Heron's formula (derivation using trig): Worked examples

This page drills Heron's formula until no case can surprise you. We deliberately hunt the edges: right triangles, obtuse ones, acute scalene, integer-perfect "Heronian" cases, degenerate (flat) triangles, forbidden side-lengths, scaling, and a real word problem.

The scenario matrix

Cell	What it tests	Under-root behaviour	Example
A. Right triangle	cross-check vs $\frac{1}{2} \times base \times height$	all $> 0$ , perfect square	Ex 1 (3-4-5)
B. Obtuse triangle	$cos C < 0$ , does Heron still work?	all $> 0$	Ex 2 (2-3-4)
C. Acute scalene	generic, all angles $< 9 0^{\circ}$ , irrational area	all $> 0$ , non-square inside	Ex 3 (4-5-6)
D. Symmetric / equilateral	general side $a$ , algebra not numbers	all $> 0$	Ex 4
E. Heronian (integer area)	messy sides, clean answer	perfect square inside	Ex 5 (5-12-13, 9-10-17)
F. Degenerate (collinear)	$a + b = c$ exactly	one factor $= 0$ → area $0$	Ex 6
G. Forbidden sides	$a + b < c$	one factor $< 0$ → no triangle	Ex 7
H. Scaling / limiting	multiply all sides by $k$	area scales by $k^{2}$	Ex 8
I. Word problem (real world)	pick sides from context, units	all $> 0$	Ex 9 (triangular field)
J. Exam twist (reverse)	given area, find missing side	solve under root	Ex 10

We now hit every cell.

Figure — Heron's formula (derivation using trig)

How to read this figure (three panels, left → right). Each panel shows the same base drawn in gray along the horizontal base axis, with the two slanted sides in orange and the apex (top vertex) marked as a coloured dot; the vertical axis is the apex's height above the base. The two orange sides are the sticks whose lengths feed the factors $(s - a)$ and $(s - b)$ .

Left panel (green apex, high up): a genuine triangle. All four under-root factors are positive, so the area (the blue shaded region) is a real positive number.

Middle panel (orange apex, on the base line): the apex has slid all the way down until it sits on the base — the shape is flat. Exactly one factor, $s - c$ , has reached $0$ , so the area is $0$ .

Right panel (red apex, red dashed gap): the two orange sticks are now too short to meet — there is a visible gap where the apex should be. One factor has gone negative, and no triangle exists.

So the single left-to-right motion of one dot — high, then on the line, then failing to meet — is the entire scenario matrix in one animation frame set.

Cell A — the right triangle (the cross-check)

Worked example Example 1 — the 3-4-5 triangle

Sides $a = 3, b = 4, c = 5$ . Find the area with Heron, then confirm with base×height.

Forecast: it's a right triangle (legs 3 and 4), so before doing any Heron algebra, guess the area from $\frac{1}{2} \times 3 \times 4$ . Hold that number.

$s = \frac{3 + 4 + 5}{2} = 6$ . Why this step? Heron needs the semi-perimeter first — every later factor $(s - a), (s - b), (s - c)$ is measured from this single reference length, so we cannot start anywhere else.
$s - a = 3, s - b = 2, s - c = 1$ . Why this step? These are the "slack" each side leaves against the semi-perimeter. All three positive is the algebraic signature of a genuine triangle (Cell A confirmed).
$Area = 6 \cdot 3 \cdot 2 \cdot 1 = 36 = 6$ . Why this step? Multiply the four factors, then undo the squaring that the derivation built in by taking the root.

Verify: the $5$ is the hypotenuse ( $3^{2} + 4^{2} = 25 = 5^{2}$ ), so the legs $3, 4$ are perpendicular and $Area = \frac{1}{2} (3) (4) = 6$ . ✅ Matches. See Law of Cosines: here $cos C = \frac{9 + 16 - 25}{24} = 0$ , so $C = 9 0^{\circ}$ — Heron doesn't need to know that, but it's consistent.

Cell B — an obtuse triangle

Worked example Example 2 — the 2-3-4 triangle (obtuse)

Sides $a = 2, b = 3, c = 4$ . Does Heron care that the biggest angle is obtuse?

