Question Number 204081 by esmaeil last updated on 05/Feb/24
$${what}\:{is}\:{the}\:{area}\:{of}\:{the}\:{largest}\:{square} \\ $$$${that}\:{can}\:{be}\:{enclosed}\:{in}\:{a}\:{triangle} \\ $$$${with}\:{an}\:{area}\:{of}\:\mathrm{1}? \\ $$
Answered by mr W last updated on 06/Feb/24
Commented by mr W last updated on 06/Feb/24
$${area}\:{of}\:{ABC}=\Delta=\mathrm{1} \\ $$$$\Delta=\frac{{ah}_{{a}} }{\mathrm{2}} \\ $$$$\frac{{s}_{{a}} }{{a}}=\frac{{h}_{{a}} −{s}_{{a}} }{{h}_{{a}} } \\ $$$${s}_{{a}} =\frac{{ah}_{{a}} }{{a}+{h}_{{a}} }=\frac{\mathrm{2}\Delta}{{a}+\frac{\mathrm{2}\Delta}{{a}}}\leqslant\frac{\mathrm{2}\Delta}{\mathrm{2}\sqrt{\mathrm{2}\Delta}}=\sqrt{\frac{\Delta}{\mathrm{2}}} \\ $$$${s}_{{a}} ^{\mathrm{2}} \leqslant\frac{\Delta}{\mathrm{2}} \\ $$$$\Rightarrow\:\left({s}_{{a}} ^{\mathrm{2}} \right)_{{max}} =\frac{\Delta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${when}\:{a}=\frac{\mathrm{2}\Delta}{{a}},\:{i}.{e}.\:{a}=\sqrt{\mathrm{2}\Delta}={h}_{{a}} \\ $$$${similarly}\:\left({s}_{{b}} ^{\mathrm{2}} \right)_{{max}} =\frac{\Delta}{\mathrm{2}},\:\left({s}_{{c}} ^{\mathrm{2}} \right)_{{max}} =\frac{\Delta}{\mathrm{2}}. \\ $$$$\Rightarrow{area}\:{of}\:{largest}\:{inscribed}\:{square}\: \\ $$$${is}\:\frac{\Delta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$
Commented by esmaeil last updated on 06/Feb/24
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$