Area of Triangle

Theorem

This page gathers a variety of formulas for the area of a triangle.

In Terms of Side and Altitude

The area of a triangle $\triangle ABC$ 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.

In Terms of Two Sides and Angle

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 $c$.

In Terms of Inradius

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Then the area $\AA$ of $\triangle ABC$ is given by:

$\AA = r s$

where:

$r$ is the inradius of $\triangle ABC$

$s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.

In Terms of Circumradius

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Then the area $\AA$ of $\triangle ABC$ is given by:

$\AA = \dfrac {a b c} {4 R}$

where $R$ is the circumradius of $\triangle ABC$.

In Terms of Exradius

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Let $\rho_a$ be the exradius of $\triangle ABC$ with respect to the excircle which is tangent to $a$.

Let $s$ be the semiperimeter of $\triangle ABC$.

Then the area $\AA$ of $\triangle ABC$ is given by:

$\AA = \rho_a \paren {s - a}$

In Terms of Inradius and Exradii

Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Then the area $\AA$ of $\triangle ABC$ is given by:

$\AA = \sqrt {r \rho_a \rho_b \rho_c}$

where:

$r$ is the inradius

$\rho_a$, $\rho_b$ and $\rho_c$ are the exradii of $\triangle ABC$ with respect to $a$, $b$ and $c$ respectively.

Heron's Formula

Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.

Then the area $\AA$ of $\triangle ABC$ is given by:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.