There are many formulas used to find the area of a triangle. These formulas are specific to a particular type of triangle. There is one formula that can be used to find the area of any triangle if the length of all three sides is known. It is called Heron’s formula.

Heron of Alexandria introduced Heron’s formula. It uses the length of all three sides and a semi-perimeter of the triangle. Let’s understand what Heron’s formula is, how it can be derived, and its application in finding the area of triangles and polygons.

## What is Heron’s Formula?

Heron’s formula can be used to find the area of any given triangle, whether it is a scalene, isosceles or equilateral, provided the sides of the triangle are known.

Consider a $\triangle \text{ABC}$, whose sides are $a$, $b$, and $c$, respectively, then the area of a triangle is given by $\text{Area of Triangle } = \sqrt{s(s – a)(s – b)(s – c)}$.

where $s = \frac{a + b + c}{2}$ is semi-perimeter of the triangle.

### Examples of Heron’s Formula

**Example 1: **Find the area of a triangle whose sides are $11$ cm, $60$ cm, and $61$ cm.

Length of three sides of a triangle are $a = 11$ cm, $b = 60$ cm, and $c = 61$ cm

Semi-perimeter of a triangle $s = \frac{11 + 60 + 61}{3} = \frac{132}{2} = 66$ cm

$\text{Area of Triangle } = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{66 (66 – 11)(66 – 60)(66 – 61)}$

$= \sqrt{66 \times 55 \times 6 \times 5} = \sqrt{11 \times 6 \times 11 \times 5 \times 6 \times 5}$

$= \sqrt{11^2 \times 6^2 \times 5^2} = \sqrt{(11 \times 6 \times 5)^2}$

$\sqrt{330^2} = 330 \text{cm}^2$

**Example 2:** Find the area of a triangle whose two sides are $18$ cm and $10$ cm respectively and the perimeter is $42$ cm.

Length of two sides of a triangle $a = 18$ cm, and $b = 10$ cm

Perimeter of triangle $\text{P} = 42$ cm

Thus semi-perimeter of triangle $s = \frac{\text{P}}{2} = \frac{42}{2} = 21$ cm

Let the length of the third side be $c$

Thus $18 + 10 + c = 42 => 28 + c = 42 => c = 42 – 28 => c = 14$ cm

$\text{Area of Triangle } = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{21 (21 – 18)(21 – 10)(21 – 14)}$

$\sqrt{21 \times 3 \times 11 \times 7} = \sqrt{21 \times 21 \times 11} = 21 \sqrt{11} \text{ cm}^2$

**Example 3:** Find the area of an equilateral triangle having a side of $6$ cm using Heron’s formula.

Length of a side of an equilateral triangle $a = 6$ cm

Semi-perimeter $s = \frac{a + a + a}{2} = \frac{6 + 6 + 6}{2} = 9$ cm

$\text{Area of Triangle } = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{9 (9 – 6)(9 – 6)(9 – 6)}$

$= \sqrt{9 \times 3 \times 3 \times 3} = \sqrt{9 \times 9 \times 3} = 9 \sqrt{3} \text{ cm}^2$

**Example 4:** The perimeter of an isosceles triangle is $42$ cm. The ratio of the equal side to its base is $3: 4$. Find the area of the triangle.

The ratio of the equal side to its base is $3: 4$.

Let the length of the equal side = $3x$

And the length of base = $4x$

Perimeter of triangle $3x + 3x + 4x = 10x$

Thus $10x = 42 => x = \frac{42}{10} = 4.2$ cm

Therefore three sides of triangle are $a = b = 3x = 3 \times 4.2 = 12.6$ cm, and $c = 4x = 4 \times 4.2 = 16.8$ cm

Semi-perimeter of triangle $s = \frac{42}{2} = 21$ cm

$\text{Area of Triangle } = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{21 (21 – 12.6)(21 – 12.6)(21 – 16.8)}$

$= \sqrt{21 \times 8.4 \times 8.4 \times 4.2} = \sqrt{21 \times \frac{84}{10} \times \frac{84}{10} \times \frac{42}{10}}$

$= \sqrt{\frac{21 \times 84 \times 84 \times 42}{10 \times 10 \times 10}}$

$= \frac{84}{10}\sqrt{\frac{21 \times 42}{10}} = 78.89 \text{ cm}^2$

**Example 5:** Sides of a triangle are in the ratio of $14 : 20: 25$ and its perimeter is $590$ cm. Find its area.

Let sides of a triangle be $14x$, $20x$, and $25x$ (in cm)

Therefore, $14x + 20x + 25x = 590 => 59x = 590 => x = \frac{590}{59} => x = 10$

Thus length of sides of triangle are $14 \times 10 = 140$ cm, $20 \times 10 = 200$ cm, and $25 \times 10 = 250$ cm

Semi-perimeter of triangle $s = \frac{590}{2} = 295$ cm

$\text{Area of Triangle } = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{295 (295 – 140)(295 – 200)(295 – 250)}$

$= \sqrt{295 \times 155 \times 95 \times 45} = 13981.21 \text{ cm}^2$

**Example 6:** A park in the shape of a quadrilateral $\text{ABCD}$ has $\angle \text{C} = 90^{\circ}$, $\text{AB} = 18$ m, $\text{BC} = 24$ m, $\text{CD} = 10$ m and $\text{AD} = 16$ m. How much area does it occupy?

Dimensions of a park $\text{ABCD}$ are $\angle \text{C} = 90^{\circ}$, $\text{AB} = 18$ m, $\text{BC} = 24$ m, $\text{CD} = 10$ m and $\text{AD} = 16$ m.

Join the points $\text{B}$ and $\text{D}$

In $\triangle \text{BCD}$, $\text{BC}^2 + \text{CD}^2 = \text{BD}^2$

$=> 24^2 + 10^2 = \text{BD}^2$

$=> \text{BD}^2 = 576 + 100 = 676$

$=> \text{BD} = \sqrt{676} = 26$ m

Area of quadrilateral $\text{ABCD}$ = Area of $\triangle \text{ABD} + $ Area of $\triangle \text{BCD}$

For $\triangle \text{ABD}$

Length of sides are $a = 18$ cm, $b = 26$ cm, $c = 16$ cm

Semi-perimeter $s = \frac{18 + 26 + 16}{2} = \frac{60}{2} = 30$ cm

Area of $\triangle \text{ABD} = \sqrt{30(30 – 18)(30 – 26)(30 – 16)}$

$= \sqrt{30 \times 12 \times 4 \times 14}$

$= \sqrt{5 \times 6 \times 6 \times 2 \times 2 \times 2 \times 7 \times 2}$

$= 6 \times 2 \times 2 \sqrt{5 \times 7}$

$ = 24 \sqrt{35} = 141.98 \text{ m}^2 $

For $\triangle \text{BCD}$

Length of sides are $a = 24$ cm, $b = 10$ cm, $c = 26$ cm

Semi-perimeter $s = \frac{24 + 10 + 26}{2} = \frac{60}{2} = 30$ cm

Area of $\triangle \text{BCD} = \sqrt{30(30 – 24)(30 – 10)(30 – 26)}$

$= \sqrt{30 \times 6 \times 20 \times 4}$

$= \sqrt{30 \times 6 \times 5 \times 4 \times 4}$

$= \sqrt{30 \times 30 \times 4 \times 4} = 30 \times 4 = 120 \text{ m}^2$

Thus area of rectangular park = $141.98 + 120 = 261.98 \text{m}^2$

## Heron’s Formula Derivation

Consider $\triangle \text{ABC}$ with sides $a$, $b$, and $c$.

Area of $\triangle \text{ABC} = \frac{1}{2} bh$ ——————————————- (1)

Draw a perpendicular $\text{BD}$ on $\text{AC}$

Consider $\triangle \text{ADB}$

By Pythagorean theorem, $x^2 + h^2 = c^2$

$=> x^2 = c^2 – h^2$ ———————————————— (2)

$=> x = \sqrt{c^2 – h^2}$ ———————————————— (3)

Consider $\triangle \text{CDB}$

$(b – x)^2 + h^2 = a^2$

$=> (b – x)^2 = a^2 – h^2$

$=> b^2 – 2bx + x^2 = a^2 – h^2$

Substituting the value of $x$ and $x^2$ from equations (2) and (3), we get

$b^2 – 2b \sqrt{c^2 – h^2}+ c^2 – h^2 = a^2 – h^2$

$=> b^2 + c^2 – a^2 = 2b \sqrt{c^2 – h^2}$

Squaring on both sides, we get

$\left(b^2+ c^2 – a^2 \right)^2 = 4b^2 \left(c^2 – h^2 \right)$

$=> \frac{\left(b^2+ c^2 – a^2 \right)^2}{4b^2} = c^2 – h^2$

$=> h^2 = c^2 – \frac{\left(b^2+ c^2 – a^2 \right)^2}{4b^2}$

$=> h^2 = \frac{4b^2c^2 – \left(b^2+ c^2 – a^2 \right)^2}{4b^2}$

$=> h^2 = \frac{(2bc)^2 – \left(b^2+ c^2 – a^2 \right)^2}{4b^2}$

$=> h^2 = \frac{(2bc)^2 – \left(b^2+ c^2 – a^2 \right)^2}{4b^2}$

$=> h^2 = \frac{(2bc – (b^2+ c^2 – a^2))(2bc + (b^2+ c^2 – a^2))}{4b^2}$

$=> h^2 = \frac{(2bc – b^2 – c^2 + a^2)(2bc + b^2+ c^2 – a^2)}{4b^2}$

$=> h^2 = \frac{(-(2bc + b^2 + c^2) + a^2)((2bc + b^2+ c^2) – a^2)}{4b^2}$

$=>h^2 = \frac{(((b + c) + a)((b + c) – a))((a + (b – c))(a – (b – c)))}{4b^2}$

$=>h^2 = \frac{(a + b + c)(b + c – a)(a + c – b)(a + b – c)}{4b^2}$

$=>h^2 = \frac{(a + b + c)(a + b + c – 2a)(a + b + c – 2b)(a + b + c – 2c)}{4b^2}$ ——————– (4)

Now the perimeter of $\triangle \text{ABC}$ is given by $\text{P} = a + b + c$

Substituting $a + b + c = \text{P}$ in equation (4)

$h^2 = \frac{\text{P}(\text{P} – 2a)(\text{P} – 2b)(\text{P} – 2c)}{4b^2}$

$=> h = \frac{\sqrt{\text{P}(\text{P} – 2a)(\text{P} – 2b)(\text{P} – 2c)}}{2b}$

Substituting the value of $h$ in equation (1), we get

Area of $\triangle \text{ABC} = \frac{1}{2} b\frac{\sqrt{\text{P}(\text{P} – 2a)(\text{P} – 2b)(\text{P} – 2c)}}{2b}$

$= \frac{1}{4}\sqrt{\text{P}(\text{P} – 2a)(\text{P} – 2b)(\text{P} – 2c)}$

$= \sqrt{\frac{1}{16} \text{P}(\text{P} – 2a)(\text{P} – 2b)(\text{P} – 2c)}$

$=\sqrt{\frac{\text{P}}{2}\frac{(\text{P} – 2a)}{2} \frac{(\text{P} – 2b)}{2} \frac{(\text{P} – 2c)}{2}}$

$ = \sqrt{\frac{\text{P}}{2} \left(\frac{\text{P}}{2} – a \right) \left(\frac{\text{P}}{2} – b \right) \left(\frac{\text{P}}{2} – c \right)}$

$= \sqrt{s(s – a)(s – b)(s – c)}$

## Applications of Heron’s Formula

There are numerous applications of Heron’s formula. They are

- It can be used to determine the area of different types of triangles if the lengths of their different sides are given.
- It can be used to find the area of the quadrilateral if the lengths of all its sides are given.

## Practice Problems

- Find the area of a triangle whose sides are $12$ cm, $6$ cm, and $15$ cm.
- Find the area of a triangle whose sides are $4.5$ cm and $10$ cm and whose perimeter is $10.5$ cm.
- Find the area of a triangle with a base = $20$ cm and a height is $10$ cm.
- Find the area of a triangle having a perimeter of $32$ cm. One side of its side is equal to $11$ cm and the difference between the other two is $5$ cm.
- The perimeter of the rhombus is $100$ m and its diagonal is $40$ m. Find the area of the rhombus.
- Two parallel sides of a trapezium are $60$ cm and $77$ cm and the other sides are $25$ cm and $26$ cm. Find the area of the trapezium.
- The perimeter of a triangular field is $135$ cm and its sides are in the ratio $25 : 17 : 12$. Find its area.

## FAQs

### What is Heron’s formula for area of triangle?

Heron’s formula is used to find the area of the triangle when the lengths of all triangles are given. It can be used to determine areas of different types of triangles, equilateral, isosceles, or scalene triangles. For a triangle having three sides, $a$, $b$, and $c$, semi-perimeter, $s$, and area of triangle A, the semi-perimeter and area of the triangle are given as $\frac{a + b + c}{2}$ and $\sqrt{s(s-a)(s-b)(s-c)}$ respectively.

### How does Heron’s formula work?

Heron’s formula depends only on the semi-perimeter of a triangle and the length of its three sides. We first determine the value of the semi-perimeter using the lengths of three sides of the triangle. Once the value of the semi-perimeter is obtained we can find the area of the shape.

### Why is it called Heron’s formula?

It is named after first-century engineer Heron of Alexandria (or Hero) who proved it in his work Metrica, though it was probably known centuries earlier.

### Does Heron’s formula work for all triangles?

Heron’s formula can be applied to any type of triangle. You can use Heron’s formula to find the area of a triangle whose all three sides are known.

### What is another name for Heron’s formula?

Heron’s formula that is used to find the area of a triangle is also termed Hero’s formula.

### What are $a$, $b$, $c$ used in Heron’s formula?

Heron’s formula $\sqrt{s(s-a)(s-b)(s-c)}$ is used to find the area of a triangle where $a$, $b$, and $c$ are the sides of a triangle and $s$ is the semi-perimeter $s = \frac{a + b + c}{2}$.

### What are the limitations of Heron’s formula?

The only limitation of Heron’s formula is that it demands the length of all three sides of a triangle.

## Conclusion

Heron’s formula is used to find the area of any given triangle, whether it is a scalene, isosceles or equilateral, provided the sides of the triangle are known. The formula used is given by $\sqrt{s(s-a)(s-b)(s-c)}$, where $a$, $b$, and $c$ are the sides of a triangle and $s$ is the semi-perimeter $s = \frac{a + b + c}{2}$.

## Recommended Reading

- Reference Books
- Sample Papers
- Area of Triangle in Coordinate Geometry – Formula, Derivation & Examples
- Perimeter of Triangle – Definition, Formula & Examples
- Altitude of a Triangle(Definition & Properties)
- What is an Isosceles Triangle – Definition, Properties & Examples
- What is a Scalene Triangle – Definition, Properties & Examples
- Angle Bisector of a Triangle – Definition, Properties & Examples
- What is an Equilateral Triangle – Definition, Properties & Examples
- Congruence of Triangles Criteria – SSS, SAS, ASA, RHS
- Similarity of Triangles Criteria – SSS, SAS, AA