Heron's formula (derivation using trig): Question bank

This deep dive is the reasoning companion to Heron's formula (derivation using trig). Nothing here is arithmetic-heavy — it is about why the machinery works and when it breaks.

True or false — justify

TF1. Heron's formula needs at least one angle of the triangle to be known.

False — it uses only the three side lengths

a, b, c

; the entire point of the derivation is to chase the angle

C

out via Law of Cosines.

TF2. Because area is always positive, the product

s (s - a) (s - b) (s - c)

is positive for any three positive numbers.

False — it is positive only when

a, b, c

actually form a triangle; if the Triangle Inequality fails, some

(s - \cdot)

factor goes negative and the product can be negative.

TF3. If you replace the semi-perimeter

s

by the full perimeter

P = a + b + c

, you just need to divide the answer by some constant.

False — there is no single constant fix; the four factors each carry their own factor of

2

, so the perimeter version has no clean closed form. You must use

s = \frac{a + b + c}{2}

TF4. Heron's formula and

Area = \frac{1}{2} ab sin C

describe two different quantities.

False — they are the same area; Heron's is literally derived from

\frac{1}{2} ab sin C

after eliminating the angle. See Area of a triangle = ½ab sin C.

TF5. Scaling every side by a factor

k

multiplies the Heron area by

k

False — area scales by

k^{2}

. Each of

s, s - a, s - b, s - c

scales by

k

, so the product under the root scales by

k^{4}

, and

k^{4} = k^{2}

TF6. If the three "sides" satisfy

a + b = c

exactly, Heron's formula still gives a positive area.

False — then

s - c = 0

, the product is

0

, so the area is exactly

0

: the "triangle" is a flat, degenerate line segment.

TF7. Heron's formula only works for right triangles.

False — it works for every triangle (acute, right, obtuse). The right-triangle example is just a convenient check because

\frac{1}{2} (legs)

is easy to verify.

TF8. Two triangles with the same three side lengths must have the same Heron area.

True — three sides fix a triangle up to congruence (SSS), so the area is completely determined; that determinism is the whole reason a side-only formula can exist.

Spot the error

SE1. "Since

sin^{2} C = 1 - cos^{2} C

, we get

sin C = 1 - cos C

, so Area

= \frac{1}{2} ab (1 - cos C)

Error:

sin C \neq = 1 - cos C

. You may only take

sin C = 1 - cos^{2} C

, and the derivation squares the area precisely to avoid this square root.

SE2. "We can drop the triangle inequality check because the square root will just warn us."

Error: for impossible sides the quantity under the root is negative, giving a non-real number — not a warning but a nonsense answer. Check

a + b > c

(and the other two) first.

SE3. "

1 - cos^{2} C = (1 - cos C) (1 + cos C)

, and

1 + cos C = \frac{2 ab + a ^{2} + b ^{2} - c ^{2}}{2 ab}

, which does not factor further."

Error: the numerator is a difference of squares:

2 ab + a^{2} + b^{2} - c^{2} = (a + b)^{2} - c^{2} = (a + b + c) (a + b - c)

. Missing that factorisation is exactly what stalls the derivation.

SE4. "

cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}

, so for an obtuse angle

C

the formula fails because

cos C

is negative."

Error: nothing fails. A negative

cos C

(obtuse

C

) is perfectly allowed;

sin^{2} C = 1 - cos^{2} C

is still valid, and the algebra carries the sign correctly.

SE5. "In Step 5 the

a^{2} b^{2}

on top cancels only one of the

2 ab

denominators, leaving a stray

2 ab

Error: there are two factors of

2 ab

downstairs and

a^{2} b^{2} = (ab) (ab)

upstairs, so

\frac{a ^{2} b ^{2}}{( 2 ab ) ( 2 ab )} = \frac{1}{4}

— both cancel cleanly, leaving the

16

in the denominator.

SE6. "Heron gives Area

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

, so Area

= \frac{( a + b + c ) ( a + b - c ) ( b + c - a ) ( c + a - b )}{4}

Error: you dropped the square root. It is

that /16

, i.e.

\frac{1}{4} (a + b + c) (a + b - c) (b + c - a) (c + a - b)

, not the product over

4

SE7. "For the equilateral triangle,

s - a = s

, since all sides are equal."

Error:

s = \frac{3 a}{2}

but

s - a = \frac{3 a}{2} - a = \frac{a}{2}

. Subtracting one side still removes an

a

; equality of sides does not make

s - a = s

Why questions

WHY1. Why does area being a function of the sides alone even make sense, before we prove any formula?

