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Precalculus

Heron's formula

You know the three sides of a triangle, but no angle and no height. Can the area come straight from the three lengths?

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Law of cosinesPrecalculus
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Applications of oblique trianglesPrecalculus
Where you are 6 in Laws of sines and cosines
Trigonometry7
Laws of sines and cosines6
Law of sinesArea from two sides and the included angleThe ambiguous caseLaw of cosines
Heron's formula
Applications of oblique triangles

When you reach for it

Three lengths are often the easiest thing to measure. A tape stretched along each edge of a triangular field gives its sides; the angles at the corners need an instrument, and the height needs a line dropped at a right angle to a side, which is awkward when the field is in the way. So the question comes up in practice: with all three sides known (SSS) and nothing else, what is the area K?

There is already a route. The law of cosines turns three sides into an angle, and the area from two sides and the included angle, K = \tfrac{1}{2}ab\sin C, turns that angle into the area. Heron’s formula does both steps once, in general, so that the angle never has to be found. It is named for Heron of Alexandria, a Greek mathematician and engineer usually dated to the first century.

As in those lessons, the triangle is ABC, and each side is named by the lowercase letter of the angle across from it.

The procedure

The semiperimeter

The Perimeter of the triangle is a + b + c. Half of it is the semiperimeter,

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

The formula uses s together with the three differences s - a, s - b and s - c. Each difference has a meaning. Since 2s = a + b + c,

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

which is half the amount by which the two other sides together exceed a. The Triangle inequality says b + c > a in every triangle, so s - a > 0, and in the same way s - b > 0 and s - c > 0.

Heron’s formula

For a triangle with sides a, b and c and semiperimeter s,

K = \sqrt{s(s - a)(s - b)(s - c)}.

Multiplying each of the four factors by 2 gives an equivalent form with no fractions, often easier when a + b + c is odd:

\begin{aligned} 16K^2 &= (a + b + c)(b + c - a) \\ &\quad\cdot (a + c - b)(a + b - c). \end{aligned}

Why it is true

C B A a b c
Figure 1The proof starts from the angle C between sides a and b, finds its cosine from the three sides, and then lets the angle disappear from the answer.

Start from the area with the angle C between sides a and b:

K = \frac{1}{2}ab\sin C,

so 16K^2 = 4a^2b^2\sin^2 C.

The Pythagorean identity gives \sin^2 C = 1 - \cos^2 C. (Because 0^\circ < C < 180^\circ, \sin C is positive, so nothing is lost by working with squares: K is the positive square root at the end.) Then

16K^2 = (2ab)^2 - (2ab\cos C)^2.

The law of cosines, c^2 = a^2 + b^2 - 2ab\cos C, solved for the product, says 2ab\cos C = a^2 + b^2 - c^2. Substituting removes the angle completely:

16K^2 = (2ab)^2 - (a^2 + b^2 - c^2)^2.

This is a difference of two squares, X^2 - Y^2 = (X - Y)(X + Y), as in Factoring special patterns. Factor it, and regroup each factor:

\begin{aligned} 16K^2 &= (2ab - a^2 - b^2 + c^2)(2ab + a^2 + b^2 - c^2) \\ &= \big(c^2 - (a - b)^2\big)\big((a + b)^2 - c^2\big). \end{aligned}

Each factor is again a difference of squares. Factoring both,

\begin{aligned} 16K^2 &= (c - a + b)(c + a - b) \\ &\quad\cdot (a + b - c)(a + b + c). \end{aligned}

Finally, a + b + c = 2s, b + c - a = 2(s - a), a + c - b = 2(s - b) and a + b - c = 2(s - c), so the right side is 16\,s(s - a)(s - b)(s - c). Dividing by 16 and taking the positive square root gives Heron’s formula. Nothing in the argument depended on C being acute: the law of cosines and \tfrac{1}{2}ab\sin C hold for every angle of a triangle, so the formula does too.

Choosing the route

Heron’s formula needs only the sides and no inverse trig function. When the sides are whole numbers and s is too, the product under the root is a whole number and the area is exact, either a whole number or a radical to simplify. When an angle is also wanted, the law of cosines followed by \tfrac{1}{2}ab\sin C gives both; the two routes always agree, since one is the other done in general.

