Area of Triangle in Terms of Side and Altitude

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Theorem

The area of a 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

Tri.PNG

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

From Halves of a Parallelogram Are Congruent:

$\triangle ABC \cong \triangle DCB$

hence their areas are equal.

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

Thus

\(\displaystyle \left({ABCD}\right)\) \(=\) \(\displaystyle c \cdot h_c\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle 2 \left({ABC}\right)\) \(=\) \(\displaystyle c \cdot h_c\) $\quad$ because congruent surfaces have equal areas $\quad$
\(\displaystyle \left({ABC}\right)\) \(=\) \(\displaystyle \frac {c \cdot h_c} 2\) $\quad$ $\quad$

where $\left({XYZ}\right)$ is the area of the plane figure $XYZ$.


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

$\blacksquare$


Note

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