Most area formulas for a triangle need a height, and the height is often the one measurement you do not have. Heron's formula skips it. Give it the three side lengths and it returns the area, with nothing to measure and no angle to find.
It is named after Heron of Alexandria, a mathematician and engineer who described it in the first century CE. Below you will see what the formula says, why it works, how to apply it step by step and where students go wrong. Start by dragging the corners of the triangle: the formula runs live as you move them.
The formula
Call the sides $latex a$, $latex b$ and $latex c$. First find the semi-perimeter, which is half the perimeter:
$$s=\frac{a+b+c}{2}$$
Then the area is:
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
Read it as four numbers multiplied under one square root: $latex s$ itself and the three gaps $latex s-a$, $latex s-b$ and $latex s-c$. Every gap has to be positive. That is the triangle inequality in disguise: a triangle exists only when each side is shorter than the sum of the other two.
If you prefer to skip $latex s$, the same formula written with the sides directly is:
$$A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$
How to use it, step by step
- Check that the sides form a triangle: the two shorter sides must add up to more than the longest side.
- Compute the semi-perimeter $latex s=(a+b+c)/2$.
- Compute the three gaps $latex s-a$, $latex s-b$ and $latex s-c$.
- Multiply the four numbers $latex s$, $latex s-a$, $latex s-b$ and $latex s-c$.
- Take the square root and write the answer in square units.
Keep the exact product until the very end and round only the final answer. Rounding the gaps early is the easiest way to lose a decimal place.
Why it works
The proof uses two facts you already know: the area $latex A=\frac{1}{2}ab\sin C$ and the law of cosines $latex c^2=a^2+b^2-2ab\cos C$. Square the area and use $latex \sin^2 C=1-\cos^2 C$:
$$16A^2=4a^2b^2\sin^2 C=4a^2b^2-(2ab\cos C)^2=4a^2b^2-(a^2+b^2-c^2)^2$$
The right side is a difference of squares, so it factors:
$$16A^2=\big(c^2-(a-b)^2\big)\big((a+b)^2-c^2\big)=(c-a+b)(c+a-b)(a+b-c)(a+b+c)$$
Each factor is twice one of our numbers: $latex a+b+c=2s$, $latex b+c-a=2(s-a)$, $latex a+c-b=2(s-b)$ and $latex a+b-c=2(s-c)$. Multiplying them gives $latex 16A^2=16\,s(s-a)(s-b)(s-c)$, and dividing by 16 and taking the root is Heron's formula.
There is also a geometric reading of the gaps: $latex s-a$ is the distance from vertex A to the points where the inscribed circle touches the two sides that meet at A. The same goes for $latex s-b$ and $latex s-c$.
Heron's formula: worked examples
Try each problem on your own, then open the solution to compare.
EXAMPLE 1
Find the area of a triangle with sides 13, 14 and 15.
Solution
The sides form a triangle because 13 + 14 is greater than 15. The semi-perimeter is:
$$s=\frac{13+14+15}{2}=\frac{42}{2}=21$$
Now the gaps and the product:
$$A=\sqrt{21\cdot(21-13)(21-14)(21-15)}=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84$$
The area is 84 square units. The numbers are clean because this is the classic 13-14-15 triangle, and you can check it in the drag tool above.
EXAMPLE 2
A triangle has sides 7 cm, 8 cm and 9 cm. Find its area.
Solution
$$s=\frac{7+8+9}{2}=12$$
$$A=\sqrt{12(12-7)(12-8)(12-9)}=\sqrt{12\cdot 5\cdot 4\cdot 3}=\sqrt{720}=12\sqrt{5}\approx 26.83$$
The area is about 26.83 cm². When the root is not exact, give the decimal and say how many places you kept.
EXAMPLE 3
A triangle has sides 10, 17 and 21. Find its area and the height drawn to the longest side.
Solution
$$s=\frac{10+17+21}{2}=24$$
$$A=\sqrt{24\cdot 14\cdot 7\cdot 3}=\sqrt{7056}=84$$
The longest side is 21, so we use it as the base in $latex A=\frac{1}{2}bh$:
$$84=\frac{1}{2}\cdot 21\cdot h\quad\Rightarrow\quad h=\frac{2\cdot 84}{21}=8$$
The height to the side of length 21 is 8. This is how Heron's formula finds a height when only the sides are known.
EXAMPLE 4
A triangular garden has sides of 52 m, 56 m and 60 m. A bag of grass seed covers 150 m². How many bags are needed?
Solution
$$s=\frac{52+56+60}{2}=84$$
$$A=\sqrt{84\cdot 32\cdot 28\cdot 24}=\sqrt{256\cdot 7056}=16\cdot 84=1344$$
The garden has an area of 1344 m². Then $latex 1344\div 150=8.96$, so you need 9 bags (round up, because 8 bags would leave part of the garden bare).
