Question-121802

Question Number 121802 by prakash jain last updated on 11/Nov/20

Commented by prakash jain last updated on 11/Nov/20

ajfour sir, i cannot find the original question but i got x∉R. Please check. I will also recheck. Thanks.

$$\mathrm{ajfour}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{cannot}\:\mathrm{find}\:\mathrm{the}\:\mathrm{original} \\ $$$$\mathrm{question}\:\mathrm{but}\:\mathrm{i}\:\mathrm{got}\:{x}\notin\mathbb{R}. \\ $$$$\mathrm{Please}\:\mathrm{check}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{also}\:\mathrm{recheck}. \\ $$$$\mathrm{Thanks}. \\ $$

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

All colored regions have equal area ((AB×AD)/5) = (a/5) If A(0,0) is the origin, E(x,y) and F(z,0) : −pink area = (1/2)(1×x) = (a/5) ⇒ x = (2/5)a −blue area = (1/2)(a×(1−y)) = (a/5) ⇒ y = (3/5) −yellow area = (1/2)(z×x) = (a/5) ⇒ z = 1 −brown area = (1/2)((a−1)×1) = (a/5) ⇒ a = (5/3) AB/AD = a = (5/3)

$$\mathrm{All}\:\mathrm{colored}\:\mathrm{regions}\:\mathrm{have}\:\mathrm{equal} \\ $$$$\mathrm{area}\:\frac{\mathrm{AB}×\mathrm{AD}}{\mathrm{5}}\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\mathrm{If}\:\mathrm{A}\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{E}\left({x},{y}\right)\:\mathrm{and}\:\mathrm{F}\left({z},\mathrm{0}\right)\:: \\ $$$$−\mathrm{pink}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}×{x}\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{x}\:=\:\frac{\mathrm{2}}{\mathrm{5}}{a} \\ $$$$−\mathrm{blue}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}×\left(\mathrm{1}−{y}\right)\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{y}\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$−\mathrm{yellow}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({z}×{x}\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{z}\:=\:\mathrm{1} \\ $$$$−\mathrm{brown}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\left({a}−\mathrm{1}\right)×\mathrm{1}\right)\:=\:\frac{{a}}{\mathrm{5}} \\ $$$$\Rightarrow\:{a}\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{AB}/\mathrm{AD}\:=\:{a}\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$

Commented by prakash jain last updated on 12/Nov/20

E=(x,y) Please check. Yellow area =(1/2)z×y? (1/2)z×(3/5)=(a/5)⇒z=((2a)/3) brown area (1/2)×(1−((2a)/3))=(a/6)≠(a/5)

$$\mathrm{E}=\left({x},{y}\right) \\ $$$$\mathrm{Please}\:\mathrm{check}. \\ $$$$\mathrm{Yellow}\:\mathrm{area}\:=\frac{\mathrm{1}}{\mathrm{2}}{z}×{y}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{z}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{{a}}{\mathrm{5}}\Rightarrow{z}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\mathrm{brown}\:\mathrm{area}\:\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{1}−\frac{\mathrm{2}{a}}{\mathrm{3}}\right)=\frac{{a}}{\mathrm{6}}\neq\frac{{a}}{\mathrm{5}} \\ $$

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