Question Number 121802 by prakash jain last updated on 11/Nov/20
Commented by prakash jain last updated on 11/Nov/20
$$\mathrm{ajfour}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{cannot}\:\mathrm{find}\:\mathrm{the}\:\mathrm{original} \\ $$$$\mathrm{question}\:\mathrm{but}\:\mathrm{i}\:\mathrm{got}\:{x}\notin\mathbb{R}. \\ $$$$\mathrm{Please}\:\mathrm{check}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{also}\:\mathrm{recheck}. \\ $$$$\mathrm{Thanks}. \\ $$
Commented by ajfour last updated on 11/Nov/20
$$ \\ $$$${I}\:{dont}\:{know}\:{what}\:{to}\:{say},\:{Prakash}\:{Sir}, \\ $$$${understanding}\:{that}\:\:{the}\:{question} \\ $$$${is}\:{faulty},\:{i}\:{deleted}\:{it}. \\ $$$${Your}\:{method}\:{suggests}\:{me}\:{to}\:{amend} \\ $$$${the}\:{question}\:{this}\:{way}... \\ $$$${Consider}\:{only}\:{E}\:{above}\:{ground}, \\ $$$${such}\:{that}\:{the}\:{pink},\:{yellow},\:{blue}, \\ $$$${and}\:{green}\:{areas}\:{are}\:{triangular} \\ $$$${faces}\:{of}\:{a}\:{tent},\:{while}\:\bigtriangleup{BCF} \\ $$$${along}\:{ground},\:{and}\:{if}\:{each}\:{of} \\ $$$${these}\:{areas}\:{are}\:{equal},\:{find} \\ $$$$\left({x}_{{E}} \:,\:{y}_{{E}} \:,\:{z}_{{E}} \:\right)\:{taking}\:{A}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$$${and}\:+{x}\:{along}\:{AD},\:\:+{y}\:{along}\:{AB}. \\ $$$$+{z}\:{vertically}\:{upwards}\:{through}\:{A}. \\ $$$$ \\ $$
Answered by Olaf last updated on 11/Nov/20
$$\mathrm{All}\:\mathrm{colored}\:\mathrm{regions}\:\mathrm{have}\:\mathrm{equal} \\ $$$$\mathrm{area}\:\frac{\mathrm{AB}×\mathrm{AD}}{\mathrm{5}}\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\mathrm{If}\:\mathrm{A}\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{E}\left({x},{y}\right)\:\mathrm{and}\:\mathrm{F}\left({z},\mathrm{0}\right)\:: \\ $$$$−\mathrm{pink}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}×{x}\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{x}\:=\:\frac{\mathrm{2}}{\mathrm{5}}{a} \\ $$$$−\mathrm{blue}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}×\left(\mathrm{1}−{y}\right)\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{y}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$−\mathrm{yellow}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({z}×{x}\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{z}\:=\:\mathrm{1} \\ $$$$−\mathrm{brown}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\left({a}−\mathrm{1}\right)×\mathrm{1}\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{a}\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{AB}/\mathrm{AD}\:=\:{a}\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$
Commented by prakash jain last updated on 12/Nov/20
$$\mathrm{E}=\left({x},{y}\right) \\ $$$$\mathrm{Please}\:\mathrm{check}. \\ $$$$\mathrm{Yellow}\:\mathrm{area}\:=\frac{\mathrm{1}}{\mathrm{2}}{z}×{y}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{z}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{{a}}{\mathrm{5}}\Rightarrow{z}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\mathrm{brown}\:\mathrm{area}\:\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{1}−\frac{\mathrm{2}{a}}{\mathrm{3}}\right)=\frac{{a}}{\mathrm{6}}\neq\frac{{a}}{\mathrm{5}} \\ $$