Math, asked by vinodnita, 1 year ago

prove heron formula

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Answered by sheru11
5
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vinodnita: you took the help of search engine
Answered by madhura41
4


{\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}

From this proof we get the algebraic statement that

{\displaystyle \sin \gamma ={\sqrt {1-\cos ^{2}\gamma }}={\frac {\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}}.}

The altitude of the triangle on base a has length b sin γ, and it follows

{\displaystyle {\begin{aligned}A&={\frac {1}{2}}({\mbox{base}})({\mbox{altitude}})\\&={\frac {1}{2}}ab\sin \gamma \\&={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\&={\frac {1}{4}}{\sqrt {(2ab-(a^{2}+b^{2}-c^{2}))(2ab+(a^{2}+b^{2}-c^{2}))}}\\&={\frac {1}{4}}{\sqrt {(c^{2}-(a-b)^{2})((a+b)^{2}-c^{2})}}\\&={\sqrt {\frac {(c-(a-b))(c+(a-b))((a+b)-c)((a+b)+c)}{16}}}\\&={\sqrt {{\frac {(b+c-a)}{2}}{\frac {(a+c-b)}{2}}{\frac {(a+b-c)}{2}}{\frac {(a+b+c)}{2}}}}\\&={\sqrt {{\frac {(a+b+c)}{2}}{\frac {(b+c-a)}{2}}{\frac {(a+c-b)}{2}}{\frac {(a+b-c)}{2}}}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}.\end{aligned}}}.......

..In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria,[1] gives the area of a triangle by requiring no arbitrary choice of side as base or vertex as origin, contrary to other formulae for the area of a triangle, such as half the base times the height or half the norm of a cross product of two sides.


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