In geometry, Heron's formula (or Hero's formula) gives the area of a triangle in terms of the three side lengths a , {\displaystyle a,} b , {\displaystyle b,} c . {\displaystyle c.} Letting s {\displaystyle s} be the semiperimeter of the triangle, s = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} , the area A {\displaystyle A} is
A = s ( s − a ) ( s − b ) ( s − c ) . {\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}.}
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.
Example
Let △ A B C {\displaystyle \triangle ABC} be the triangle with sides a = 4 {\displaystyle a=4} , b = 13 {\displaystyle b=13} , and c = 15 {\displaystyle c=15} . This triangle's semiperimeter is s = 1 2 ( a + b + c ) =
{\displaystyle s={\tfrac {1}{2}}(a+b+c)={}}
1 2 ( 4 + 13 + 15 ) = 16 {\displaystyle {\tfrac {1}{2}}(4+13+15)=16} therefore s − a = 12 {\displaystyle s-a=12} , s − b = 3 {\displaystyle s-b=3} , s − c = 1 {\displaystyle s-c=1} , and the area is
A = s ( s − a ) ( s − b ) ( s − c ) = 16 ⋅ 12 ⋅ 3 ⋅ 1 ) = 24. {\displaystyle {\begin{aligned}A&={\textstyle {\sqrt {s(s-a)(s-b)(s-c)}}}\\[3mu]&={\textstyle {\sqrt {16\cdot 12\cdot 3\cdot 1{\vphantom {)}}}}}\\[3mu]&=24.\end{aligned}}}
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.
Alternative expressions Heron's formula can also be written in terms of just the side lengths instead of using the semiperimeter, in several ways,
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