Forecast: $c = 4$ is longer than what a right triangle would allow ( $2^{2} + 3^{2} = 13 \approx 3.6$ ), so the angle opposite $c$ is obtuse and $cos C < 0$ . Guess: does the formula flinch? (It won't.)

$s = \frac{2 + 3 + 4}{2} = \frac{9}{2} = 4.5$ . Why this step? Build $s$ as always; note it need not be an integer — Heron works over the reals.
$s - a = 2.5, s - b = 1.5, s - c = 0.5$ . Why this step? The smallest slack $s - c = 0.5$ is tiny because $c = 4$ is the "greedy" long side — but it is still positive, so the triangle is real. Obtuseness lives in the angle, not in the sign of any factor.
$Area = 4.5 \cdot 2.5 \cdot 1.5 \cdot 0.5 = 8.4375 \approx 2.9047$ . Why this step? Plug in and root; the result is irrational, which is perfectly normal.

Verify: with Law of Cosines, $cos C = \frac{4 + 9 - 16}{2 \cdot 2 \cdot 3} = \frac{- 3}{12} = - 0.25 < 0$ — confirmed obtuse. Then $sin C = 1 - 0.0625 = 0.9375$ and $\frac{1}{2} ab sin C = \frac{1}{2} (2) (3) 0.9375 = 3 0.9375 \approx 2.9047$ . ✅ Identical. Heron never asked about the sign of $cos C$ ; the derivation already swallowed it.

Cell C — a generic acute scalene triangle

Worked example Example 3 — the 4-5-6 triangle (all angles acute, all sides different)

Sides $a = 4, b = 5, c = 6$ . Find the area, then confirm every angle is genuinely acute.

Forecast: no side is "special," no right angle, no symmetry — this is the everyday triangle. Predict whether the area comes out as a clean number or an irrational one.

$s = \frac{4 + 5 + 6}{2} = \frac{15}{2} = 7.5$ . Why this step? Build $s$ ; here it is a half-integer, reminding us Heron isn't limited to integer inputs.
$s - a = 3.5, s - b = 2.5, s - c = 1.5$ . Why this step? All three slacks are comfortably positive, and none is tiny — the hallmark of a "fat," well-proportioned triangle rather than a thin one.
$Area = 7.5 \cdot 3.5 \cdot 2.5 \cdot 1.5 = 98.4375 = \frac{15 7}{4} \approx 9.9216$ . Why this step? The product under the root is $\frac{15}{2} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} = \frac{1575}{16}$ , and $1575 = 157$ , so the exact area is $\frac{15 7}{4}$ — irrational, exactly as a generic triangle should be.

Verify (all angles acute): by the Law of Cosines, an angle is acute exactly when the square of its opposite side is less than the sum of the other two squares. Largest side is $c = 6$ : $6^{2} = 36$ vs $4^{2} + 5^{2} = 41$ , and $36 < 41$ , so even the biggest angle is acute — hence all angles are acute. Cross-check the area with $\frac{1}{2} ab sin C$ : $cos C = \frac{16 + 25 - 36}{2 \cdot 4 \cdot 5} = \frac{5}{40} = 0.125$ , $sin C = 1 - 0.12 5^{2} = 0.984375$ , and $\frac{1}{2} (4) (5) 0.984375 = 10 0.984375 \approx 9.9216$ . ✅ Matches Heron.

Cell D — the symmetric / equilateral case (algebra, not numbers)

Worked example Example 4 — equilateral of side

$a$ , and isosceles $a, a, b$ Derive the equilateral area, then generalise to any isosceles triangle with equal sides $a$ and base $b$ .

Forecast: you probably remember $\frac{3}{4} a^{2}$ for equilateral. Can Heron reproduce it and give the isosceles version in one shot?

