude of a square is the same as its base.

Hence the result.

$\blacksquare$


Rectangle

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:

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

$\blacksquare$


Parallelogram

Area-of-Parallelogram.png

Let $ABCD$ be the parallelogram whose area is being sought.

Let $F$ be the foot of the altitude from $C$

Also construct the point $E$ such that $DE$ is the altitude from $D$ (see figure above).

Extend $AB$ to $F$.

Then:

\(\ds AD\) \(\cong\) \(\ds BC\)
\(\ds \angle AED\) \(\cong\) \(\ds \angle BFC\)
\(\ds DE\) \(\cong\) \(\ds CF\)

Thus:

$\triangle AED \cong \triangle BFC \implies \map \Area {AED} = \map \Area {BFC}$

So:

\(\ds \map \Area {ABCD}\) \(=\) \(\ds EF \cdot FC\)
\(\ds \) \(=\) \(\ds AB \cdot DE\)
\(\ds \) \(=\) \(\ds b h\)
\(\ds \) \(=\) \(\ds a b \sin \theta\) Definition of Sine of Angle: $h = a \sin \theta$

$\blacksquare$


Sources

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