Heron's formula

Heron's formula is a two-step recipe. First reduce the three sides to a single number, the semi-perimeter; then feed that number, together with the sides, into a square root.

Step A — the semi-perimeter. Half the perimeter of the triangle is called $s$:

where: $a$, $b$, $c$ are the three side lengths of the triangle (all in the same unit), and $s$ is the semi-perimeter, in that same unit.

Step B — the area. Heron's formula then gives the area directly:

where: $A$ is the area of the triangle, and $s-a$, $s-b$, $s-c$ are the semi-perimeter reduced by each side in turn. NCERT states this result without proof at Class 9 level — you are expected to apply it fluently, not derive it, though the derivation section below sketches why it is trustworthy. Two features are worth noticing before you ever plug in numbers. First, the formula is completely symmetric in $a$, $b$, $c$ — swap any two sides and the answer does not change, which is a quick way to sanity-check that you have not mislabelled a side. Second, for $s-a$, $s-b$, $s-c$ to all come out positive — which they must, for the square root to make sense — the three lengths $a$, $b$, $c$ must satisfy the triangle inequality (each side shorter than the sum of the other two). If your arithmetic ever produces a negative value under the root, the given "sides" cannot form a real triangle, and that is the signal to recheck the numbers rather than push on. Notice too that Heron's formula does not care what kind of triangle you are given. Plug in an isosceles triangle's sides and it still works; plug in an equilateral triangle's side, repeated three times, and it still works, collapsing to the familiar $A = \frac{\sqrt{3}}{4} a^2$ once you simplify $s = \frac{3a}{2}$ inside the root. This generality — one formula, every triangle, no case-by-case rules for "is it right-angled" or "is it isosceles" — is exactly why it is worth learning as a single reliable procedure rather than as a special-case trick.

Problem. Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm.

Step 1. Compute the semi-perimeter: $s = \frac{13+14+15}{2} = \frac{42}{2} = 21$ cm.

Step 2. Compute $s-a$, $s-b$, $s-c$: $s-13 = 8$, $s-14 = 7$, $s-15 = 6$.

Step 3. Substitute into Heron's formula: $A = \sqrt{21 \times 8 \times 7 \times 6}$.

Step 4. Multiply inside the root, then simplify: $21 \times 8 = 168$, $168 \times 7 = 1176$, $1176 \times 6 = 7056$, so $A = \sqrt{7056}$. Since $84 \times 84 = 7056$, $A = 84 \text{ cm}^2$. The area of the triangle is 84 cm².

Competitive stretch

The same method scales to any triangle, including ones where the numbers are less friendly. For a triangular park with sides 40 m, 24 m and 32 m: $s = \frac{40+24+32}{2} = 48$ m, and $s-a, s-b, s-c = 8, 24, 16$. Then $A = \sqrt{48 \times 8 \times 24 \times 16} = \sqrt{147456}$. Rather than multiply everything out and guess, prime-factorise: $48 = 2^4 \times 3$, $8 = 2^3$, $24 = 2^3 \times 3$, $16 = 2^4$, so the product is $2^{14} \times 3^2$. Taking the square root, $2^7 \times 3 = 128 \times 3 = 384$. So $A = 384 \text{ m}^2$, confirming the direct multiplication.

• Forgetting to halve the perimeter. Using $a+b+c$ directly instead of $s = \frac{a+b+c}{2}$ inside the formula is the single most common slip — always compute $s$ as a clearly labelled first step.
• Subtracting in the wrong order. Writing $a-s$ instead of $s-a$ produces a negative number under the root; a valid triangle never gives a negative $s-a$, $s-b$, or $s-c$, so a negative value is a signal to recheck your subtraction, not the triangle.
• Guessing the square root instead of factorising. For large products like $147456$, multiplying blindly and estimating the root invites error — prime-factorise and cancel matching pairs systematically instead.
• Mixing units. If one side is given in metres and another in centimetres, convert everything to one unit before computing $s$; otherwise both $s$ and the final area are meaningless.

Ask the tutor: Want to try Heron's formula on a triangle with sides that don't give a perfect square, or see the full height-elimination algebra behind Step 2 of the reasoning section? Ask the AI tutor to work through it with you.