Notice the sides are exactly four times the 13-14-15 triangle, so the area is 16 times larger: 16 × 84 = 1344.
EXAMPLE 5
Can a triangle have sides 3, 4 and 8? What does Heron's formula say?
Solution
No. The two shorter sides add up to 3 + 4 = 7, which is less than 8. The formula warns you: $latex s=7.5$ and $latex s-c=7.5-8=-0.5$, so the number under the square root is negative.
A negative result under the root always means the three lengths cannot form a triangle.
Heron's formula or another formula?
| What you know | Formula to use |
|---|---|
| Base and height | $latex A=\frac{1}{2}bh$ |
| Two sides and the angle between them | $latex A=\frac{1}{2}ab\sin C$ |
| All three sides | Heron's formula |
| Coordinates of the three vertices | The shoelace formula |
| Equilateral triangle with side $latex a$ | $latex A=\frac{\sqrt{3}}{4}a^2$ |
Shortcuts and real uses
Heron's formula also gives shortcuts. For an isosceles triangle with two equal sides of length $latex a$ and a base $latex b$, the gaps simplify and the area becomes:
$$A=\frac{b}{4}\sqrt{4a^2-b^2}$$
With sides 10, 10 and 12 this gives $latex \frac{12}{4}\sqrt{400-144}=3\cdot 16=48$, the same triangle as the third preset in the drag tool above.
Surveyors rely on the same idea. An irregular plot of land is cut into triangles, the three lengths of each triangle are measured with a tape or a laser, and Heron's formula returns each area without anyone needing to stand at a right angle to a side. Add the pieces and you have the area of the whole plot.
The order of the sides does not matter. The formula is symmetric in $latex a$, $latex b$ and $latex c$, so you can label them in any order and get the same area.
Common mistakes
- Using the perimeter instead of the semi-perimeter. For the 13-14-15 triangle that would put 42 in place of 21 and give a wrong answer.
- Forgetting the square root at the end, and reporting 7056 instead of 84.
- Rounding $latex s$ or the gaps early. Keep fractions or full decimals until the last step.
- Mixing units, for example one side in meters and another in centimeters.
- Skipping the triangle check. If any gap is zero or negative, there is no triangle (a zero gap gives a flat triangle with area 0).
Heron's formula: practice problems
Use the steps above. Open each answer when you are ready.
1. Find the area of a triangle with sides 5, 12 and 13.
Answer
$latex s=15$ and $latex A=\sqrt{15\cdot 10\cdot 3\cdot 2}=\sqrt{900}=30$. The area is 30 square units (it is a right triangle, and $latex \frac{1}{2}\cdot 5\cdot 12=30$ agrees).
2. Find the area of a triangle with sides 9, 10 and 17.
Answer
$latex s=18$ and $latex A=\sqrt{18\cdot 9\cdot 8\cdot 1}=\sqrt{1296}=36$. The area is 36 square units.
3. Find the area of an equilateral triangle with sides of 6 cm.
Answer
$latex s=9$ and $latex A=\sqrt{9\cdot 3\cdot 3\cdot 3}=\sqrt{243}=9\sqrt{3}\approx 15.59$. The area is about 15.59 cm².
4. Find the area of a triangle with sides 25, 29 and 36.
Answer
$latex s=45$ and $latex A=\sqrt{45\cdot 20\cdot 16\cdot 9}=\sqrt{129600}=360$. The area is 360 square units.
5. The triangle with sides 13, 14 and 15 has area 84. What is the height drawn to the side of length 15?
Answer
$latex h=\frac{2A}{15}=\frac{168}{15}=11.2$. The height is 11.2 units.
6. Does a triangle with sides 4, 5 and 10 exist?
Answer
No. $latex 4+5=9$ is less than 10. In the formula, $latex s=9.5$ and $latex s-c=-0.5$, so the product under the root is negative.
Frequently asked questions
Does Heron's formula work for every triangle?
Yes: acute, right or obtuse. The only requirement is that the three lengths can form a triangle.
Is the result exact or approximate?
The formula itself is exact. Any approximation comes from rounding a square root. For nearly flat triangles with very long sides, calculators use an equivalent form that sorts the sides first, so that subtracting nearly equal numbers does not lose digits.
Can I use it for a four-sided shape?
Not directly. Split the shape into two triangles with a diagonal, use Heron's formula on each, and add the areas. You need the diagonal length for that.
Why is it called the semi-perimeter?
Because $latex s$ is half of the perimeter. It is the natural number to use: $latex s-a$ is also the distance from a vertex to the points where the inscribed circle touches the sides.
Related calculators
Check your answers with these tools:
See also
More on triangles and their areas: