Triangles – the three-sided polygons – are among the simplest and most important shapes in geometry. And in this tutorial, we’ll learn how to find the area of a triangle and solve several example problems using the different formulas.

## Area of a Triangle

The area of a triangle is a measure of the region (in the plane) enclosed within the triangle.

For example, the area of the triangle above is the quantity that gives an accurate measure of the yellow region.

1. The most fundamental formula for the area of a triangle is –

$A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

2. For a triangle with adjacent sides a and b and included angle C,

$A = \frac{ab \sin C}{2}$

3. [Heron’s Formula] For a triangle with sides a, b, and c,

$A = \sqrt{s(s-a)(s-b)(s-c)}$

Here s is the semi-perimeter given by

$s = \frac{a + b + c}{2}$

4. For an equilateral triangle with side a,

$A = \frac{\sqrt{3}}{4}a^2$

Skip to examples We have a few different formulas to look at for the area of a triangle. Let’s start with the simplest and most popular one.

The formula most commonly used to find the area of a triangle is –

$A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

We can take any side of the triangle as the base. The corresponding height would be the perpendicular drawn to the base from the opposite vertex.

Here, AD is the height corresponding to the base BC. So the area would be –

$\text{Area}(\triangle ABC) = \frac{1}{2} \cdot BC \cdot AD$

Here's how we get this formula.

If you take a triangle (pink), make a copy of it (yellow), and join the two side by side as shown below, you will end up with a parallelogram.

As you can see above, the base and height of the parallelogram would be the same as those of the triangle.

Also, the area of the parallelogram would just be two times the area of the triangle – it’s made of two (same) triangles. So,

$A(\triangle) = \frac{A(▱)}{2} \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (1)$

Using the formula for the area of a parallelogram –

$A(▱) = bh \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (2)$

From (1) and (2), we have

$A(\triangle) = \frac{bh}{2}$

And that's our formula.

If you know all three sides of a triangle, you can find its area using Heron’s formula.

The area of a triangle with sides a, b, and c is given by –

$A = \sqrt{s(s-a)(s-b)(s-c)}$ Heron's formula

Where a, b, and c are the three sides of the triangle. And s is the semi-perimeter given by –

$s = \frac{a + b + c}{2}$

The derivation of Heron’s formula is beyond the scope of this tutorial.

Note – If you don’t have a basic understanding of trigonometric ratios, feel free to skip the rest of this section.

If you know two sides and the included angle (the angle formed by those two sides), you can find the area using the formula –

$\text{Area} = \frac{ab \sin C}{2}$

Let's see how we can derive this formula.

Say, in △ ABC, we know the sides a and b, and the angle C. How do we find its area?

Again, we’ll use our favorite formula for the area of a triangle.

$A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$

We can take a to be the base. But we don’t know the height (h). So let’s find that.

$\begin{align*} \sin C &= \frac{h}{b} \\[1.3em] h &= b \sin C \end{align*}$

Now that we have h in terms b and C (which are known), we can plug the values of base and height into the area formula.

$\begin{align*} A &= \frac{1}{2} \cdot a \cdot b \sin C \\[1.3em] &= \frac{ab \sin C}{2} \end{align*}$

Derived.

##### The Formula for Equilateral Triangles

In an equilateral triangle, all three sides are equal in length (usually denoted by $\hspace{0.2em} a \hspace{0.2em}$) and each angle is $\hspace{0.2em} 60 \degree \hspace{0.2em}$.

The area of such a triangle is given by –

$A = \frac{\sqrt{3}}{4}a^2$

And here's the derivation.

In the figure below, we have an equilateral triangle with side a. AD is a perpendicular drawn from A on BC – so is the height.

Now, all sides have the same length, a. So our base would be a.

Also, in △ ADC,

$\begin{align*} \sin 60 \degree &= \frac{h}{a} \\[1.3em] \frac{\sqrt{3}}{2} &= \frac{h}{a} \\[1.3em] h &= \frac{\sqrt{3}a}{2} \end{align*}$

So the area of the triangle would be –

$\begin{align*} \text{Area} &= \frac{1}{2} \cdot \text{base} \cdot \text{height} \\[1em] A &= \frac{1}{2} \cdot a \cdot \frac{\sqrt{3}a}{2} \\[1em] &= \frac{\sqrt{3}}{4}a^2 \end{align*}$

## How to Find the Area of a Triangle | Examples

In this section, we’ll use the formulas discussed above to solve problems involving the area of a triangle.

### When One Side and the Corresponding Height Is Known

Example

Find the area of the triangle, also find the length $\hspace{0.2em} BE \hspace{0.2em}$.

Solution

For a triangle with a base $\hspace{0.2em} b \hspace{0.2em}$ and corresponding height $\hspace{0.2em} h$, the area is

$\text{Area} = \frac{bh}{2}$

If we take the side $\hspace{0.2em} BC \hspace{0.2em}$ as the base, the height would be $\hspace{0.2em} AD \hspace{0.2em}$. So,

$\begin{align*} \text{Area} &= \frac{BC \cdot AD}{2} \\[1.3em] &= \frac{30 \cdot 20}{2} \\[1.3em] &= 300 \end{align*}$

Now, of course, the area would be the same even if we took $\hspace{0.2em} AC \hspace{0.2em}$ as the base. Only that the corresponding height would be $\hspace{0.2em} BE \hspace{0.2em}$ (which we need to find). So,

$\begin{align*} \text{Area} &= \frac{AC \cdot BE}{2} \\[1.3em] 300 &= \frac{25 \cdot BE}{2} \\[1.3em] BE &= 24 \end{align*}$

### When All Three Sides Are Known

Example

Find the area of a triangle with side lengths of $\hspace{0.2em} 5$, $\hspace{0.2em} 6$, and $\hspace{0.2em} 7$.

Solution

When we want to find the area of a triangle using its side-lengths, the first step is to find the semi-perimeter, $\hspace{0.2em} s$.

$\begin{align*} s &= \frac{a + b + c}{2} \\[1.3em] &= \frac{5 + 6 + 7}{2} = 9 \end{align*}$

Next, we can use the formula for the area.

$\begin{align*} A &= \sqrt{s(s-a)(s-b)(s-c)} \\[1em] &= \sqrt{9(9-5)(9-6)(9-7)} \\[1em] &\approx 14.70 \end{align*}$

### When It's an Equilateral Triangle

Example

Find the area of an equilateral triangle with a side length of $\hspace{0.2em} 14 \hspace{0.2em}$ cm.

Solution

The area of an equilateral triangle with a side-length $\hspace{0.2em} a \hspace{0.2em}$ is –

$A = \frac{\sqrt{3}}{4}a^2$

Substituting the side-length into the formula, we get

$\begin{align*} A &= \frac{\sqrt{3}}{4} \cdot 14^2 \\[1em] &\approx 84.87 \end{align*}$

So the area is $\hspace{0.2em} 84.87 \text{ cm}^2 \hspace{0.2em}$.

### When Two Sides and the Included Angle Are Known

Example

Find the area of the triangle below.

Solution

When we know the lengths of two sides of a triangle and the included angle, we can use the following formula for the area –

$A = \frac{ab \sin C}{2}$

Plugging in the values into the formula,

$\begin{align*} A &= \frac{15 \cdot 18 \cdot \sin 50 \degree}{2} \\[1.3em] &\approx 103.42 \end{align*}$

And that brings us to the end of this tutorial on how to find the area of a triangle. Until next time.