Keep full values and round only at the end; decimal areas here are given to two decimal places.

Worked examples

Example 1

A whole-number area

Find the area of the triangle with sides 10, 17 and 21, and the height to the longest side.

Solution

The semiperimeter is s = \dfrac{10 + 17 + 21}{2} = 24, and the differences are 24 - 10 = 14, 24 - 17 = 7 and 24 - 21 = 3. All three are positive, so the sides make a triangle. Then

K = \sqrt{24 \cdot 14 \cdot 7 \cdot 3} = \sqrt{7056} = 84.

The area formula K = \tfrac{1}{2}(\text{base})(\text{height}) with base 21 gives the height: h = \dfrac{2(84)}{21} = 8.

Check: the height splits the side of 21 into two pieces, which by the The Pythagorean theorem are \displaystyle \sqrt{17^2 - 8^2} = 15 and \displaystyle \sqrt{10^2 - 8^2} = 6, and 15 + 6 = 21.

Example 2

An exact radical

Find the area of the triangle with sides 6, 9 and 11, exactly and to two decimal places.

Solution

Here s = \dfrac{26}{2} = 13, and the differences are 7, 4 and 2. The product is 13 \cdot 7 \cdot 4 \cdot 2 = 728. Since 728 = 4 \cdot 182 and 182 = 2 \cdot 7 \cdot 13 has no square factor,

K = \sqrt{728} = 2\sqrt{182} \approx 26.98.
Example 3

A field measured with a tape

Temitope walks the edges of a triangular field on her family’s homestead and measures them as 48.6 meters, 71.2 meters and 65.0 meters. What is its area?

Solution

The semiperimeter is

s = \frac{48.6 + 71.2 + 65.0}{2} = 92.4,

and the differences are 43.8, 21.2 and 27.4. Then

\begin{aligned} K &= \sqrt{92.4 \cdot 43.8 \cdot 21.2 \cdot 27.4} \\ &\approx 1533.26. \end{aligned}

The field covers about 1533.26 square meters. No corner angle was measured, and none was needed.

Example 4

The same area two ways

Find the area of the triangle with sides 5, 16 and 19 by Heron’s formula, then again by finding the largest angle and using \tfrac{1}{2}ab\sin C.

Solution

Heron first: s = 20, the differences are 15, 4 and 1, and

K = \sqrt{20 \cdot 15 \cdot 4 \cdot 1} = \sqrt{1200} = 20\sqrt{3}.

Now the long way. Take a = 5, b = 16, c = 19, so C is across from the longest side. The law of cosines gives

\cos C = \frac{25 + 256 - 361}{2(5)(16)} = \frac{-80}{160} = -\frac{1}{2},

so C = 120^\circ and \displaystyle \sin C = \dfrac{\sqrt{3}}{2}. Then

K = \frac{1}{2}(5)(16)\cdot\frac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.64.

The two routes agree, as the proof says they must. Heron’s formula took one line and never met the obtuse angle; the second route took longer but also found C.

Example 5

When the sides do not close

Yetunde has three fence rails of lengths 7, 10 and 17 feet, and Adebola suggests swapping the longest for one of 18 feet. What does Heron’s formula say about each set?

Solution

For 7, 10 and 17: s = 17, and the differences are 10, 7 and 0. The product under the root is 0, so K = 0. The rails of 7 and 10 add to exactly 17; they can only lie flat along the long one, and the “triangle” is a segment.

For 7, 10 and 18: s = 17.5, and the differences are 10.5, 7.5 and -0.5. The product is

17.5 \cdot 10.5 \cdot 7.5 \cdot (-0.5) \approx -689.06,

a negative number, which has no real square root. No triangle has these sides: 7 + 10 < 18, so the two short rails cannot reach across. The law of cosines agrees, giving the largest angle a cosine of \dfrac{49 + 100 - 324}{140} = -1.25, outside [-1, 1].

Example 6

A four-sided field

A B C D 42 55 38 47 63
Figure 2Example 6. One measured diagonal turns a four-sided field into two triangles whose sides are all known, in meters.

