# Heron’s Formula Explained

By Kathleen Knowles, 03 Apr 2021

The perimeter of a triangle is its three sides added together. The area of a triangle is quite interesting. It will usually be expressed as area = (1/2) x base x height, `A = (1/2)b*h`

, where height is measured as a vertical line from the base to the opposite vertex. This formula makes calculating the area of a triangle relatively easy, but it is quite difficult to naturally find a triangle that has at least one known side (base) and a known height. Obtaining the length of all three sides of the triangle is often much easier. Fortunately, a formula does exist for calculating the area of a triangle when all three sides are known.

Heron's formula allows us to find the area of a triangle when we know the length of all three sides of that triangle. It turns out that we can calculate its area without proving any additional theorems or using trigonometry. The following equation is Heron’s formula.

`A = sqrt(s(s-a)(s-b)(s-c))`

Where `s = (a+b+c)/2`

or `1/2`

the perimeter, and `A`

stands for area

**Derivation of Herron's formula**

The triangle `DeltaABC`

has sides `a`

, `b`

, and `c`

. We can create a formula for the area of triangle ABC using these three sides.

First, we'll recall the standard equation to find the area of a triangle - it is the base multiplied by the height of the triangle, all divided by two `(1/2)bh`

. We know all three side lengths, so we can use any of them as the base - in this case, we choose side `b`

. Since we don't have the height of the triangle, we draw a line for the height, `h`

, at a right angle to side `b`

up to the vertex of `a`

and `c`

. Drawing line `h`

creates two right triangles - one on the right and one on the left of `h`

.

Labeling the section of side `b`

from `h`

to the vertex of `b`

and `c`

as `x`

, we now have two right triangles. The left-hand triangle has sides `h`

, `(b-x)`

, and `a`

, and the right-hand triangle has sides `h`

, `x`

, and `c`

.

Using the Pythagorean theorem, we can derive expressions for `x`

and `h`

by simply using `a`

, `b`

, and `c`

.

**The solution for x:**

From the right triangle on the right side, we get

`h^2 + x^2 = c^2`

, or `h = sqrt(c^2 - x^2)`

It is also possible to start with the left side of the right triangle.

`h^2 + (b-x)^2 = a^2`

, or `h = sqrt(a^2 - (b-x)^2) = sqrt(a^2 - b^2 + 2bx - x^2)`

Since `h`

is the same in both these triangle we can set the two equations for `h`

equal to each other.

`sqrt(c^2-x^2) = sqrt(a^2-b^2+2bx-x^2)`

Since the lengths here are all positive, we can square both sides and get `c^2-x^2 = a^2-b^2+2bx-x^2`

.

By simplifying and rearranging, we can solve for `x`

and get

`x = (-a^2+b^2+c^2)/(2b)`

.

**The solution for h:**

Now we can substitute the equation for `x`

into one of our original right triangle equations above to solve for `h`

. Using the right-hand side triangle, `h^2 + x^2 = c^2`

, so `h^2 = c^2 - x^2`

. Plugging in the equation for `x`

, we get `h^2 = c^2-((-a^2 + b^2 + c^2)/(2b))^2`

.

The squared difference identity states that `x^2-y^2 = (x+y)(x-y)`

. Replacing `x`

and `y`

in this identity with `b`

and `(-a^2+b^2+c^2)/(2b)`

, respectively, we get

`h^2 = (c+(-a^2+b^2+c^2)/(2b))(c-(-a^2+b^2+c^2)/(2b))`

`= ((2bc-a^2+b^2+c^2)/(2b))((2bc+a^2-b^2-c^2)/(2b))`

. Now let's simplify and rearrange it a bit. Multiplying across and factoring out the denominator, we get

`h^2 = (1/(4b^2))(2bc-a^2+b^2+c^2)(2bc+a^2-b^2-c^2)`

.

From here we can factor and then apply the square difference identity again to get the following.

`h^2 = (1/(4b^2))((b+c)^2-a^2)(a^2-(b-c)^2)`

`= (1/(4b^2)) ((b+c)+a)((b+c)-a)(a+(b-c))(a-(b-c))`

It follows that `h = (1/(2b))sqrt((b+c+a)(b+c-a)(a+b-c)(a+c-b)`

.

Since the area of the triangle is `(1/2)bh`

, we can simply multiply them together to get

`A = (1/4)sqrt((b+c+a)(b+c-a)(a+b-c)(a+c-b)`

.

**Simplifying Heron's Formula to its Final Form**

So, let's play with the lengths to get a simpler formula and an interesting result.

Since `a`

, `b`

, and `c`

are the sides of a triangle, `a+b+c`

is its perimeter. Defining `s`

as half of the perimeter we get `s = (a+b+c)/2`

, or `2s = b+c+a`

.

From this we can find that `b+c-a = 2(s-a)`

; `a+b-c = 2(s-c)`

, and `a+c-b = 2(s-b)`

. Plugging these into the equation for the area of `DeltaABC`

above, we get

`A = (1/4)sqrt((2s)(2(s-a))(2(s-b))(2(s-c))`

.

Then, with some simplification and a little rearranging we get

`A = sqrt(s(s-a)(s-b)(s-c))`

.

Now we have reached the final state of Heron’s formula, and we can use it to find the area of a triangle if the height is unknown, but all three sides are.

**Examples of Using Heron’s Formula**

While Heron's formula is used to find the area of a triangle from just the lengths of three sides, it can also be used to find the height of a triangle.

**Using Heron’s Formula to Find Area**

If you have all three sides, use Herron's formula. Herron's formula has two parts. First, you must find the variable `s`

, which is equal to half the triangle’s perimeter. This formula is: `s = (a+b+c)/2`

.

For a triangle with side lengths `a = 3`

, `b = 4`

, and `c = 5`

,

`s = (3+4+5)/2`

`s = (12)/2`

`s = 6`

Then use the second part of Heron's equation, `A = sqrt(s(s-a)(s-b)(s-c))`

. `A = sqrt(6(6-3)(6-4)(6-5))`

`A = sqrt(6(3)(2)(1)) = 6`

The area of the triangle is 6 square units.

**Using Heron’s Formula to Find Height**

In addition to finding the area of a triangle when only the three sides are known, this formula also allows us to calculate the height of a triangle. To accomplish this, set the area to `1/2bh`

and then solving for the height. This gives us the equation `1/2bh = sqrt(s(s-a)(s-b)(s-c))`

or `h = (2sqrt(s(s-a)(s-b)(s-c)))/b`

For our example triangle with side lengths 3, 4, and 5, it is solved as follows.

`(1/2)(4)h = sqrt(6(6-3)(6-4)(6-5)`

`2h = 6`

`h = 3`

So, with side `b`

as the base, the height is 3.

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