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Question Number 216239 by mr W last updated on 01/Feb/25
Commented by mr W last updated on 01/Feb/25
find area of triangle
$${find}\:{area}\:{of}\:{triangle} \\ $$
Commented by SonGoku last updated on 01/Feb/25
A_△ = ((AB×CA×sin(cos^(−1) (−((BC^2 − CA^2 − AB^2 )/(2×CA×AB)))))/2) A_△ = (((√4)×(√5)×sin(cos^(−1) (−((((√3))^2 − ((√5))^2 − ((√4))^2 )/(2×(√5)×(√4))))))/2) A_△ = ((2(√5)sin(cos^(−1) (−((3 − 5 − 4)/(2×(√5)×2)))))/2) A_△ = (√5)sin(cos^(−1) (−((−6)/( 4(√5))))) A_△ = (√5)sin(cos^(−1) ((3/( 2(√5))))) A_△ = (√5)sin(cos^(−1) (((3(√5))/( 2×5)))) A_△ = (√5)sin(cos^(−1) (((3(√5))/(10)))) A_△ = (√5)sin(47.867585°) A_△ = (√5)×0.741596 A_△ ≈1.658259 determinant (((A_△ ≈1.66 u^2 )))
$$\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\frac{\boldsymbol{{AB}}×\boldsymbol{{CA}}×\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(−\frac{\boldsymbol{{BC}}^{\mathrm{2}} \:−\:\boldsymbol{{CA}}^{\mathrm{2}} \:−\:\boldsymbol{{AB}}^{\mathrm{2}} }{\mathrm{2}×\boldsymbol{{CA}}×\boldsymbol{{AB}}}\right)\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\frac{\sqrt{\mathrm{4}}×\sqrt{\mathrm{5}}×\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(−\frac{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\:\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \:−\:\left(\sqrt{\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{2}×\sqrt{\mathrm{5}}×\sqrt{\mathrm{4}}}\right)\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\frac{\cancel{\mathrm{2}}\sqrt{\mathrm{5}}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(−\frac{\mathrm{3}\:−\:\mathrm{5}\:−\:\mathrm{4}}{\mathrm{2}×\sqrt{\mathrm{5}}×\mathrm{2}}\right)\right)}{\cancel{\mathrm{2}}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\sqrt{\mathrm{5}}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(−\frac{−\mathrm{6}}{\:\mathrm{4}\sqrt{\mathrm{5}}}\right)\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\sqrt{\mathrm{5}}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\:\mathrm{2}\sqrt{\mathrm{5}}}\right)\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\sqrt{\mathrm{5}}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\:\mathrm{2}×\mathrm{5}}\right)\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\sqrt{\mathrm{5}}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\sqrt{\mathrm{5}}\boldsymbol{\mathrm{sin}}\left(\mathrm{47}.\mathrm{867585}°\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:=\:\sqrt{\mathrm{5}}×\mathrm{0}.\mathrm{741596} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:\approx\mathrm{1}.\mathrm{658259} \\ $$$$\begin{array}{|c|}{\boldsymbol{\mathrm{A}}_{\bigtriangleup} \:\approx\mathrm{1}.\mathrm{66}\:\boldsymbol{\mathrm{u}}^{\mathrm{2}} }\\\hline\end{array} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Feb/25
please stand in the queue of “answers”.
$${please}\:{stand}\:{in}\:{the}\:{queue}\:{of}\:“{answers}''. \\ $$
Answered by ajfour last updated on 01/Feb/25
a+b=(√5) (√(4−c^2 ))+(√(3−c^2 ))=(√5) 7−2c^2 +2(√(4−c^2 ))(√(3−c^2 ))=5 (4−c^2 )(3−c^2 )=(c^2 −1)^2 −7c^2 +12=1−2c^2 5c^2 =11 △=((√5)/2)c = ((√5)/2)×((√(11))/5)=((√(11))/2)
$${a}+{b}=\sqrt{\mathrm{5}} \\ $$$$\sqrt{\mathrm{4}−{c}^{\mathrm{2}} }+\sqrt{\mathrm{3}−{c}^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$$\mathrm{7}−\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{4}−{c}^{\mathrm{2}} }\sqrt{\mathrm{3}−{c}^{\mathrm{2}} }=\mathrm{5} \\ $$$$\left(\mathrm{4}−{c}^{\mathrm{2}} \right)\left(\mathrm{3}−{c}^{\mathrm{2}} \right)=\left({c}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$−\mathrm{7}{c}^{\mathrm{2}} +\mathrm{12}=\mathrm{1}−\mathrm{2}{c}^{\mathrm{2}} \\ $$$$\mathrm{5}{c}^{\mathrm{2}} =\mathrm{11} \\ $$$$\bigtriangleup=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{c}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{11}}}{\mathrm{5}}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$
