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Question Number 188119 by mr W last updated on 25/Feb/23

Commented by mr W last updated on 26/Feb/23

other forms of Heron′s formua.  maybe one day you may use them.

$${other}\:{forms}\:{of}\:{Heron}'{s}\:{formua}. \\ $$$${maybe}\:{one}\:{day}\:{you}\:{may}\:{use}\:{them}. \\ $$

Answered by manxsol last updated on 26/Feb/23

other way    a,b,c  sides triangle  A=(1/2)absinθ  4A^2 =a^2 b^2 sin^2 θ  4A^2 =a^2 b^2 (1−cos^2 θ)  c^2 =a^2 +b^2 −2abcosθ  cosθ=((a^2 +b^2 −c^2 )/(2ab))  4A^2 =a^2 b^2 −a^2 b^2 (((a^2 +b^2 −c^2 )/(2ab)))^2   4A^2 =a^2 b^2 −a^2 b^2 (((a^2 +b^2 −c^2 )^2 )/(4a^2 b^2 ))  16A^2 =4a^2 b^2 −(a^2 +b^2 −c^2 )  4A=(√(4a^2 b^2 −(a^2 +b^2 −c^2 )^2 ))  A=(1/4)(√(4a^2 b^2 −(a^2 +b^2 −c^2 )^2 ))

$${other}\:{way}\:\:\:\:{a},{b},{c}\:\:{sides}\:{triangle} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}{absin}\theta \\ $$$$\mathrm{4}{A}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\ $$$$\mathrm{4}{A}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right) \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta \\ $$$${cos}\theta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\mathrm{4}{A}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{A}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} −\cancel{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}\cancel{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }} \\ $$$$\mathrm{16}{A}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$$\mathrm{4}{A}=\sqrt{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$ \\ $$

Commented by mr W last updated on 26/Feb/23

correct!  this formula is also given in the  collection i posted.

$${correct}! \\ $$$${this}\:{formula}\:{is}\:{also}\:{given}\:{in}\:{the} \\ $$$${collection}\:{i}\:{posted}. \\ $$

Commented by manxsol last updated on 26/Feb/23

Sir W. can you give me the   publication number.than you

$${Sir}\:{W}.\:{can}\:{you}\:{give}\:{me}\:{the} \\ $$$$\:{publication}\:{number}.{than}\:{you} \\ $$

Commented by mr W last updated on 26/Feb/23

what is a publication number?  so if i even don′t know what a   publication number is, i wont also   have one.

$${what}\:{is}\:{a}\:{publication}\:{number}? \\ $$$${so}\:{if}\:{i}\:{even}\:{don}'{t}\:{know}\:{what}\:{a}\: \\ $$$${publication}\:{number}\:{is},\:{i}\:{wont}\:{also}\: \\ $$$${have}\:{one}. \\ $$

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