Area of a Triangle

This page contains a variety of formulas for the 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:

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

In Terms of Circumradius

Theorem

The area of $\triangle ABC$ is given by the formula:

$(ABC) = \dfrac {a \cdot b \cdot c} {4r}$

where $r$ is the circumradius and $a, b, c$ are the sides.

In Terms of Exradii

Theorem

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

$(ABC) = \rho_a \left({s - a}\right) = \rho_b \left({s - b}\right) = \rho_c \left({s - c}\right) = \rho s = \sqrt {\rho_a \rho_b \rho_c \rho}$

In this formula:

• $s$ is the semiperimeter

• $I$ is the incenter

• $\rho$ is the inradius

• $I_a, I_b, I_c$ are the excenters

• $\rho_a, \rho_b, \rho_c$ are the exradii from $I_a, I_b, I_c$, respectively.

Heron's Formula

Theorem

Given a triangle $\triangle ABC$ with sides $a$, $b$, and $c$ opposite points $A$, $B$, and $C$, respectively.

Let $s$ be the semiperimeter, so $s = \dfrac{a + b + c} 2$.

Then the area $A$ of the triangle is given by the formula $A = \sqrt{s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right)}$.