Question Number 164154 by HongKing last updated on 14/Jan/22
$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{xy}\:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{9}}\\{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{yz}\:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{16}}\\{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{xz}\:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{25}}\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{S}\:=\:\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{xz} \\ $$
Answered by behi834171 last updated on 14/Jan/22
$${p}=\frac{\mathrm{3}+\mathrm{4}+\mathrm{5}}{\mathrm{2}}=\mathrm{6} \\ $$$${A}=\sqrt{\mathrm{6}×\mathrm{1}×\mathrm{2}×\mathrm{3}}=\mathrm{6} \\ $$$$\mathrm{6}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\boldsymbol{{xy}}+\boldsymbol{{yz}}+\boldsymbol{{zx}}\right)\Rightarrow \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{xy}}+\boldsymbol{{yz}}+\boldsymbol{{zx}}=\mathrm{8}\sqrt{\mathrm{3}}\:\:\:.\blacksquare \\ $$
Commented by mr W last updated on 15/Jan/22
$${very}\:{nice}\:{solution}! \\ $$
Commented by HongKing last updated on 15/Jan/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{Sir} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jan/22
$$\mathcal{E}{xplain}\:\mathrm{3}{rd}\:{line}\:{please}! \\ $$
Commented by mr W last updated on 15/Jan/22
$${geometry}\:{meaning}\:{of}\:{the}\:{solution} \\ $$$${see}\:{Q}\mathrm{164174} \\ $$
Commented by Tawa11 last updated on 15/Jan/22
$$\mathrm{Nice}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jan/22
$$\mathcal{T}{han}\mathcal{X}\:{mr}\:{W}\:\boldsymbol{{sir}}! \\ $$
Commented by mr W last updated on 15/Jan/22
$${we}\:{get}\:{further} \\ $$$${x}=\frac{\mathrm{4}\sqrt{\mathrm{6}}×\mathrm{6}+\mathrm{3}\sqrt{\mathrm{2}}\left(−\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \right)}{\mathrm{6}\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}×\mathrm{6}}}=\frac{\mathrm{16}\sqrt{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{50}+\mathrm{24}\sqrt{\mathrm{3}}}} \\ $$$${y}=\frac{\mathrm{4}\sqrt{\mathrm{6}}×\mathrm{6}+\mathrm{3}\sqrt{\mathrm{2}}\left(\mathrm{3}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \right)}{\mathrm{6}\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}×\mathrm{6}}}=\frac{\mathrm{9}\sqrt{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{50}+\mathrm{24}\sqrt{\mathrm{3}}}} \\ $$$${z}=\frac{\mathrm{4}\sqrt{\mathrm{6}}×\mathrm{6}+\mathrm{3}\sqrt{\mathrm{2}}\left(\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} \right)}{\mathrm{6}\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}×\mathrm{6}}}=\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{50}+\mathrm{24}\sqrt{\mathrm{3}}}} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jan/22
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