Menu Close

Question-28597




Question Number 28597 by beh.i83417@gmail.com last updated on 27/Jan/18
Commented by beh.i83417@gmail.com last updated on 27/Jan/18
also largest square ,inscribed in  equilateral triangle of side=1  unit.
$$\boldsymbol{{also}}\:\boldsymbol{{largest}}\:\boldsymbol{{square}}\:,\boldsymbol{{inscribed}}\:\boldsymbol{{in}} \\ $$$$\boldsymbol{{equilateral}}\:\boldsymbol{{triangle}}\:\boldsymbol{{of}}\:\boldsymbol{{side}}=\mathrm{1}\:\:\boldsymbol{{unit}}. \\ $$
Answered by mrW2 last updated on 29/Jan/18
max. equilateral triangle:  a=(1/(cos 15°))=(4/( (√2)+(√6)))=(√6)−(√2)  A=((√3)/4)a^2 =((√3)/4)(8−4(√3))=2(√3)−3    max. square in equilateral triangle:  (b/1)=((((√3)/2)−b)/((√3)/2))=(((√3)−2b)/( (√3)))  ⇒b=((√3)/(2+(√3)))=2(√3)−3  A=b^2 =21−12(√3)
$${max}.\:{equilateral}\:{triangle}: \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{15}°}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}=\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}} \\ $$$${A}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{8}−\mathrm{4}\sqrt{\mathrm{3}}\right)=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3} \\ $$$$ \\ $$$${max}.\:{square}\:{in}\:{equilateral}\:{triangle}: \\ $$$$\frac{{b}}{\mathrm{1}}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{b}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{3}}−\mathrm{2}{b}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{b}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3} \\ $$$${A}={b}^{\mathrm{2}} =\mathrm{21}−\mathrm{12}\sqrt{\mathrm{3}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *