**Heron’s Formula** can be used to find the area of a triangle given the lengths of the three sides. A triangle with side lengths , , and , its area can be calculated using the Heron’s formula

where

is the semiperimeter (half the perimeter) of the triangle.

In this post, I will provide a detailed derivation of this formula.

##### Some Preliminaries

The area of a triangle is half the product of its base and its altitude. In the figure below, is the altitude of triangle . If the length of the altitude is not given, and an angle measure is given, we can use Trigonometry to calculate the altitude.

In the figure above, the altitude forms right triangle . We know that

= **(length of opposite side)/(length of hypotenuse) **

so,

.

Simplifying, we have which is equivalent to

Since the base of triangle is and its altitude is , its area is given by the formula

. (**1**)

##### The Derivation

Now we apply the preceding formula, the Cosine Law and the Pythagorean identity to derive the Heron’s formula.

Using the Pythagorean identity, and manipulating algebraically

(**2**)

By the Cosine Law, in a triangle with side lengths , , and

.

Calculating for , we have

.

Substituting the preceding equation to (2), we have

.

.

.

.

Getting the square root of both sides, we have

(**3**).

Using (1) and (3), we calculate the area of the triangle ,

(**4**)

Now, if we let be the semiperimeter (half the perimeter) of triangle , then

.

Now, .

Also, and .

Substituting the expressions with s to (4), we have

which is equivalent to

.

Simplifying, we have , the Heron’s formula.

Very nice proof. I’ve seen some really ugly ones to verify the Heron’s formula. This one relies on the formula for the area of a triangle, the law of cosines, and the Pythagorean Identity. Straightforward and easily digestible. Thank you

Thanks Jack. 🙂