Because SSS fixes a triangle up to congruence — the three sticks can't flex — so every measurable quantity, area included, is already determined by

a, b, c

WHY2. Why do we square the area during the derivation instead of working with the area directly?

Squaring turns

sin C

into

sin^{2} C

, which the identity

sin^{2} C = 1 - cos^{2} C

converts to pure

cos C

; without squaring we'd be stuck with a square-rooted sine.

WHY3. Why is the Law of Cosines the right tool to eliminate the angle, rather than the Law of Sines?

Law of Cosines gives

cos C

directly in terms of all three sides

a, b, c

; the Law of Sines would reintroduce another angle or the circumradius, not remove angles.

WHY4. Why does the factor

(a + b - c)

signal the triangle inequality?

The inequality says

a + b > c

, i.e.

a + b - c > 0

; that same quantity equals

2 (s - c)

under the root, so a valid triangle is exactly what keeps the root's factors positive.

WHY5. Why does the denominator become

16

and not, say,

4

Each of the four side factors is twice an

(s - \cdot)

term, contributing

2^{4} = 16

when we switch to

s

; that

16

is precisely what the semi-perimeter absorbs to leave the clean product

s (s - a) (s - b) (s - c)

WHY6. Why doesn't Heron's formula need to know which vertex is the angle

C

The final expression is fully symmetric in

a, b, c

; the choice of included angle in Step 1 was arbitrary and washes out, so the answer can't depend on it.

WHY7. Why does

s

reappear when we later study the incircle radius

r

Because Area

= r s

, so

r = Area / s

— the same semi-perimeter that builds the area also scales the incircle. See Semi-perimeter and Incircle (r = Area/s).

WHY8. Why can integer sides give an integer area (a Heronian triangle) even though a square root is involved?

When

s (s - a) (s - b) (s - c)

happens to be a perfect square, the root is an integer; such lucky side-sets are exactly the Heronian Triangles.

Edge cases

EC1. What does Heron's formula return for

a = 1, b = 1, c = 2

s = 2

, so

s - c = 0

and Area

= 0

— the degenerate "triangle" is a flat segment where the two unit sticks lie along the base.

EC2. What happens for

a = 1, b = 1, c = 5

s = \frac{7}{2}

, giving

s - c = \frac{7}{2} - 5 = - \frac{3}{2} < 0

; the product under the root goes negative, correctly signalling no real triangle exists (fails

a + b > c

EC3. As one vertex angle shrinks toward

0°

(the triangle gets very thin), what does the Heron area approach?

It approaches

0

: the triangle flattens, one

(s - \cdot)

factor tends to

0

, and the area smoothly vanishes rather than jumping.

EC4. Does Heron's formula distinguish between an acute and an obtuse triangle with the same three sides?

No — but there's nothing to distinguish: three fixed sides give exactly one triangle shape, whose angles (acute or obtuse) are already determined, and its single area is what Heron returns.

EC5. If all three sides are equal to

a

, why is it valid that

s - a = s - b = s - c

With

a = b = c

, subtracting any one side removes the same length, so all three difference terms coincide at

\frac{a}{2}

; symmetry of equal sides forces symmetric factors.

EC6. For a valid triangle can

s - a

ever exceed

s

itself?

No —

s - a < s

always, since

a > 0

; and it also stays positive by the triangle inequality, so every

(s - \cdot)

sits strictly between

0

and

s

Active recall

Recall One-line litmus test for "is this a real triangle?"

Check that the largest side is less than the sum of the other two; equivalently, that all of $s - a, s - b, s - c$ are strictly positive. Equal-to means degenerate (area $0$ ); greater-than means impossible.

Recall The three traps that catch most people

(1) Using perimeter instead of semi-perimeter. (2) Writing $sin C = 1 - cos C$ instead of $1 - cos^{2} C$ . (3) Forgetting to check the triangle inequality and trusting a negative-under-root answer.

Connections

Heron's formula (derivation using trig) — the parent derivation these traps probe.
Law of Cosines — why the angle can be eliminated at all.
Area of a triangle = ½ab sin C — the starting formula behind TF4.
Triangle Inequality — the gate for every edge case here.
Difference of Squares Factoring — the move SE3 tests.
Semi-perimeter and Incircle (r = Area/s) — where $s$ returns (WHY7).
Heronian Triangles — integer-area edge case (WHY8).

Question bank — Heron's formula (derivation using trig)