Area of Triangle in Terms of Side and Altitude

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Theorem

The area of a triangle $\triangle ABC$ is given by:

$\dfrac {c \cdot h_c} 2 = \dfrac {b \cdot h_b} 2 = \dfrac {a \cdot h_a} 2$

where:

$a, b, c$ are the sides
$h_a, h_b, h_c$ are the altitudes from $A$, $B$ and $C$ respectively.


Corollary

The area of a triangle $ABC$ is given by:

$\dfrac 1 2 a b \sin C$

where:

$a, b$ are two of the sides
$C$ is the angle of the vertex opposite its other side $c$.


Proof 1

Area-of-Triangle.png

Construct a point $D$ so that $\Box ABDC$ is a parallelogram.

From Opposite Sides and Angles of Parallelogram are Equal:

$\triangle ABC \cong \triangle DCB$

hence their areas are equal.

The Area of Parallelogram is equal to the product of one of its bases and the associated altitude.

Thus

\(\ds \paren {ABCD}\) \(=\) \(\ds c \cdot h_c\)
\(\ds \leadsto \ \ \) \(\ds 2 \paren {ABC}\) \(=\) \(\ds c \cdot h_c\) because congruent surfaces have equal areas
\(\ds \paren {ABC}\) \(=\) \(\ds \frac {c \cdot h_c} 2\)

where $\paren {XYZ}$ is the area of the plane figure $XYZ$.


A similar argument can be used to show that the statement holds for the other sides.

$\blacksquare$


Dissection Proof

Area-of-Triangle-Dissection-Proof.png


Motivation

This formula is perhaps the best-known and most useful for determining a triangle's area.

It is usually remembered, and quoted, as half base times height.


Sources

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