Area of a Triangle in Terms of Side and Altitude

Theorem

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

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

where:

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

Corollary

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

$\displaystyle \frac {a b \sin C} 2 = \frac {a c \sin B} 2 = \frac {b c \sin A} 2$

Proof

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

Then we have $\triangle ABC \cong \triangle DCB$, hence their areas are equal.

The area of a parallelogram is equal to the product of one of its bases and the associated altitude.

Thus

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

where $(XYZ)$ is the area of the plane figure $XYZ$.

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

$\blacksquare$

Proof of Corollary

Follows from the (geometric) definition of sine.

Using the notation of the above triangle:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle h_a\)	\(=\)	\(\displaystyle c \sin B\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle h_b\)	\(=\)	\(\displaystyle a \sin C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle h_c\)	\(=\)	\(\displaystyle b \sin A\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Substituting in:

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

... gives:

$\displaystyle \frac {a b \sin C} 2 = \frac {a c \sin B} 2 = \frac {b c \sin A} 2$

$\blacksquare$

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

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $4.5$