# Area of Triangle in Terms of Side and Altitude

## 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

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

$\triangle ABC \cong \triangle DCB$

hence their areas are equal.

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

Thus

 $\displaystyle \paren {ABCD}$ $=$ $\displaystyle c \cdot h_c$ $\displaystyle \leadsto \ \$ $\displaystyle 2 \paren {ABC}$ $=$ $\displaystyle c \cdot h_c$ because congruent surfaces have equal areas $\displaystyle \paren {ABC}$ $=$ $\displaystyle \frac {c \cdot h_c} 2$

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.

$\blacksquare$

## 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.