Equilateral: sides $a, a, a$ , so $s = \frac{3 a}{2}$ and $s - a = \frac{a}{2}$ (three times). Why this step? Symmetry means all three $(s - \cdot)$ factors are equal, so the four-factor product collapses to $s \cdot (s - a)^{3}$ — a huge simplification we get for free from the symmetry.
$Area = \frac{3 a}{2} (\frac{a}{2})^{3} = \frac{3 a ^{4}}{16} = \frac{3}{4} a^{2}$ . Why this step? Collect powers of $a$ under the root; $a^{4} = a^{2}$ pulls the length-squared out, leaving the pure constant $\frac{3}{4}$ .
Isosceles $a, a, b$ : $s = \frac{2 a + b}{2} = a + \frac{b}{2}$ . Then $s - a = \frac{b}{2}$ (twice) and $s - b = a - \frac{b}{2}$ . Why this step? Two equal sides force two equal factors, so again the product partly collapses — this is why symmetric cases are algebraically easy.
$Area = (a + \frac{b}{2}) (\frac{b}{2})^{2} (a - \frac{b}{2}) = \frac{b}{2} a^{2} - \frac{b ^{2}}{4} = \frac{b}{4} 4 a^{2} - b^{2}$ . Why this step? Pull the repeated $\frac{b}{2}$ out of the root; the remaining two factors are a difference of squares $(a + \frac{b}{2}) (a - \frac{b}{2}) = a^{2} - \frac{b ^{2}}{4}$ — the same Difference of Squares Factoring move that powered the original derivation.

Verify: set $b = a$ in step 4: $Area = \frac{a}{4} 4 a^{2} - a^{2} = \frac{a}{4} 3 a = \frac{3}{4} a^{2}$ . ✅ The isosceles formula collapses to the equilateral one. Also the classic height check: isosceles height $h = a^{2} - (b /2)^{2}$ , area $= \frac{1}{2} bh = \frac{b}{2} a^{2} - b^{2} /4$ — same thing.

Cell E — Heronian triangles (integer sides, integer area)

Worked example Example 5 — 5-12-13 and 9-10-17

Show both give whole-number areas (that's what makes them Heronian Triangles).

Forecast: $5, 12, 13$ is another Pythagorean triple — bet on a clean area. $9, 10, 17$ is not right-angled; will it still be clean?

Triangle 5-12-13:

$s = \frac{5 + 12 + 13}{2} = 15$ . Why? Build $s$ ; it lands on an integer because the perimeter $30$ is even — a good omen for a clean area.
$s - a = 10, s - b = 3, s - c = 2$ . Why? All positive → a real triangle, and all integers, so the under-root product will be an integer.
$Area = 15 \cdot 10 \cdot 3 \cdot 2 = 900 = 30$ . Why? $900 = 3 0^{2}$ is a perfect square, which is the precise condition for an integer area.

Triangle 9-10-17:

$s = \frac{9 + 10 + 17}{2} = 18$ . Why? Build $s$ ; perimeter $36$ is even so $s$ is an integer.
$s - a = 9, s - b = 8, s - c = 1$ . Why? All positive, so valid — but $s - c = 1$ is small because $c = 17$ is nearly $9 + 10 = 19$ , i.e. a "thin" triangle sitting close to degenerate.
$Area = 18 \cdot 9 \cdot 8 \cdot 1 = 1296 = 36$ . Why? $1296 = 3 6^{2}$ → integer area, even without a right angle.

Verify: 5-12-13: $5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2}$ , so it's right-angled, $\frac{1}{2} (5) (12) = 30$ . ✅ 9-10-17: not a triple, but $18 \cdot 9 = 162$ , $8 \cdot 1 = 8$ , $162 \cdot 8 = 1296 = 3 6^{2}$ . ✅ Both integer — genuinely Heronian.

Cell F — the degenerate (flat) triangle

Worked example Example 6 — sides 3, 4, 7 (a squashed triangle)

Compute the "area" of a triangle with $a = 3, b = 4, c = 7$ .

Forecast: notice $3 + 4 = 7 = c$ exactly. The two short sticks laid end-to-end reach the far endpoint — the triangle is flat as paper. Guess the area before computing.

$s = \frac{3 + 4 + 7}{2} = 7$ . Why? Build $s$ ; note $s$ equals $c$ here, which is the tell-tale of the boundary case.
$s - a = 4, s - b = 3, s - c = 0$ . Why this step? The factor $s - c = 0$ is the alarm bell — it happens precisely when $c$ equals the sum of the other two, the equality edge of the triangle inequality.
$Area = 7 \cdot 4 \cdot 3 \cdot 0 = 0 = 0$ . Why this step? Anything times $0$ is $0$ ; the root of $0$ is $0$ , so Heron reports the flat shape's true area cleanly.

Verify: geometrically the three points are collinear, enclosing no space — area must be $0$ , and Heron returns exactly $0$ . This is the boundary case set by the Triangle Inequality ( $a + b \geq c$ with equality). ✅ Look at the middle panel of the figure: the apex has landed on the base.

Cell G — forbidden sides (no triangle exists)

Worked example Example 7 — sides 1, 1, 5

Try Heron on $a = 1, b = 1, c = 5$ .

Forecast: two 1-length sticks can never span a gap of 5. Predict what the formula does when handed impossible data.

$s = \frac{1 + 1 + 5}{2} = \frac{7}{2} = 3.5$ . Why? Build $s$ mechanically — the formula doesn't yet "know" anything is wrong.
$s - c = 3.5 - 5 = - 1.5 < 0$ . Why this step? One factor is negative, which is the exact algebraic fingerprint of the Triangle Inequality $a + b > c$ being violated ( $2 < 5$ ).
$Area = 3.5 \cdot 2.5 \cdot 2.5 \cdot (- 1.5) = - 32.8125$ → not a real number. Why this step? A negative under the square root signals "no real triangle exists," not "small triangle." The formula is honestly refusing.

Verify: the product $s (s - a) (s - b) (s - c) = 3.5 \cdot 2.5 \cdot 2.5 \cdot (- 1.5) = - 32.8125 < 0$ . ✅ Negative ⇒ Heron correctly reports impossibility. Always test $a + b > c$ before trusting a numerical output.

Cell H — scaling / limiting behaviour

Worked example Example 8 — double every side of the 3-4-5

Take the 3-4-5 triangle and scale all sides by $k = 2$ to get 6-8-10. Predict the new area.

Forecast: lengths double, but area is a two-dimensional quantity — guess whether it doubles or quadruples.

New sides $6, 8, 10$ , so $s = \frac{6 + 8 + 10}{2} = 12$ . Why this step? Because $s$ is a sum of lengths, scaling every side by $k$ scales $s$ by $k$ too — here $s$ went from $6$ to $12$ , exactly $\times 2$ . This linearity is the seed of the whole scaling law.
$s - a = 6, s - b = 4, s - c = 2$ . Why this step? Each slack is also exactly $k = 2$ times the original ( $3, 2, 1$ ), because subtraction of two scaled lengths scales by $k$ as well — so all four under-root factors scale by $k$ together.
$Area = 12 \cdot 6 \cdot 4 \cdot 2 = 576 = 24$ . Why this step? Four factors each $\times k$ multiply the product by $k^{4}$ , so the root grows by $k^{2}$ ; taking $576 = 24$ confirms it.

Verify: original area was $6$ ; since the root scales by $k^{2} = 4$ , we expect $6 \times 4 = 24$ . ✅ Area scales as the square of the length factor — a universal fact, visible directly in Heron's four-factor product.

Cell I — a real-world word problem

Worked example Example 9 — a triangular field

A farmer's field has straight edges of $13 m$ , $14 m$ , and $15 m$ . Seed costs $ $2$ per square metre. What's the seeding bill?

Forecast: the sides $13, 14, 15$ are a famous Heronian set — you may recall the area is $84$ . Predict the cost.

$s = \frac{13 + 14 + 15}{2} = 21 m$ . Why? Semi-perimeter, in metres; keeping units on $s$ means the final area lands automatically in $m^{2}$ .
$s - a = 8, s - b = 7, s - c = 6$ (all in metres). Why? All positive → a valid, buildable field.
$Area = 21 \cdot 8 \cdot 7 \cdot 6 = 7056 = 84 m^{2}$ . Why? $21 \cdot 8 = 168$ , $7 \cdot 6 = 42$ , $168 \cdot 42 = 7056 = 8 4^{2}$ — a perfect square, so a clean integer area.
Cost $=84\text{ m}^2\times\$ 2/\text{m}^2=$168 $. * W h y t hi ss t e p ? * M u l t i pl y a r e ab y u ni tp r i ce; t h e$ \text{m}^2 $in t h e a r e a c an ce l s t h e$ \text{m}^2$ in the denominator of the price, leaving pure dollars.