Adebola measures a four-sided field ABCD on the homestead: AB = 42, BC = 55, CD = 38 and DA = 47 meters, and the diagonal AC = 63 meters. Find its area.

Solution

The diagonal cuts the field into triangles ABC and ACD, and each has three known sides.

Triangle ABC has sides 42, 55 and 63, so s = 80 with differences 38, 25 and 17:

\begin{aligned} K_1 &= \sqrt{80 \cdot 38 \cdot 25 \cdot 17} \\ &= \sqrt{1{,}292{,}000} \approx 1136.66. \end{aligned}

Triangle ACD has sides 47, 38 and 63, so s = 74 with differences 27, 36 and 11:

\begin{aligned} K_2 &= \sqrt{74 \cdot 27 \cdot 36 \cdot 11} \\ &= \sqrt{791{,}208} \approx 889.50. \end{aligned}

The field covers about 2026.16 square meters. The diagonal is needed: four sides alone do not fix a four-sided shape, since a frame of four hinged rods can be pushed flatter, changing its area without changing a side.

Example 7

The inscribed circle

Find the radius of the circle that fits inside the triangle with sides 9, 10 and 17, touching all three sides.

Solution

The center of that circle is the Incenter, at the same distance r from all three sides. Joining it to the three corners cuts the triangle into three triangles. Each has one side of the big triangle as its base and r as its height, so

K = \frac{1}{2}ar + \frac{1}{2}br + \frac{1}{2}cr = r\cdot\frac{a + b + c}{2} = rs.

Here s = 18 with differences 9, 8 and 1, and

K = \sqrt{18 \cdot 9 \cdot 8 \cdot 1} = \sqrt{1296} = 36.

So r = \dfrac{K}{s} = \dfrac{36}{18} = 2. The semiperimeter appears twice here: once in Heron’s formula and once as the factor in K = rs.

Boundaries

The three lengths must make a triangle. The product s(s - a)(s - b)(s - c) is positive exactly when every side is less than the sum of the other two. If one side equals the sum of the others, one factor is 0 and the formula gives 0: the points are on one line. If one side is longer than the sum of the others, one factor is negative, and the root is not a real number. At most one factor can fail, because any two of them add to a side, for example (s - a) + (s - b) = c > 0.

The square root is the positive one. The proof squares K, and an area is positive, so K is the positive root.

It gives the area, not the angles. Heron’s formula never finds an angle; for those, use the law of cosines.

Decimal answers are approximations, exact only when the product under the root is a perfect square or the radical is kept.

Where this goes next

Heron’s formula finishes the toolkit for a triangle given any three independent parts: sides and angles by the two laws, and area by \tfrac{1}{2}ab\sin C or by this formula. Applications of oblique triangles puts the same tools to work in surveying, where a plot of land is cut into triangles by measured lines and each triangle’s area is found from its sides, as in Example 6. With K = rs it gives the inscribed circle’s radius from the sides alone.

Exercises

Heron’s formula

Find each area, exactly where the root is exact or a simple radical, otherwise to two decimal places.

  1. Sides 5, 12 and 13. Check your answer another way.
  2. Sides 13, 20 and 21.
  3. Sides 4, 5 and 6.
  4. Sides 8, 8 and 8.
  5. Sides 2.5, 6.1 and 4.8.
  6. Sides \tfrac{3}{2}, 2 and \tfrac{5}{2}.

Do the sides close?

  1. Use Heron’s formula on the lengths 3, 4 and 8. What does it show?
  2. Use Heron’s formula on the lengths 6, 10 and 16. What does it show?
  3. Two sides of a triangle are 9 and 12. For which third sides x does Heron’s formula give a positive area?

Heron’s formula and the angles

  1. A triangle has sides 7, 15 and 20. Find its area by Heron’s formula, then use K = \tfrac{1}{2}ab\sin C to find the sine of the largest angle. Decide whether that angle is acute or obtuse, and find it to two decimal places.
  2. Find the area of the triangle with sides 8, 11 and 13 by Heron’s formula, and again by finding the cosine of the largest angle and using \tfrac{1}{2}ab\sin C. Give the exact area.
  3. Using your area from Exercise 2, find all three heights of the triangle with sides 13, 20 and 21.