Answered by efronzo1 last updated on 01/Feb/25
cos θ = ((9−3)/(4(√5))) = (3/(2(√5))) sin θ =(√(1−(9/(20)))) =(√((11)/(20))) = ((√(55))/(10)) Area of △ = (1/2) .2.(√5). ((√(55))/(10)) = ((√(11))/2)
$$\:\mathrm{cos}\:\theta\:=\:\frac{\mathrm{9}−\mathrm{3}}{\mathrm{4}\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\:\mathrm{sin}\:\theta\:=\sqrt{\mathrm{1}−\frac{\mathrm{9}}{\mathrm{20}}}\:=\sqrt{\frac{\mathrm{11}}{\mathrm{20}}}\:=\:\frac{\sqrt{\mathrm{55}}}{\mathrm{10}} \\ $$$$\:\:\mathrm{Area}\:\mathrm{of}\:\bigtriangleup\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:.\mathrm{2}.\sqrt{\mathrm{5}}.\:\frac{\sqrt{\mathrm{55}}}{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$
Answered by Frix last updated on 01/Feb/25
((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/4)= =((√(2(a^2 b^2 +a^2 c^2 +b^2 c^2 )−(a^4 +b^4 +c^4 )))/4)= =((√(2(12+15+20)−(9+16+25)))/4)= =((√(94−50))/4)=((√(44))/4)=((√(11))/2)
$$\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{2}\left(\mathrm{12}+\mathrm{15}+\mathrm{20}\right)−\left(\mathrm{9}+\mathrm{16}+\mathrm{25}\right)}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{94}−\mathrm{50}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{44}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}} \\ $$
Answered by MathematicalUser2357 last updated on 02/Feb/25
s=(((√3)+(√4)+(√5))/2)=1+(((√3)+(√5))/2)≈2.984 A=(√(s(s−(√3))(s−2)(s−(√5))))=(√((1+(((√3)+(√5))/2))(1+(((√5)−(√3))/2))((((√3)+(√5))/2)−1)(1+(((√3)−(√5))/2)))) Using this, (1+(((√A)−(√B))/2))(1+(((√B)−(√A))/2))=1−((((√A)−(√B))^2 )/4) Continuing, =(√({((((√3)+(√5))^2 )/4)−1}{1−((((√3)−(√5))^2 )/4)}))=(√(−{1−((((√3)+(√5))^2 )/4)}{1−((((√3)−(√5))^2 )/4)})) =(√(−[1−{((((√3)+(√5))^2 )/4)+((((√3)−(√5))^2 )/4)}+(({((√3)+(√5))((√3)−(√5))}^2 )/(16))])) =(√(−[1−((((√3)+(√5))^2 +((√3)−(√5))^2 )/4)+(({((√3))^2 −((√5))^2 }^2 )/(16))])) =(√(−{1−((16)/4)+(((−2)^2 )/(16))}))=(√(−(1−4+(1/4))))=(√(−(−((11)/4))))=(√((11)/4))=((√(11))/2)≈1.658 ✓
$${s}=\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{1}+\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx\mathrm{2}.\mathrm{984} \\ $$$${A}=\sqrt{{s}\left({s}−\sqrt{\mathrm{3}}\right)\left({s}−\mathrm{2}\right)\left({s}−\sqrt{\mathrm{5}}\right)}=\sqrt{\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{1}\right)\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} \\ $$$$\mathrm{Using}\:\mathrm{this}, \\ $$$$\left(\mathrm{1}+\frac{\sqrt{{A}}−\sqrt{{B}}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\sqrt{{B}}−\sqrt{{A}}}{\mathrm{2}}\right)=\mathrm{1}−\frac{\left(\sqrt{{A}}−\sqrt{{B}}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{Continuing}, \\ $$$$=\sqrt{\left\{\frac{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\right\}\left\{\mathrm{1}−\frac{\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}}=\sqrt{−\left\{\mathrm{1}−\frac{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}\left\{\mathrm{1}−\frac{\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}} \\ $$$$=\sqrt{−\left[\mathrm{1}−\left\{\frac{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}+\frac{\left\{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}\right)\right\}^{\mathrm{2}} }{\mathrm{16}}\right]} \\ $$$$=\sqrt{−\left[\mathrm{1}−\frac{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left\{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \right\}^{\mathrm{2}} }{\mathrm{16}}\right]} \\ $$$$=\sqrt{−\left\{\mathrm{1}−\frac{\mathrm{16}}{\mathrm{4}}+\frac{\left(−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{16}}\right\}}=\sqrt{−\left(\mathrm{1}−\mathrm{4}+\frac{\mathrm{1}}{\mathrm{4}}\right)}=\sqrt{−\left(−\frac{\mathrm{11}}{\mathrm{4}}\right)}=\sqrt{\frac{\mathrm{11}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{658}\:\checkmark \\ $$

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