Heron's formula
Formula for calculating the area of a triangle
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 M...
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Heron's formula
Formula for calculating the area of a triangle
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 M...
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From Wikipedia
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.
Text: Wikipédia, CC BY-SA 4.0. · Image: David Weisman (Dweisman) (Public domain) ·
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