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