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Question-179440




Question Number 179440 by mr W last updated on 29/Oct/22
Commented by mr W last updated on 29/Oct/22
find the largest equilateral triangle  which can be placed inside a square  with side length 1.
$${find}\:{the}\:{largest}\:{equilateral}\:{triangle} \\ $$$${which}\:{can}\:{be}\:{placed}\:{inside}\:{a}\:{square} \\ $$$${with}\:{side}\:{length}\:\mathrm{1}. \\ $$
Commented by Beginner last updated on 29/Oct/22
Approx 1.0353
$${Approx}\:\mathrm{1}.\mathrm{0353} \\ $$
Commented by nikif99 last updated on 29/Oct/22
x^2 =a^2 +1^2 =b^2 +1^2  ⇒a=b ⇒ϑ=15° ⇒  a=(tan 15)/1=2−(√3)  x=(√((2−(√3))^2 +1))=2(√(2−(√3)))=1.0353 approx.
$${x}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} ={b}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \:\Rightarrow{a}={b}\:\Rightarrow\vartheta=\mathrm{15}°\:\Rightarrow \\ $$$${a}=\left(\mathrm{tan}\:\mathrm{15}\right)/\mathrm{1}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${x}=\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{1}}=\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}=\mathrm{1}.\mathrm{0353}\:{approx}. \\ $$
Commented by nikif99 last updated on 29/Oct/22
Commented by mr W last updated on 29/Oct/22
thanks sirs!
$${thanks}\:{sirs}! \\ $$

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