# Area of Triangle in Terms of Side and Altitude

## Contents

## 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:

### Corollary

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

- $\dfrac 1 2 a b \sin C$

where:

## Proof

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

From Halves of Parallelogram Are Congruent Triangles:

- $\triangle ABC \cong \triangle DCB$

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**.

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 4$: Geometric Formulas: $4.5$