Question Number 211340 by Tawa11 last updated on 06/Sep/24
Commented by Frix last updated on 06/Sep/24
$$\mathrm{Question}\:\mathrm{211294} \\ $$
Answered by mr W last updated on 06/Sep/24
$${s}=\frac{{a}+{b}+{c}+{d}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}}{\mathrm{2}}=\mathrm{5} \\ $$$$\left[{ABCD}\right]=\sqrt{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)\left({s}−{d}\right)−{abcd}\:\mathrm{cos}^{\mathrm{2}} \:\left(\frac{{x}+{y}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:=\sqrt{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}−\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{60}°} \\ $$$$\:\:\:\:=\sqrt{\mathrm{24}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)}=\mathrm{3}\sqrt{\mathrm{2}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 06/Sep/24
$$\mathrm{Wow}. \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
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
$${see}\:{also}\:{Q}\mathrm{30233} \\ $$
Answered by mahdipoor last updated on 06/Sep/24
$${BD}^{\mathrm{2}} ={a}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ad}.{cosx}={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}.{cos}\left(\mathrm{120}−{x}\right) \\ $$$$\Rightarrow\mathrm{17}−\mathrm{8}{cosx}=\mathrm{13}−\mathrm{12}{cos}\left(\mathrm{120}−{x}\right) \\ $$$$\Rightarrow\mathrm{4}=\mathrm{8}{cosx}−\mathrm{12}\left(−\frac{\mathrm{1}}{\mathrm{2}}{cosx}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sinx}\right) \\ $$$$\Rightarrow\mathrm{2}=\mathrm{7}{c}−\mathrm{3}\sqrt{\mathrm{3}}{s}\:\:\:\:\:{eq}\:\:\:{i} \\ $$$${S}=\frac{{ad}}{\mathrm{2}}{sin}\left({x}\right)+\frac{{bc}}{\mathrm{2}}{sin}\left(\mathrm{120}−{x}\right)= \\ $$$$\mathrm{2}{sin}\left({x}\right)+\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left({x}\right)\right)= \\ $$$$\Rightarrow\mathrm{2}{S}={s}+\mathrm{3}\sqrt{\mathrm{3}}{c}\:\:\:\:\:\:\:\:{eq}\:\:{j} \\ $$$${i}\:,\:{j}\:\Rightarrow\begin{cases}{{c}=\frac{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}{S}}{\mathrm{17}}}\\{{s}=\frac{\mathrm{7}{S}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{17}}}\end{cases}\:\Rightarrow\: \\ $$$$\mathrm{1}={c}^{\mathrm{2}} +{s}^{\mathrm{2}} =\frac{\mathrm{28}+\mathrm{76}{S}^{\mathrm{2}} −\mathrm{36}\sqrt{\mathrm{3}}{S}}{\mathrm{17}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{76}{S}^{\mathrm{2}} −\mathrm{36}\sqrt{\mathrm{3}}{S}−\mathrm{261}=\mathrm{0}\Rightarrow{S}=\frac{\mathrm{9}\sqrt{\mathrm{3}}+\mathrm{8}\sqrt{\mathrm{129}}}{\mathrm{38}}\:!? \\ $$
Commented by Frix last updated on 06/Sep/24
$${S}=\mathrm{2sin}+\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{2}{S}=\mathrm{7}{s}+\mathrm{3}\sqrt{\mathrm{3}}{c} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{{c}=\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{3}}{S}}{\mathrm{38}}}\\{{s}=\frac{\mathrm{7}{S}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{38}}}\end{cases} \\ $$$$\Rightarrow\:{S}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$