Question Number 186551 by ajfour last updated on 05/Feb/23
Commented by mr W last updated on 06/Feb/23
$${square}\:{with}\:{side}\:{length}\:\mathrm{1}? \\ $$$${A}_{{shade}} =\frac{{t}^{\mathrm{2}} \left({t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+{t}\right)\left[\left(\mathrm{1}+{t}\right)^{\mathrm{2}} +\mathrm{1}\right]} \\ $$
Answered by ajfour last updated on 06/Feb/23
$${Left}\:{bottom}\:{corner}\:{be}\:{Origin}. \\ $$$${q}=\frac{{p}}{\mathrm{1}+{t}}\:\:\:{and}\:\:\:{q}=\left({t}+\mathrm{1}\right)\left(\mathrm{1}−{p}\right) \\ $$$$\Rightarrow\:\:{p}=\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{q}=\frac{{t}+\mathrm{1}}{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${k}=\frac{{t}+\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:;\:\:{h}={p}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Area}={kh} \\ $$$$\:\:\:=\left(\frac{{t}+\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\right)\left(\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:=\frac{\left\{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}}{\mathrm{2}\left({t}+\mathrm{1}\right)}×\frac{\left\{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$$${A}=\frac{\left\{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}^{\mathrm{2}} }{\mathrm{4}\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$$${A}=\frac{{t}^{\mathrm{2}} \left({t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$$$\:\:\:\: \\ $$
Answered by mr W last updated on 06/Feb/23
Commented by mr W last updated on 06/Feb/23
$${h}_{\mathrm{2}} =\frac{\mathrm{1}+{t}}{\mathrm{2}} \\ $$$${h}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{t}\right)} \\ $$$$\frac{\mathrm{1}−{b}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}+{t}}{\mathrm{1}}\:\Rightarrow\frac{\mathrm{1}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}+{t}}{\mathrm{1}−{b}} \\ $$$$\frac{{b}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{1}+{t}}\:\Rightarrow\frac{\mathrm{1}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}}{{b}\left(\mathrm{1}+{t}\right)} \\ $$$$\frac{\mathrm{1}+{t}}{\mathrm{1}−{b}}=\frac{\mathrm{1}}{{b}\left(\mathrm{1}+{t}\right)} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${A}_{{shade}} =\left({h}_{\mathrm{2}} −{h}_{\mathrm{1}} \right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{b}\right) \\ $$$$\:\:\:\:\:=\left[\frac{\mathrm{1}+{t}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{t}\right)}\right]\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$\:\:\:\:\:=\frac{{t}^{\mathrm{2}} \left({t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+{t}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$
Commented by ajfour last updated on 06/Feb/23
$${Thanks}\:{you}\:{Sir},\:{I}\:{corrected}! \\ $$