Triangle area and perimeter

A triangular plot of land is surveyed by walking its three boundary lines and recording their lengths — nobody climbs inside to measure a perpendicular height. Yet a surveyor can still hand you the exact area, because of a result more than two thousand years old: Heron's formula, which turns three side lengths alone into an area, with no angle and no altitude required. This is the leap Class 9 Mensuration makes beyond the familiar "half base times height" rule. You already know the perimeter of a triangle is just the sum of its sides, and that its area is half the base times the height when that height is given. The new tool is for the very common case where it is not: a scalene triangle described only by $a$ , $b$ , $c$ . Heron's formula, combined with the Pythagoras theorem for right triangles, covers every triangle mensuration question this chapter asks.

The core insight

The area of any triangle can be found from its three sides alone using Heron's formula, $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ with semi-perimeter $s = \tfrac{a+b+c}{2}$ , while the perimeter is simply $P = a+b+c$ .

The core idea

Perimeter — always the same rule. However a triangle is described, its perimeter is the sum of its three sides:

P = a + b + c

where: $a$ , $b$ , $c$ are the three side lengths, all in the same unit.

Area when a height is known. If a base and the perpendicular height to that base are given, the familiar formula applies directly:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

where: base is any one side, and height is the perpendicular distance from that side to the opposite vertex — not a slanting side.

Area when only the three sides are known — Heron's formula. First find the semi-perimeter, then use it as follows:

s = \frac{a+b+c}{2}, \qquad \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}

where: $s$ is half the perimeter, and $a$ , $b$ , $c$ are the three sides. This is the star tool of the chapter: it needs no height at all, only the boundary lengths.

Right-angled triangles — a shortcut, not a new rule. When the triangle has a right angle, the two sides forming that angle already act as base and perpendicular height, so

\text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2

and if one side is missing, the Pythagoras theorem, $(\text{hypotenuse})^2 = (\text{leg}_1)^2 + (\text{leg}_2)^2$ , finds it first. Heron's formula still works on a right triangle too — it must agree with this shortcut, since both describe the same triangle.

Figure

Two triangles side by side one right angled with legs used directly as base and height and one scalene with only its three sides marked, illustrating when to use the base height rule and when to use Herons formula

Derivation AWhy Heron's formula only needs the three sides

The goal is to see why a formula built purely from $a$ , $b$ , $c$ and their sum can correctly reproduce the ordinary area $\tfrac12 \times \text{base} \times \text{height}$ , and to show it gives a consistent, checkable number for a real triangle rather than an arbitrary combination of the sides.

Step 1. Take any triangle with sides $a$ , $b$ , $c$ and drop a perpendicular height $h$ from one vertex onto the opposite side (the base). By the ordinary area rule, $\text{Area} = \tfrac12 \times \text{base} \times h$ . The trouble is that $h$ is usually not given directly — it depends on the triangle's shape in a way that mixes all three sides together through the Pythagoras theorem applied twice, once for each of the two smaller right triangles the height creates.

Step 2. Working through that algebra (splitting the base into two parts at the foot of the perpendicular, and applying Pythagoras theorem to each half-triangle) eventually expresses $h$ entirely in terms of $a$ , $b$ and $c$ , with no leftover unknown. Substituting that expression for $h$ back into $\tfrac12 \times \text{base} \times h$ and simplifying the resulting expression is what produces Heron's formula. The heavy algebra is not needed for exam use — what matters is that the height is fully determined by the three sides, which is precisely why an area formula using only $a$ , $b$ , $c$ can exist at all.

Step 3. Package the result using the semi-perimeter $s = \tfrac{a+b+c}{2}$ , which makes the formula compact: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ . Introducing $s$ is purely for tidiness — it is nothing more than half the perimeter, computed first so the four factors under the root can be written quickly.

