Area of a triangle
The familiar formula for the area of a triangle is one half times the base times the height. But the height is often not given. This page gives two formulas that use the sides and angles instead: one for two sides and the angle between them, and one, called Heron's formula, for three sides.
Two sides and the angle between them
Take side ${b}$ as the base. The height ${h}$ goes from ${B}$ straight down to the base. In the right triangle it makes, ${h}$ is opposite the angle ${C}$ and ${a}$ is the hypotenuse:
So ${h=a\sin C}$. The area is one half times the base ${b}$ times the height ${h}$. Putting in ${h=a\sin C}$ gives
${\text{Area}=\dfrac{1}{2}ab\sin C}$
In words: half the product of two sides times the sine of the angle between them. Any two sides work, as long as you use the angle between them: the area is also ${\dfrac{1}{2}bc\sin A}$ or ${\dfrac{1}{2}ac\sin B}$.
Example 1: Find the area of a triangle with sides ${8}$ and ${11}$ and an angle of ${50^\circ}$ between them.
Solution:
$\begin{align*}\text{Area}&=\dfrac{1}{2}(8)(11)\sin 50^\circ\\&\approx 33.71\end{align*}$
The area is about ${33.71}$ square units.
Example 2: In Example 1 of The law of sines, the triangle had ${A=40^\circ}$, ${B=60^\circ}$, ${C=80^\circ}$, ${a=10}$, and ${b\approx 13.47}$. Find its area.
Solution:
The angle between sides ${a}$ and ${b}$ is ${C}$:
$\begin{align*}\text{Area}&=\dfrac{1}{2}(10)(13.47)\sin 80^\circ\\&\approx 66.34\end{align*}$
A useful fact follows: for two given sides, the area is largest when ${\sin C=1}$, that is, when the angle between them is ${90^\circ}$.
Three sides: Heron's formula
If you know all three sides, you could find an angle with the law of cosines and then use the formula above. Heron's formula does it in one step. First find the semiperimeter ${s}$, which is half the perimeter:
${s=\dfrac{a+b+c}{2}}$
${\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}}$
The formula is named after Heron of Alexandria, who wrote it down almost ${2000}$ years ago. It can be proved from the law of cosines with a lot of algebra.
Example 3: Find the area of a triangle with sides ${7}$, ${8}$, and ${9}$.
Solution:
The semiperimeter is ${s=\dfrac{7+8+9}{2}=12}$. Then ${s-a=5}$, ${s-b=4}$, and ${s-c=3}$:
$\begin{align*}\text{Area}&=\sqrt{12\cdot 5\cdot 4\cdot 3}\\&=\sqrt{720}\approx 26.83\end{align*}$
Example 4: A farmer's field is a triangle with sides ${120}$ m, ${150}$ m, and ${210}$ m. Find its area.
Solution:
${s=\dfrac{120+150+210}{2}=240}$
So ${s-a=120}$, ${s-b=90}$, and ${s-c=30}$:
$\begin{align*}\text{Area}&=\sqrt{240\cdot 120\cdot 90\cdot 30}\\&=\sqrt{77{,}760{,}000}\approx 8818\text{ m}^2\end{align*}$
That is about ${2.2}$ acres (an acre is about ${4047}$ m${^2}$).
Watch out: In Heron's formula, ${s}$ is half the perimeter. Using the whole perimeter is a common mistake.
Summary
- Two sides and the angle between them: ${\text{Area}=\dfrac{1}{2}ab\sin C}$.
- Three sides: with ${s=\dfrac{a+b+c}{2}}$, the area is ${\sqrt{s(s-a)(s-b)(s-c)}}$.
- For any other given parts, solve the triangle first with the law of sines or cosines.