Fields and other uses

  1. Yetunde’s triangular garden plot on the homestead has sides of 18.5 meters, 24.0 meters and 30.2 meters. Find its area.
  2. Temitope measures a four-sided plot PQRS: PQ = 30, QR = 34, RS = 20 and SP = 26 meters, with diagonal PR = 40 meters. Find its area.
  3. Adebola’s triangular hay field has sides of 120, 150 and 210 meters. Seed costs 0.12 dollars per square meter. Find the area and the cost of seeding the whole field.
  4. Find the area of the triangle with sides 6, 25 and 29, and the radius of the circle inscribed in it.
  5. Find the area of the triangle with sides 25, 25 and 14 by Heron’s formula, and check it by finding the height to the side of 14.

Reasoning

  1. Use Heron’s formula to show that an equilateral triangle with side t has area \displaystyle \dfrac{\sqrt{3}}{4}t^2.
  2. An isosceles triangle has two sides of length p and a base of length q. Use Heron’s formula to show that its area is \displaystyle \dfrac{q}{4}\sqrt{4p^2 - q^2}.
  3. Explain why at most one of s - a, s - b and s - c can be zero or negative.
  4. For a right triangle with legs a and b and hypotenuse c, use the form 16K^2 = \big((a + b)^2 - c^2\big)\big(c^2 - (a - b)^2\big) to show that Heron’s formula gives K = \tfrac{1}{2}ab.
  5. Every side of a triangle is multiplied by k > 0. Use Heron’s formula to show that the area is multiplied by k^2.
  6. Explain why a four-sided field needs a diagonal (or an angle) measured before its area is known, while a triangular field needs only its three sides.
Answers
  1. 30. s = 15, differences 10, 3, 2: \displaystyle \sqrt{15 \cdot 10 \cdot 3 \cdot 2} = \sqrt{900} = 30. Since 5^2 + 12^2 = 13^2, it is a right triangle with legs 5 and 12, and \tfrac{1}{2}(5)(12) = 30.
  2. 126. s = 27, differences 14, 7, 6: \displaystyle \sqrt{27 \cdot 14 \cdot 7 \cdot 6} = \sqrt{15{,}876} = 126.
  3. \displaystyle \dfrac{15\sqrt{7}}{4} \approx 9.92. s = \tfrac{15}{2}, differences \tfrac{7}{2}, \tfrac{5}{2}, \tfrac{3}{2}: the product is \dfrac{1575}{16}, and 1575 = 225 \cdot 7.
  4. \displaystyle 16\sqrt{3} \approx 27.71. s = 12, differences 4, 4, 4: \displaystyle \sqrt{12 \cdot 64} = \sqrt{768} = 16\sqrt{3}.
  5. About 5.66. s = 6.7, differences 4.2, 0.6, 1.9: \displaystyle \sqrt{6.7 \cdot 4.2 \cdot 0.6 \cdot 1.9} = \sqrt{32.0796}.
  6. \tfrac{3}{2}. s = 3, differences \tfrac{3}{2}, 1, \tfrac{1}{2}: \displaystyle \sqrt{3 \cdot \tfrac{3}{2} \cdot 1 \cdot \tfrac{1}{2}} = \sqrt{\tfrac{9}{4}}. (The sides are 3, 4, 5 halved, a right triangle with legs \tfrac{3}{2} and 2.)
  7. No triangle. s = 7.5, differences 4.5, 3.5, -0.5: the product 7.5 \cdot 4.5 \cdot 3.5 \cdot (-0.5) \approx -59.06 is negative, because 3 + 4 < 8.
  8. Area 0: no triangle, only a segment. s = 16, and s - 16 = 0, because 6 + 10 = 16.
  9. 3 < x < 21. Here s = \dfrac{21 + x}{2}, so s - 9 = \dfrac{3 + x}{2}, s - 12 = \dfrac{x - 3}{2} and s - x = \dfrac{21 - x}{2}. All three are positive exactly when x > 3 and x < 21.
