Heron's formula (derivation using trig): Visual walkthrough

Prerequisites we lean on: Area of a triangle = ½ab sin C, Law of Cosines, Difference of Squares Factoring, Triangle Inequality. This page is the visual companion to the parent derivation.

Step 0 — The characters in our story

WHAT: we label a triangle so every symbol has a home. WHY: the entire proof is a hunt to remove the angle $C$ and keep only the lengths $a, b, c$ . You must first see which side is which. PICTURE: the figure below. Notice how sides $a$ and $b$ squeeze the angle $C$ between them — $C$ is the "included angle" of $a$ and $b$ . Side $c$ sits across from it.

Figure — Heron's formula (derivation using trig)

The big idea in one sentence: three sticks fix one shape, so the area is already decided by $a, b, c$ alone. Our job is to prove what that number is.

Step 1 — Start with the one area formula that uses an angle

WHAT: we write the area using two sides and the angle wedged between them.

WHY this formula and not another? Because two sides $a, b$ plus the angle $C$ between them pin the triangle down completely — nothing else can wiggle. So the area must be a formula in exactly those three quantities. This is our launching pad. Its only flaw: it carries the angle $C$ , the very thing we want gone.

WHY does $sin C$ appear at all? Look at the picture. Drop a straight line from the top corner down to side $b$ — that's the height $h$ . The area of any triangle is $\frac{1}{2} \times base \times height$ . Here the base is $b$ . The height is not $a$ itself, because side $a$ leans over at angle $C$ . The vertical part of that leaning stick is $a sin C$ — that's what $sin$ measures: the fraction of a leaning length that points straight up. So $h = a sin C$ , and area $= \frac{1}{2} b (a sin C)$ .

The enemy is now clearly visible: that $sin C$ . Everything below is the chase to replace it with $a, b, c$ .

Step 2 — Square both sides to make the chase easier

WHAT: we multiply the whole equation by itself.

WHY: angles love to hide inside square roots and $\pm$ signs. Squaring pushes all of that into one clean place. And there's a golden identity waiting for us: $sin^{2} C = 1 - cos^{2} C$ . Cosine, unlike sine, is something the Law of Cosines can hand us purely in terms of sides. So squaring is the doorway from "sine world" (hard) into "cosine world" (easy).

PICTURE: think of $Area^{2}$ as the area of a literal square whose side is the triangle's area. Squaring is not scary — it's just a bookkeeping move that unlocks the identity below.

$sin^{2} C = 1 - cos^{2} C ⟹ Area^{2} = \frac{1}{4} a^{2} b^{2} (1 - cos^{2} C)$

The identity $sin^{2} + cos^{2} = 1$ is just the Pythagoras theorem living on a circle of radius $1$ — the figure shows why. Now every $sin$ is gone; only $cos C$ remains.

Step 3 — Trade the angle for sides with the Law of Cosines

WHAT: we replace $cos C$ by a pure ratio of side lengths.

WHY this tool? The Law of Cosines is the only rule that connects an angle directly to all three sides. It's the bridge out of "angle land." Rearranging it isolates $cos C$ — now a fraction with only $a, b, c$ in it. The enemy angle is officially expelled: no $C$ survives on the right.

Reading the fraction term by term:

top $a^{2} + b^{2} - c^{2}$ — measures how "sharp" the angle is; if $c$ is small, the top is large and positive (acute); if $c$ is large, the top goes negative (obtuse).
bottom $2 ab$ — a scaling factor so the ratio always lands between $- 1$ and $+ 1$ .

PICTURE: the figure shows the same triangle three ways — acute, right, obtuse — with the sign of $cos C$ flipping as side $c$ grows. This is important: our formula must survive all these cases, and it will, because we never assumed $C$ was acute.

Step 4 — Split $1 - cos^{2} C$ into two friendly halves

WHAT: we factor using $x^{2} - y^{2} = (x - y) (x + y)$ with $x = 1$ , $y = cos C$ .

WHY: see Difference of Squares Factoring. A single messy square becomes a product of two simple pieces. This matters enormously, because when we plug in the Law-of-Cosines fraction, each piece turns into its own difference of squares in the sides — and those crack open into the exact factors Heron needs.

PICTURE: literally cut a square of area $1$ and remove a smaller square of area $cos^{2} C$ . The leftover L-shape rearranges into a rectangle with sides $(1 - cos C)$ and $(1 + cos C)$ . That rearrangement is the factoring.

Step 5 — Compute each half; watch the sides appear

Now substitute $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ into each piece.

WHAT: each half becomes a fraction whose top is another difference of squares, which factors again.

WHY it works — read the middle step:

In the first line, $2 ab + a^{2} + b^{2}$ is exactly $(a + b)^{2}$ (the perfect square). Subtracting $c^{2}$ gives $(a + b)^{2} - c^{2}$ — a difference of squares! It splits into $(a + b + c) (a + b - c)$ .
In the second line, $- a^{2} - b^{2} + 2 ab = - (a - b)^{2}$ , so we get $c^{2} - (a - b)^{2}$ — again a difference of squares — splitting into $(c + a - b) (c - a + b)$ .

PICTURE: each colored bracket $(a + b + c)$ , $(a + b - c)$ , $(c + a - b)$ , $(c - a + b)$ is a length built from the three sticks. The figure lines them up so you can literally see $(a + b - c)$ as "how much the two shorter sticks overshoot the third" — and this must be positive for a real triangle (Triangle Inequality).

