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Question Number 211340 by Tawa11 last updated on 06/Sep/24

Commented by Frix last updated on 06/Sep/24

Question 211294

$$\mathrm{Question}\:\mathrm{211294} \\ $$

Answered by mr W last updated on 06/Sep/24

s=((a+b+c+d)/2)=((1+2+3+4)/2)=5 [ABCD]=(√((s−a)(s−b)(s−c)(s−d)−abcd cos^2 (((x+y)/2)))) =(√(4×3×2×1−1×2×3×4 cos^2 60°)) =(√(24×(1−(1/4))))=3(√2) ✓

$${s}=\frac{{a}+{b}+{c}+{d}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}}{\mathrm{2}}=\mathrm{5} \\ $$$$\left[{ABCD}\right]=\sqrt{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)\left({s}−{d}\right)−{abcd}\:\mathrm{cos}^{\mathrm{2}} \:\left(\frac{{x}+{y}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:=\sqrt{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}−\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{60}°} \\ $$$$\:\:\:\:=\sqrt{\mathrm{24}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)}=\mathrm{3}\sqrt{\mathrm{2}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 06/Sep/24

Wow. Thank you sir. I really appreciate.

$$\mathrm{Wow}. \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by mr W last updated on 06/Sep/24

Commented by mr W last updated on 06/Sep/24

Commented by mr W last updated on 06/Sep/24

see also Q30233

$${see}\:{also}\:{Q}\mathrm{30233} \\ $$

Answered by mahdipoor last updated on 06/Sep/24

BD^2 =a^2 +d^2 −2ad.cosx=b^2 +c^2 −2bc.cos(120−x) ⇒17−8cosx=13−12cos(120−x) ⇒4=8cosx−12(−(1/2)cosx+((√3)/2)sinx) ⇒2=7c−3(√3)s eq i S=((ad)/2)sin(x)+((bc)/2)sin(120−x)= 2sin(x)+3(((√3)/2)cos(x)−(1/2)sin(x))= ⇒2S=s+3(√3)c eq j i , j ⇒ { ((c=((1+3(√3)S)/(17)))),((s=((7S−3(√3))/(17)))) :} ⇒ 1=c^2 +s^2 =((28+76S^2 −36(√3)S)/(17^2 )) ⇒76S^2 −36(√3)S−261=0⇒S=((9(√3)+8(√(129)))/(38)) !?

$${BD}^{\mathrm{2}} ={a}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ad}.{cosx}={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}.{cos}\left(\mathrm{120}−{x}\right) \\ $$$$\Rightarrow\mathrm{17}−\mathrm{8}{cosx}=\mathrm{13}−\mathrm{12}{cos}\left(\mathrm{120}−{x}\right) \\ $$$$\Rightarrow\mathrm{4}=\mathrm{8}{cosx}−\mathrm{12}\left(−\frac{\mathrm{1}}{\mathrm{2}}{cosx}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sinx}\right) \\ $$$$\Rightarrow\mathrm{2}=\mathrm{7}{c}−\mathrm{3}\sqrt{\mathrm{3}}{s}\:\:\:\:\:{eq}\:\:\:{i} \\ $$$${S}=\frac{{ad}}{\mathrm{2}}{sin}\left({x}\right)+\frac{{bc}}{\mathrm{2}}{sin}\left(\mathrm{120}−{x}\right)= \\ $$$$\mathrm{2}{sin}\left({x}\right)+\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left({x}\right)\right)= \\ $$$$\Rightarrow\mathrm{2}{S}={s}+\mathrm{3}\sqrt{\mathrm{3}}{c}\:\:\:\:\:\:\:\:{eq}\:\:{j} \\ $$$${i}\:,\:{j}\:\Rightarrow\begin{cases}{{c}=\frac{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}{S}}{\mathrm{17}}}\\{{s}=\frac{\mathrm{7}{S}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{17}}}\end{cases}\:\Rightarrow\: \\ $$$$\mathrm{1}={c}^{\mathrm{2}} +{s}^{\mathrm{2}} =\frac{\mathrm{28}+\mathrm{76}{S}^{\mathrm{2}} −\mathrm{36}\sqrt{\mathrm{3}}{S}}{\mathrm{17}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{76}{S}^{\mathrm{2}} −\mathrm{36}\sqrt{\mathrm{3}}{S}−\mathrm{261}=\mathrm{0}\Rightarrow{S}=\frac{\mathrm{9}\sqrt{\mathrm{3}}+\mathrm{8}\sqrt{\mathrm{129}}}{\mathrm{38}}\:!? \\ $$

Commented by Frix last updated on 06/Sep/24

S=2sin+3(((√3)/2)cos x +(1/2)sin x) 2S=7s+3(√3)c ⇒ { ((c=((7+3(√3)S)/(38)))),((s=((7S+3(√3))/(38)))) :} ⇒ S=3(√2)

$${S}=\mathrm{2sin}+\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{2}{S}=\mathrm{7}{s}+\mathrm{3}\sqrt{\mathrm{3}}{c} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{{c}=\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{3}}{S}}{\mathrm{38}}}\\{{s}=\frac{\mathrm{7}{S}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{38}}}\end{cases} \\ $$$$\Rightarrow\:{S}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$

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