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Question-188119

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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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