what is heron's formula??
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If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; uopposite to U and so on), then[14]
{\displaystyle {\text{volume}}={\frac {\sqrt {\,(-a+b+c+d)\,(a-b+c+d)\,(a+b-c+d)\,(a+b+c-d)}}{192\,u\,v\,w}}}
where
{\displaystyle {\begin{aligned}a&={\sqrt {xYZ}}\\b&={\sqrt {yZX}}\\c&={\sqrt {zXY}}\\d&={\sqrt {xyz}}\\X&=(w-U+v)\,(U+v+w)\\x&=(U-v+w)\,(v-w+U)\\Y&=(u-V+w)\,(V+w+u)\\y&=(V-w+u)\,(w-u+V)\\Z&=(v-W+u)\,(W+u+v)\\z&=(W-u+v)\,(u-v+W).\end{aligned}}}
{\displaystyle {\text{volume}}={\frac {\sqrt {\,(-a+b+c+d)\,(a-b+c+d)\,(a+b-c+d)\,(a+b+c-d)}}{192\,u\,v\,w}}}
where
{\displaystyle {\begin{aligned}a&={\sqrt {xYZ}}\\b&={\sqrt {yZX}}\\c&={\sqrt {zXY}}\\d&={\sqrt {xyz}}\\X&=(w-U+v)\,(U+v+w)\\x&=(U-v+w)\,(v-w+U)\\Y&=(u-V+w)\,(V+w+u)\\y&=(V-w+u)\,(w-u+V)\\Z&=(v-W+u)\,(W+u+v)\\z&=(W-u+v)\,(u-v+W).\end{aligned}}}
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Hola Mate:-
Here's your answer:-
In geometry, Heron's formula is named after hero of Alexandria.
It is based on how to find out the area of a triangle using three sides.
where s = semi perimeter
& a , b & c = 3 sides of triangle.
Hope u understand.
Mark me as brainliest if it helped;-)
Here's your answer:-
In geometry, Heron's formula is named after hero of Alexandria.
It is based on how to find out the area of a triangle using three sides.
where s = semi perimeter
& a , b & c = 3 sides of triangle.
Hope u understand.
Mark me as brainliest if it helped;-)
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