An Intuitive Proof of Heron’s Formula

Heron’s formula finds a triangle’s area from its three side lengths, without needing an angle or a height. Its familiar product of four terms can seem mysterious; the geometric idea is to drop an altitude, use the right-triangle relations it gives, and then factor the area equation.

The picture: one altitude, two right triangles

Let the triangle have base length $a$ and other side lengths $b$ and $c$. Drop a perpendicular from the top vertex to the line containing the base. Call the height $h$, and let the foot of the perpendicular have coordinate $x$ when the base runs from 0 to $a$.

             C
            /|\\
         c / |h \\ b
          /  |   \\
         A---D----B
          x   a-x
             a

The diagram shows the foot $D$ inside the base, as it is for an acute triangle. For an obtuse triangle, the foot can fall outside the base; using a signed coordinate $x$ makes the same equations work.

The area is $K=ah/2$. The right triangles determined by the altitude give, by the Pythagorean theorem,

$$c^2=x^2+h^2, \qquad b^2=(a-x)^2+h^2.$$

Express the height using the sides

Subtract the second equation from the first. The shared height cancels, leaving

$$c^2-b^2=x^2-(a-x)^2=2ax-a^2.$$

So the location of the altitude’s foot is

$$x=\frac{a^2+c^2-b^2}{2a}.$$

Now use $h^2=c^2-x^2$. Instead of substituting $x$ and expanding everything, multiply by $4a^2$ and recognize a difference of squares:

$$\begin{aligned} 16K^2 &= 4a^2h^2 \\ &=4a^2c^2-(a^2+c^2-b^2)^2 \\ &=\big((a+c)^2-b^2\big)\big(b^2-(a-c)^2\big). \end{aligned}$$

The last line is the key factorization. Each factor is itself a difference of squares, so

$$16K^2=(a+c-b)(a+c+b)(b-a+c)(b+a-c).$$

See the semiperimeter appear

The semiperimeter is $s=(a+b+c)/2$. Each factor in the product is twice either $s$ or the amount by which $s$ exceeds one side:

$$a+c+b=2s, \quad a+c-b=2(s-b), \quad b-a+c=2(s-a), \quad b+a-c=2(s-c).$$

Therefore,

$$16K^2=16s(s-a)(s-b)(s-c),$$

and dividing by 16 gives

$$K^2=s(s-a)(s-b)(s-c).$$

Since area is positive, take the positive square root:

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

The geometric intuition is that the altitude encodes the triangle’s shape, while the factorization turns that information into the four side-based quantities in Heron’s formula. The terms $s-a$, $s-b$, and $s-c$ are half the gaps in the triangle inequalities; for example, $s-a=(b+c-a)/2$.

Worked example: sides 13, 14, and 15

For a triangle with side lengths 13, 14, and 15, the semiperimeter is

$s=\frac{13+14+15}{2}=21.$

The three differences are $21-13=8$, $21-14=7$, and $21-15=6$. Thus

$$K=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7056}=84.$$

As a check using the altitude picture, take $a=14$, $b=13$, and $c=15$. The altitude foot is at $x=(14^2+15^2-13^2)/(2\cdot14)=9$, so $h=\sqrt{15^2-9^2}=12$. The area is $14\cdot12/2=84$, matching Heron’s formula.

Common mistakes

• Using the wrong semiperimeter: $s$ is half the sum of all three sides, not the sum itself.
• Forgetting the square root: The product $s(s-a)(s-b)(s-c)$ equals the area squared.
• Assuming the altitude always lands inside the base: It may fall outside for an obtuse triangle. The coordinate derivation still applies.
• Ignoring whether the sides make a triangle: For a nondegenerate triangle, each side is less than the sum of the other two, so all four factors in the factored expression are positive. If equality holds, the triangle is degenerate and its area is zero.