  10. Area 42; \sin C = \tfrac{4}{5}; obtuse, C \approx 126.87^\circ. s = 21, differences 14, 6, 1: \displaystyle \sqrt{1764} = 42. With a = 7, b = 15, \sin C = \dfrac{2(42)}{7 \cdot 15} = \dfrac{84}{105}. Since 7^2 + 15^2 = 274 < 400 = 20^2, C is obtuse, so C = 180^\circ - \sin^{-1}(0.8), by inverse sine. (Indeed \cos C = -\tfrac{3}{5}.)
  11. \displaystyle 8\sqrt{30} \approx 43.82. Heron: s = 16, differences 8, 5, 3, product 1920 = 64 \cdot 30. Sine route: \cos C = \dfrac{64 + 121 - 169}{176} = \dfrac{1}{11}, \displaystyle \sin C = \sqrt{1 - \tfrac{1}{121}} = \dfrac{2\sqrt{30}}{11}, and \displaystyle \tfrac{1}{2}(8)(11)\cdot\dfrac{2\sqrt{30}}{11} = 8\sqrt{30}.
  12. About 19.38, 12.6 and 12. Each height is \dfrac{2K}{\text{base}} = \dfrac{252}{\text{base}}: \dfrac{252}{13}, \dfrac{252}{20} and \dfrac{252}{21}.
  13. About 221.99 square meters. s = 36.35, differences 17.85, 12.35, 6.15.
  14. About 721.68 square meters. Triangle PQR (sides 30, 34, 40): s = 52, \displaystyle \sqrt{52 \cdot 22 \cdot 18 \cdot 12} \approx 497.10. Triangle PRS (sides 40, 20, 26): s = 43, \displaystyle \sqrt{43 \cdot 3 \cdot 23 \cdot 17} \approx 224.59.
  15. About 8818.16 square meters and 1058.18 dollars. s = 240, differences 120, 90, 30: \displaystyle \sqrt{77{,}760{,}000}; then multiply the area by 0.12.
  16. Area 60, radius 2. s = 30, differences 24, 5, 1: \displaystyle \sqrt{3600} = 60. Then r = \dfrac{K}{s} = \dfrac{60}{30}.
  17. 168. s = 32, differences 7, 7, 18: \displaystyle \sqrt{28{,}224} = 168. The height to the base of 14 meets it at its midpoint, so it is \displaystyle \sqrt{25^2 - 7^2} = 24, and \tfrac{1}{2}(14)(24) = 168.
  18. s = \tfrac{3t}{2} and each difference is \tfrac{t}{2}, so K^2 = \tfrac{3t}{2}\cdot\tfrac{t^3}{8} = \tfrac{3t^4}{16} and \displaystyle K = \dfrac{\sqrt{3}}{4}t^2.
  19. s = p + \tfrac{q}{2}; s - p = \tfrac{q}{2} twice, and s - q = p - \tfrac{q}{2}. So K^2 = \left(p + \tfrac{q}{2}\right)\left(p - \tfrac{q}{2}\right)\tfrac{q^2}{4}, which is \left(p^2 - \tfrac{q^2}{4}\right)\tfrac{q^2}{4} = \dfrac{q^2(4p^2 - q^2)}{16}, and the positive root is \displaystyle \dfrac{q}{4}\sqrt{4p^2 - q^2}.
  20. Any two of them add to a side: (s - a) + (s - b) = 2s - a - b = c, and likewise for the other pairs. A side is positive, so two of the differences cannot both be zero or negative.
  21. (a + b)^2 - c^2 = a^2 + 2ab + b^2 - c^2 = 2ab and c^2 - (a - b)^2 = c^2 - a^2 + 2ab - b^2 = 2ab, using c^2 = a^2 + b^2. So 16K^2 = 4a^2b^2, K^2 = \tfrac{a^2b^2}{4} and K = \tfrac{1}{2}ab.
  22. The new semiperimeter is ks, and each difference becomes ks - ka = k(s - a). The product under the root gains four factors of k, so it is multiplied by k^4, and its square root by k^2.
  23. Three sides fix a triangle (SSS), so they fix its area, and Heron’s formula computes it. Four sides do not fix a four-sided figure: a frame of four hinged rods can be pushed from a square toward a flat rhombus, keeping every side while the area shrinks toward 0. A diagonal splits the figure into two triangles whose sides are all known, which fixes the shape and the area.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0