Question Number 53071 by behi83417@gmail.com last updated on 16/Jan/19 Commented by behi83417@gmail.com last updated on 16/Jan/19 $${as}\:{shown}\:{in}\:{fig}: \\ $$$${prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\rightarrow\:\frac{\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{of}}\:\:{outer}\:\:\boldsymbol{\mathrm{hexagon}}}{\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{of}}\:\:{inner}\:\boldsymbol{\mathrm{triangle}}}\:\:\:\geqslant\:\mathrm{13} \\ $$ Commented by…
Question Number 53066 by ajfour last updated on 16/Jan/19 Commented by ajfour last updated on 17/Jan/19 $${Determine}\:{radius}\:{R}\:{of}\:{larger}\:{circle} \\ $$$${in}\:{terms}\:{of}\:{a},{b},{c}\:.\:\:\:\: \\ $$ Commented by ajfour last…
Question Number 184135 by ajfour last updated on 03/Jan/23 Commented by mr W last updated on 03/Jan/23 $${PQ}_{{min}} ={b}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)} \\ $$ Answered by…
Question Number 184105 by Acem last updated on 02/Jan/23 $$\:{We}\:{cut}\:{a}\:{square}\:{sheet}\:{of}\:{metal}\:{into}\:{equal}\:{thirteen} \\ $$$$\:{cross}\:{slices},\:{we}\:{measured}\:{its}\:{perimeter}\:{of}\:\mathrm{168}\:{cm} \\ $$$$\:{How}\:{much}\:{was}\:{the}\:{perimeter}\:{of}\:{the}\:{plate} \\ $$$$\:{before}\:{cutting}? \\ $$ Answered by manxsol last updated on 03/Jan/23…
Question Number 53031 by behi83417@gmail.com last updated on 16/Jan/19 Commented by behi83417@gmail.com last updated on 16/Jan/19 $${AB}<{CB}. \\ $$$$\measuredangle{ABE}=\measuredangle{EBF}=\measuredangle{FBC} \\ $$$$……{show}\:{that}:\:\:\:\:\frac{\boldsymbol{\mathrm{BE}}}{\boldsymbol{\mathrm{BF}}}\:>\mathrm{1}\:\:. \\ $$ Answered by…
Question Number 118545 by I want to learn more last updated on 18/Oct/20 Commented by Tawa11 last updated on 15/Sep/21 $$\mathrm{nice} \\ $$ Answered by…
Question Number 184041 by Acem last updated on 02/Jan/23 Answered by HeferH last updated on 02/Jan/23 $$\overset{\frown} {{TN}}\:=\:\frac{\pi}{\mathrm{3}}\:=\:\frac{\mathrm{180}°}{\mathrm{3}}\:=\:\mathrm{60}°\:\:\Rightarrow\:\angle{TAN}\:=\:\frac{\overset{\frown} {{TN}}}{\mathrm{2}}\:=\:\mathrm{30}° \\ $$$$\:\angle{AOT}\:=\:\mathrm{120}°\:\left(\bigtriangleup{AOT}\:{isosceles}\right)\: \\ $$$$\:\angle{ATC}\:=\:\frac{\overset{\frown} {{AT}}}{\mathrm{2}}\:=\:\mathrm{60}\:\:\Rightarrow\:\angle{ACT}\:=\:\mathrm{90}°\: \\…
Question Number 52936 by ajfour last updated on 15/Jan/19 Commented by ajfour last updated on 15/Jan/19 $${base}\:{line}\:{is}\:{of}\:{length}\:\boldsymbol{{b}}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 183991 by HeferH last updated on 01/Jan/23 Commented by HeferH last updated on 01/Jan/23 $${If}\:“{I}''\:{is}\:{the}\:{incenter},\:{find}\:{x} \\ $$ Answered by mr W last updated…
Question Number 52911 by ajfour last updated on 15/Jan/19 Commented by ajfour last updated on 15/Jan/19 $${let}\:{for}\:{example}:\:\:{q}=\mathrm{5},\:{p}=\mathrm{7},\:{r}=\mathrm{4}\:. \\ $$ Commented by MJS last updated on…