Heron's Formula

Explained with examples and pictures

The Formula

Heron's formula is named after Hero of Alexendria, a Greek Engineer and Mathematician in 10 - 70 AD. You can use this formula to find the area of a triangle using the 3 side lengths.

Therefore, you do not have to rely on the formula for area that uses base and height. Diagram 1 below illustrates the general formula where S represents the semi-perimeter of the triangle.

semi-perimeter is just the perimeter divided by 2 : $$ \frac{perimeter}{2} $$ .

Diagram 1
Picture of Herons Formula
Diagram 2

A specific example

Heron's Formula Applet

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See applet on its own page

Side Lengths of Triangle

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Examples

Example 1
(Straight forward example)

Use Heron's formula to find the area of triangle ABC, if $$ AB= 3, BC = 2 , CA = 4 $$ .

triangle Step 1
Calculate the semi perimeter, S

$ s = \frac{ 3 + 2 +4}{2 } \\ s = 4.5 $

Step 2
Substitute S into the formula.

Round answer to nearest tenth.

$ A= \sqrt{ 4.5( 4.5 - 3 ) (4.5-2)( 4.5-4) } \\ A= \sqrt{8.4375 } \\ A \approx 2.9 $

challenge problem Since Heron's formula relates the side lengths, perimeter and area of a triangle, you might need to answer more challenging question types like the following example.

Example 2

Given $$ \triangle ABC $$, with an area of $$ 8.94 $$ square units, a perimeter of $$ 16 $$ units and side lengths $$AB = 3 $$ and $$ CA = 7 $$, what is $$ \red { BC }$$ ?

Step 1

Calculate the semi perimeter, S.

$ S =\frac{ perimeter}{2} \\ S = \frac{16}{2} \\ S = 8 $

Step 2

Substitute known values into the formula . Let $$ \red x = \red{ BC} $$

$ A= \sqrt{ S(S - AB)(S - \red { BC} )(S- CA) } \\ 8.94 = \sqrt{ 8(8 - 3)(8 - \red x )(8- 7) } \\ 8.94 = \sqrt{ 8(5)(8 - \red x )(1) } $

Step 3

Solve for $$ \red x $$ (square both sides and go from there).

$ 8.94^{\blue 2} = \left( \sqrt{ 8(5)(8 - \red x )(1) } ) \right) ^{\blue 2} \\ 79.9236 = 8(5)(8 -\red x)(1) \\ 79.9236 = 40(8 -\red x) \\ \frac{ 79.9236}{40} = 8 -\red x \\ 1.999809 = 8 -\red x \\ x \approx 6.0 $

Practice Problems

Problem 1
triangle

Use Heron's formula to find the area of the triangle pictured with the following side lengths.

$ AB = 8 \\ BC = 41 \\ CA = 44 $

This problem is similar to example 1.

Step 1

Calculate the semi perimeter, S.

$ S = \frac{8+41+44}{2} \\ S= \frac{93}{2} \\ S = \blue { 46.5 } $

Step 2

Substitute S into the formula. Round answer to nearest tenth.

$ A = \sqrt{ \blue{ 46.5} \cdot ( \blue{ 46.5} -8) \cdot ( \blue{ 46.5}- 41) \cdot ( \blue{ 46.5} - 44) } \\ A = \sqrt{ \blue{ 46.5} \cdot 38.5 \cdot 2.5 \cdot 5.5 } A \approx 156.9 $

Problem 2
triangle

Determine the area of the triangle using Heron's formula to find the area of the triangle pictured with the following side lengths.

This problem is similar to example 1.

Step 1

Calculate the semi perimeter, S.

$ S = \frac{ 7+6+ 8}{2} \\ S = \frac{21}{2} \\ S = \blue {10.5 } $

Step 2

Substitute S into the formula . Round answer to nearest tenth.

$ A = \sqrt{\blue {10.5 } (\blue{10.5} - 7) (\blue{10.5} - 6 ))(\blue{10.5} - 8 )} \\ A \approx 20.3 $

Problem 3
triangle

Determine the area of the triangle using Heron's formula to find the area of the triangle pictured with the following side lengths.

This problem is similar to example 1.

Step 1

Calculate the semi perimeter, S.

$ S = \frac{ 11 + 12 + 5}{2 } \\ S = \frac{ 28}{2} \\ S = \blue { 14 } $

Step 2

Substitute S into the formula.

Round answer to nearest tenth.

$ A = \sqrt{ \blue{14} (\blue{14} - 11) (\blue{14} - 12) ( \blue{14} - 5) } \\ A \approx 27.5 $

challenge problem Problems 2 and onward are similar to example 2 .
Problem 4

If the perimeter of $$ \triangle ABC $$ is $$32$$ units, its area is $$ 35.8 $$ units squared, and $$ AB= 14 $$ and $$ BC = 12 $$, what is the length of the third side, side $$ \red {CA} $$ ?

This problem is like example 2.

Step 1

Calculate the semi perimeter, S.

$ S = \frac{perimeter}{2} \\ S = \frac{32}{2} \\ S = \blue{16} $

Step 2

Substitute known values into the formula. Let $$ \red {x} = \red {CA} $$ .

$ A =\sqrt{ S (S- AB)( S- BC )( S - CA)} \\ 35.8 =\sqrt{ \blue{16} (\blue{16}- 14)( \blue{16}- 12 )( \blue{16} - \red x)} $

Step 3

Solve for x (square both sides and go from there).

$ 35.8^{\blue{2}} =\sqrt{ 16 ( 16- 14)( 16- 12 )( 16 - \red x)}^{\blue{2}} \\ 1281.64 =16 ( 16- 14)( 16- 12 )( 16 - \red x) \\ 1281.64 =16 (2 )( 4 )( 16 - \red x) \\ 1281.64 = 128( 16 - \red x) \\ \frac{ 1281.64}{128} = 16 - \red x \\ \frac{ 1281.64}{128} = 16 - \red x \\ 10.0128125 = 16 - \red x \\ \red x \approx 6 $

Problem 5

If the perimeter of a triangle is 26 units, its area is 18.7 units squared, and the lengths of AB = 12 and BC = 4, what is the length of the third side, side CA?

This problem is like example 2.

Step 1

Calculate the semi perimeter, S.

S = perimeter / 2
S = 26 / 2 = 13

Step 2

Substitute known values into the formula. Let x equal side length CA.

heron's formula
Step 3

Solve for x (square both sides and go from there).

heron's formula
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