Step 6 — Multiply everything back and cancel

WHAT: we drop the two halves back into $Area^{2} = \frac{1}{4} a^{2} b^{2} (1 - cos C) (1 + cos C)$ .

WHY: now the bookkeeping pays off. The $\frac{1}{4} a^{2} b^{2}$ on the left has an $a^{2} b^{2}$ ; the two denominators together give $2 ab \times 2 ab = 4 a^{2} b^{2}$ . Top and bottom cancel perfectly:

$Area^{2} = \frac{a ^{2} b ^{2}}{4} \cdot \frac{1}{4 a ^{2} b ^{2}} \cdot (a + b + c) (a + b - c) (b + c - a) (c + a - b)$

$Area^{2} = \frac{( a + b + c ) ( a + b - c ) ( b + c - a ) ( c + a - b )}{16}$

PICTURE: the figure shows the $a^{2} b^{2}$ "boxes" annihilating, leaving four side-factors over $16$ . No angle anywhere — the chase is complete.

Step 7 — Fold the four factors into the semi-perimeter

WHAT: rewrite each of the four factors using $s$ .

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

WHY: each factor is twice a clean $(s - side)$ term. There are four of them, so four factors of $2$ appear — that's $2^{4} = 16$ , which exactly cancels the 16 downstairs.

$Area^{2} = \frac{( 2 s ) 2 ( s - a ) 2 ( s - b ) 2 ( s - c )}{16} = \frac{16 s ( s - a ) ( s - b ) ( s - c )}{16} = s (s - a) (s - b) (s - c)$

PICTURE: the figure shows a segment of length $a + b + c$ ; folding it in half marks $s$ . Then $s - a$ , $s - b$ , $s - c$ are the little overshoots — each visibly positive for a genuine triangle.

Step 8 — The degenerate cases (never skip these)

Three cases to see (figure below):

Genuine triangle ( $a + b > c$ and the two other inequalities): all of $s - a, s - b, s - c > 0$ → positive product → real area. ✅
Flat / degenerate ( $a + b = c$ exactly): then $s - c = 0$ , so the product is $0$ → Area $= 0$ . The triangle has collapsed into a straight line — correct!
Impossible ( $a + b < c$ ): with $a = 1, b = 1, c = 5$ we get $s = \frac{7}{2}$ , and $s - c = \frac{7}{2} - 5 = - \frac{3}{2} < 0$ (the single lone negative factor). The whole product $s (s - a) (s - b) (s - c)$ then comes out negative → the root is imaginary → no real triangle. Heron warns you.

WHY it matters: Heron's formula is honest. It returns the right area, returns zero for a squashed triangle, and refuses (goes imaginary) for impossible sticks. Always check the Triangle Inequality first.

The one-picture summary

This final figure compresses all eight steps into a single flow: start with $\frac{1}{2} ab sin C$ (angle present) → square → identity → Law of Cosines → double difference-of-squares → cancel → semi-perimeter → Heron. The angle $C$ enters bright and leaves the picture entirely.

Recall Feynman: retell the whole walkthrough to a friend

We had a triangle made of three sticks $a$ , $b$ , $c$ . We knew one honest area formula, but it had an angle $C$ hiding inside a $sin$ . That angle was the troublemaker — we wanted an area recipe using only the stick lengths. So: first we squared the formula, which let us swap $sin^{2}$ for $1 - cos^{2}$ (that's just Pythagoras on a circle). Then the Law of Cosines handed us $cos C$ as a pure fraction of the sides — the angle was gone. Then we factored $1 - cos^{2}$ into $(1 - cos) (1 + cos)$ , and each of those turned out to be its own difference of squares in the sides, cracking into four neat brackets: $(a + b + c)$ , $(a + b - c)$ , $(b + c - a)$ , $(c + a - b)$ . The messy $a^{2} b^{2}$ 's cancelled, leaving those four brackets over $16$ . Finally we noticed each bracket is twice an $(s - side)$ term, the four $2$ 's make $16$ , the $16$ 's cancel, and out pops $Area = s (s - a) (s - b) (s - c)$ . And it's honest: a squashed triangle gives area zero, an impossible triangle gives an imaginary answer. No protractor was ever needed.

Recall Rebuild the picture-chain from memory

$\frac{1}{2} ab sin C$ → square → $\frac{1}{4} a^{2} b^{2} (1 - cos^{2} C)$ → Law of Cosines for $cos C$ → factor into two halves → each half is a difference of squares in sides → multiply, cancel $a^{2} b^{2}$ → $\frac{( a + b + c ) ( a + b - c ) ( b + c - a ) ( c + a - b )}{16}$ → each factor $= 2 (s - \cdot)$ → the four $2$ 's give $16$ , cancels → $s (s - a) (s - b) (s - c)$ .

Connections

Area of a triangle = ½ab sin C — Step 1, the launch pad.
Law of Cosines — Step 3, the angle-remover.
Difference of Squares Factoring — Steps 4–5, the twice-used trick.
Triangle Inequality — Step 8, why the root stays real.
Semi-perimeter and Incircle (r = Area/s) — Step 7, where $s$ is born.
Heronian Triangles — when all this gives whole-number areas.
Parent: Heron's formula — the algebra-first version.

Visual walkthrough — Heron's formula (derivation using trig)