Verify: units — $\text{m}^2\cdot(\$ /\text{m}^2)=$ $.✅ N u m er i c —$ 84^2=7056 $. ✅ The bill is **\$ 168**.

Cell J — the exam twist (reverse-engineer a side)

Worked example Example 10 — given the area, find the third side

An isosceles triangle has two sides of $10$ and area $48$ . Find the base $b$ (there are two answers!).

Forecast: from Example 4, isosceles area $= \frac{b}{4} 4 a^{2} - b^{2}$ . Setting this to $48$ with $a = 10$ is a hidden quadratic in $b^{2}$ — expect possibly two valid bases.

$Area = \frac{b}{4} 4 (10)^{2} - b^{2} = \frac{b}{4} 400 - b^{2} = 48$ . Why this step? Use the isosceles Heron result so only one unknown $b$ remains — turning a triangle problem into an algebra problem.
Square both sides: $\frac{b ^{2}}{16} (400 - b^{2}) = 4 8^{2} = 2304$ , i.e. $b^{2} (400 - b^{2}) = 36864$ . Why this step? Squaring removes the root; introduce $u = b^{2}$ so the equation becomes a plain quadratic.
$u (400 - u) = 36864 \Rightarrow u^{2} - 400 u + 36864 = 0$ . Solve with the quadratic formula: $u = \frac{400 \pm 40 0 ^{2} - 4 \cdot 36864}{2} = \frac{400 \pm 160000 - 147456}{2} = \frac{400 \pm 12544}{2}$ . Why this step? Standard quadratic; the whole difficulty is now hidden inside one square root.
Find $12544$ by factoring, not guessing: $12544 = 2^{7} \cdot 98 = \dots$ more cleanly, $12544 = 64 \cdot 196$ , and both are perfect squares, so $12544 = 64 \cdot 196 = 8 \cdot 14 = 112$ . Why this step? Splitting a big radicand into a product of known squares ( $64 = 8^{2}$ , $196 = 1 4^{2}$ ) is the reliable way to extract a perfect square by hand.
$u = \frac{400 \pm 112}{2} = 256$ or $144$ , so $b = 256 = 16$ or $b = 144 = 12$ . Why this step? $b = u$ ; both are positive and satisfy the Triangle Inequality ( $b < 2 a = 20$ ), so both are legitimate.

Verify: for $b = 12$ : $\frac{12}{4} 400 - 144 = 3256 = 3 \cdot 16 = 48$ . ✅ For $b = 16$ : $\frac{16}{4} 400 - 256 = 4144 = 4 \cdot 12 = 48$ . ✅ Two triangles (a "tall thin" one and a "short wide" one) share the same two legs and the same area — exactly the twist an exam loves.

Active recall

Recall Which single quantity decides every scenario?

The sign of the product $s (s - a) (s - b) (s - c)$ . Positive ::: real triangle, real area. Zero ::: degenerate (collinear, area $0$ ). Negative ::: no triangle exists (triangle inequality violated).

Recall How does area respond to scaling all sides by

$k$ ? It multiplies by $k^{2}$ (area is 2-dimensional; the four under-root factors each scale by $k$ , so the root scales by $k^{2}$ ).

Recall Why can one area value give two different triangles (Ex 10)?

Because Heron squares to a quadratic in $b^{2}$ ; two positive roots can both satisfy the triangle inequality — a "tall" and a "wide" solution.

Recall How do you know a triangle is acute from its sides alone (Ex 3)?

Square the longest side and compare to the sum of the other two squares. Less than ::: all angles acute. Equal ::: right angle. Greater than ::: obtuse.

Connections

Heron's formula (derivation using trig) — the parent formula being drilled.
Law of Cosines — checks acute (Ex 3), obtuse (Ex 2) and right-angle (Ex 1) cases.
Area of a triangle = ½ab sin C — the sanity cross-check in Ex 2 and Ex 3.
Triangle Inequality — governs Cells F and G (zero / forbidden).
Semi-perimeter and Incircle (r = Area/s) — every example computes $s$ first.
Difference of Squares Factoring — the isosceles collapse in Ex 4.
Heronian Triangles — Ex 5 and Ex 9 land on integer areas.