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Question Number 203480 by mr W last updated on 20/Jan/24
Commented by mr W last updated on 20/Jan/24
is it possible that the red lines divide the square into 4 parts with given areas? if yes, find the length of the red lines.
$${is}\:{it}\:{possible}\:{that}\:{the}\:{red}\:{lines}\:{divide} \\ $$$${the}\:{square}\:{into}\:\mathrm{4}\:{parts}\:{with}\:{given}\: \\ $$$${areas}?\:{if}\:{yes},\:{find}\:{the}\:{length}\:{of}\:{the} \\ $$$${red}\:{lines}. \\ $$
Commented by behi834171 last updated on 20/Jan/24
1)from the area of sqare: s^2 =20+30+40+70=160⇒s=4(√(10)) 2)from the area of trapeziuss, p:is the vertical part of section with, 20 area q:is the horizontal part of section with,20 area ((20p+40p)/2).s=20+40 ((20q+30q)/2).s=20+30 ((30q+70q)/2).s=30+70 ((40p+70p)/2).s=40+70 [𝚺(rhs)].s=2×320⇒4s.s=640⇒s=4(√(10)) from 1&2, yes. it is possible.
$$\left.\mathrm{1}\right){from}\:{the}\:{area}\:{of}\:{sqare}: \\ $$$${s}^{\mathrm{2}} =\mathrm{20}+\mathrm{30}+\mathrm{40}+\mathrm{70}=\mathrm{160}\Rightarrow\boldsymbol{{s}}=\mathrm{4}\sqrt{\mathrm{10}} \\ $$$$\left.\mathrm{2}\right){from}\:{the}\:{area}\:{of}\:{trapeziuss}, \\ $$$${p}:{is}\:{the}\:{vertical}\:{part}\:{of}\:{section}\:{with},\:\mathrm{20}\:{area} \\ $$$${q}:{is}\:{the}\:{horizontal}\:{part}\:{of}\:{section}\:{with},\mathrm{20}\:{area} \\ $$$$\frac{\mathrm{20}\boldsymbol{{p}}+\mathrm{40}\boldsymbol{{p}}}{\mathrm{2}}.\boldsymbol{{s}}=\mathrm{20}+\mathrm{40} \\ $$$$\frac{\mathrm{20}\boldsymbol{{q}}+\mathrm{30}\boldsymbol{{q}}}{\mathrm{2}}.\boldsymbol{{s}}=\mathrm{20}+\mathrm{30} \\ $$$$\frac{\mathrm{30}\boldsymbol{{q}}+\mathrm{70}\boldsymbol{{q}}}{\mathrm{2}}.\boldsymbol{{s}}=\mathrm{30}+\mathrm{70} \\ $$$$\frac{\mathrm{40}\boldsymbol{{p}}+\mathrm{70}\boldsymbol{{p}}}{\mathrm{2}}.\boldsymbol{{s}}=\mathrm{40}+\mathrm{70} \\ $$$$\left[\boldsymbol{\Sigma}\left({rhs}\right)\right].\boldsymbol{{s}}=\mathrm{2}×\mathrm{320}\Rightarrow\mathrm{4}\boldsymbol{{s}}.\boldsymbol{{s}}=\mathrm{640}\Rightarrow\boldsymbol{{s}}=\mathrm{4}\sqrt{\mathrm{10}} \\ $$$$\boldsymbol{{from}}\:\mathrm{1\&2},\:{yes}.\:{it}\:{is}\:{possible}. \\ $$
Answered by esmaeil last updated on 20/Jan/24
a^2 =160→a=4(√(10)) (((m+n)/2))×4(√(10))=50(Area of the top trapizoid)→ m+n=((5(√(10)))/2) rsinα=n−m<m+n<((5(√(10)))/2) α=(angle of red line with the horizon) r>a>4(√(10)) max(sinα)=1→rsinα≮((5(√(10)))/2)→ not possible
$${a}^{\mathrm{2}} =\mathrm{160}\rightarrow{a}=\mathrm{4}\sqrt{\mathrm{10}} \\ $$$$\left(\frac{{m}+{n}}{\mathrm{2}}\right)×\mathrm{4}\sqrt{\mathrm{10}}=\mathrm{50}\left({Area}\:{of}\:{the}\:\:{top}\:{trapizoid}\right)\rightarrow \\ $$$${m}+{n}=\frac{\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{2}} \\ $$$${rsin}\alpha={n}−{m}<{m}+{n}<\frac{\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{2}} \\ $$$$\alpha=\left({angle}\:{of}\:{red}\:{line}\:{with}\:{the}\:\right. \\ $$$$\left.{horizon}\right) \\ $$$${r}>{a}>\mathrm{4}\sqrt{\mathrm{10}} \\ $$$${max}\left({sin}\alpha\right)=\mathrm{1}\rightarrow{rsin}\alpha\nless\frac{\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{2}}\rightarrow \\ $$$${not}\:{possible} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 21/Jan/24
what if one area is not 70 but 50? 20, 30, 40 remain.
$${what}\:{if}\:{one}\:{area}\:{is}\:{not}\:\mathrm{70}\:{but}\:\mathrm{50}? \\ $$$$\mathrm{20},\:\mathrm{30},\:\mathrm{40}\:{remain}. \\ $$
Commented by mr W last updated on 21/Jan/24
Commented by esmaeil last updated on 21/Jan/24
a^2 =140 (((m+n)/2))×2(√(35))=50→m+n=((10(√(35)))/7)→ n−m<((10(√(35)))/7) sinα=((n−m)/r)→rsinα=n−m<((10(√(35)))/7) ≈8.4515 but r>(√(140))=a≈12 i think only for(α=0)→ m=n=(a/2) is possible
$${a}^{\mathrm{2}} =\mathrm{140} \\ $$$$\left(\frac{{m}+{n}}{\mathrm{2}}\right)×\mathrm{2}\sqrt{\mathrm{35}}=\mathrm{50}\rightarrow{m}+{n}=\frac{\mathrm{10}\sqrt{\mathrm{35}}}{\mathrm{7}}\rightarrow \\ $$$${n}−{m}<\frac{\mathrm{10}\sqrt{\mathrm{35}}}{\mathrm{7}} \\ $$$${sin}\alpha=\frac{{n}−{m}}{{r}}\rightarrow{rsin}\alpha={n}−{m}<\frac{\mathrm{10}\sqrt{\mathrm{35}}}{\mathrm{7}} \\ $$$$\approx\mathrm{8}.\mathrm{4515} \\ $$$${but}\:\:{r}>\sqrt{\mathrm{140}}={a}\approx\mathrm{12} \\ $$$${i}\:{think}\:{only}\:{for}\left(\alpha=\mathrm{0}\right)\rightarrow \\ $$$${m}={n}=\frac{{a}}{\mathrm{2}}\:{is}\:{possible} \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 21/Jan/24
Commented by mr W last updated on 21/Jan/24
side length of square =s s^2 =20+30+40+70 ⇒s=(√(20+30+40+70)) xy−((x^2 tan θ)/2)+((y^2 tan θ)/2)=20 2xy−(x^2 −y^2 )tan θ=40 ...(i) similarly 2y(s−x)−[y^2 −(s−x)^2 ]tan θ=60 ...(ii) 2x(s−y)−[(s−y)^2 −x^2 ]tan θ=80 ...(iii) from (i): tan θ=((2xy−40)/(x^2 −y^2 )) put this into (ii) and (iii): 2y(s−x)−[y^2 −(s−x)^2 ](((2xy−40)/(x^2 −y^2 )))=60 ...(I) 2x(s−y)−[(s−y)^2 −x^2 ](((2xy−40)/(x^2 −y^2 )))=80 ...(II) if the area of the fourth part is 70, there is no solution for the equation system. if the fourth part is 60, we get a solution: x≈4.3225, y≈5.0431, θ≈−28.06°
$${side}\:{length}\:{of}\:{square}\:={s} \\ $$$${s}^{\mathrm{2}} =\mathrm{20}+\mathrm{30}+\mathrm{40}+\mathrm{70} \\ $$$$\Rightarrow{s}=\sqrt{\mathrm{20}+\mathrm{30}+\mathrm{40}+\mathrm{70}} \\ $$$${xy}−\frac{{x}^{\mathrm{2}} \mathrm{tan}\:\theta}{\mathrm{2}}+\frac{{y}^{\mathrm{2}} \mathrm{tan}\:\theta}{\mathrm{2}}=\mathrm{20} \\ $$$$\mathrm{2}{xy}−\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\mathrm{tan}\:\theta=\mathrm{40}\:\:\:…\left({i}\right) \\ $$$${similarly} \\ $$$$\mathrm{2}{y}\left({s}−{x}\right)−\left[{y}^{\mathrm{2}} −\left({s}−{x}\right)^{\mathrm{2}} \right]\mathrm{tan}\:\theta=\mathrm{60}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{2}{x}\left({s}−{y}\right)−\left[\left({s}−{y}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right]\mathrm{tan}\:\theta=\mathrm{80}\:\:\:…\left({iii}\right) \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{xy}−\mathrm{40}}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$${put}\:{this}\:{into}\:\left({ii}\right)\:{and}\:\left({iii}\right): \\ $$$$\mathrm{2}{y}\left({s}−{x}\right)−\left[{y}^{\mathrm{2}} −\left({s}−{x}\right)^{\mathrm{2}} \right]\left(\frac{\mathrm{2}{xy}−\mathrm{40}}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }\right)=\mathrm{60}\:\:\:…\left({I}\right) \\ $$$$\mathrm{2}{x}\left({s}−{y}\right)−\left[\left({s}−{y}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right]\left(\frac{\mathrm{2}{xy}−\mathrm{40}}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }\right)=\mathrm{80}\:\:\:…\left({II}\right) \\ $$$${if}\:{the}\:{area}\:{of}\:{the}\:{fourth}\:{part}\:{is}\:\mathrm{70}, \\ $$$${there}\:{is}\:{no}\:{solution}\:{for}\:{the}\:{equation} \\ $$$${system}.\:{if}\:{the}\:{fourth}\:{part}\:{is}\:\mathrm{60},\:{we} \\ $$$${get}\:{a}\:{solution}: \\ $$$${x}\approx\mathrm{4}.\mathrm{3225},\:{y}\approx\mathrm{5}.\mathrm{0431},\:\theta\approx−\mathrm{28}.\mathrm{06}° \\ $$
Commented by mr W last updated on 21/Jan/24
Answered by ajfour last updated on 21/Jan/24
Commented by ajfour last updated on 21/Jan/24
let 2s=a+b=c+d=(√(P+Q+R+S)) tan θ=2t Q−a^2 t+c^2 t=ac P−c^2 t+b^2 t=bc S−b^2 t+d^2 t=bd R−d^2 t+a^2 t=da
$${let}\:\:\mathrm{2}{s}={a}+{b}={c}+{d}=\sqrt{{P}+{Q}+{R}+{S}} \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{t} \\ $$$${Q}−{a}^{\mathrm{2}} {t}+{c}^{\mathrm{2}} {t}={ac} \\ $$$${P}−{c}^{\mathrm{2}} {t}+{b}^{\mathrm{2}} {t}={bc} \\ $$$${S}−{b}^{\mathrm{2}} {t}+{d}^{\mathrm{2}} {t}={bd} \\ $$$${R}−{d}^{\mathrm{2}} {t}+{a}^{\mathrm{2}} {t}={da} \\ $$

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