Step 4. Check that the formula behaves sensibly. Each factor $s-a$ , $s-b$ , $s-c$ must be positive for a genuine triangle, because the triangle inequality (each side less than the sum of the other two) guarantees $s$ is bigger than any single side. If the three given lengths cannot actually form a triangle, one of these factors becomes zero or negative, the product under the root is not a valid positive area, and this is a built-in check on whether the data given even describes a real triangle.

Worked example

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

Step 1. Find the perimeter by adding the three sides directly:

P = 13 + 14 + 15 = 42 \text{ cm}

Step 2. Find the semi-perimeter, half of what was just found:

s = \frac{42}{2} = 21 \text{ cm}

Step 3. Substitute into Heron's formula:

\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6}

Step 4. Multiply the four factors carefully in pairs rather than all at once, to keep the arithmetic manageable:

21 \times 8 = 168, \qquad 7 \times 6 = 42, \qquad 168 \times 42 = 7056

\text{Area} = \sqrt{7056} = 84 \text{ cm}^2

The perimeter is 42 cm and the area is 84 cm².

Competitive stretch

A right-angled triangle has legs 9 cm and 12 cm. Verify Heron's formula against the direct base-height rule. By the Pythagoras theorem the hypotenuse is $\sqrt{9^2+12^2} = \sqrt{225} = 15$ cm, giving perimeter $9+12+15 = 36$ cm. Direct area using the two legs as base and height is $\tfrac12 \times 9 \times 12 = 54\ \text{cm}^2$ . Checking with Heron's formula, $s = \tfrac{36}{2} = 18$ , and $\text{Area} = \sqrt{18 \times 9 \times 6 \times 3} = \sqrt{2916} = 54\ \text{cm}^2$ — the two methods agree, confirming both the missing side and the area.

Common traps

Reaching for Heron's formula when a height is already stated — if the perpendicular height is given directly, $\tfrac12 \times \text{base} \times \text{height}$ is faster and less error-prone; keep Heron's formula for when only the three sides are known.
Confusing $s$ with the full perimeter — $s$ is the semi-perimeter, half the sum of the sides; forgetting to halve is one of the most common slips in the chapter.
Multiplying two non-perpendicular sides as if they were base and height — the shortcut $\tfrac12 \times \text{leg}_1 \times \text{leg}_2$ only works when those two sides meet at the right angle; for any other pair of sides this gives a wrong answer.
Not checking the triangle inequality first — if one given side is greater than or equal to the sum of the other two, no such triangle exists, and Heron's formula will produce zero or an undefined value under the root instead of a real area.

Ask the tutor: Want to try Heron's formula on a triangle with surd-valued sides, or check whether a set of three lengths can even form a triangle? Ask the AI tutor to work through it with you step by step.

How do I know whether to use base-height or Heron's formula for a given question?

Look at what is given. If a perpendicular height is stated or can be read directly from the figure, use $\tfrac12 \times \text{base} \times \text{height}$ — it is quicker and avoids the square root. If only the three side lengths are given, with no height and no right angle to exploit, Heron's formula is the only direct route, since finding the height yourself would need extra construction and Pythagoras theorem work that Heron's formula already packages for you.

Why does Heron's formula need a square root at all?

Area has units of length squared, but the four factors $s$ , $s-a$ , $s-b$ , $s-c$ inside the root are each a length, so their product has units of length to the fourth power. The square root brings that back down to length squared, matching an area. This is also a quick sanity check on your working: if you forget the square root, your answer will be numerically far too large and have the wrong units.

What happens if I plug in side lengths that do not actually form a triangle?

One of the bracketed terms $s-a$ , $s-b$ or $s-c$ becomes zero or negative, so the product under the square root is zero or negative and the formula gives no valid real area. This is not a flaw in the formula — it is the formula correctly telling you the triangle inequality has failed, meaning a triangle with those three side lengths cannot exist. Always check that no side is greater than or equal to the sum of the other two before trusting your answer.

Triangle area and perimeter

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads