Category: Mensuration

Question Number 188078 by Rupesh123 last updated on 25/Feb/23 Answered by a.lgnaoui last updated on 25/Feb/23 $${from}\:{Graphe}:\bigtriangleup{ABC}\:\:{Equilaterale} \\ $$$${big}\:{circle}\left({radius}\:{OF}=\sqrt{\mathrm{4}+\frac{{EF}^{\mathrm{2}} }{\mathrm{4}}\:}\:\right) \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}}{{OF}}\Rightarrow\left({OF}=\mathrm{4};{EF}=\mathrm{4}\sqrt{\mathrm{3}.}\:\:\right) \\ $$$${Area}\left({big}\:{circleC}_{{B}} =\mathrm{16}\pi\right)…

Question Number 122443 by ajfour last updated on 17/Nov/20 Commented by ajfour last updated on 17/Nov/20 $${Find}\:\left(\frac{{r}}{{R}}\right)_{{max}} . \\ $$ Answered by ajfour last updated…

Question Number 187979 by Mingma last updated on 24/Feb/23 Answered by HeferH last updated on 24/Feb/23 $$\mathrm{80}°\:=\:\mathrm{40}°\:+\:{x} \\ $$$$\mathrm{40}°\:=\:{x} \\ $$ Terms of Service Privacy…

Question Number 187953 by Mingma last updated on 24/Feb/23 Answered by som(math1967) last updated on 24/Feb/23 $$\bigtriangleup{AOB}=\bigtriangleup{AOC}=\bigtriangleup{AOC}=\frac{\mathrm{1}}{\mathrm{3}}\bigtriangleup{ABC} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{48}=\mathrm{16}{cm}^{\mathrm{2}} \\ $$$$\bigtriangleup{AOM}=\frac{\mathrm{1}}{\mathrm{2}}\bigtriangleup{AOB}=\mathrm{8}{cm}^{\mathrm{2}} \\ $$$$\left[{OM}\:{is}\:{median}\:{of}\bigtriangleup{AOB}\right] \\ $$$$\bigtriangleup{AOK}=\frac{\mathrm{1}}{\mathrm{2}}\bigtriangleup{AOC}=\mathrm{8}{cm}^{\mathrm{2}}…

Question Number 187932 by Rupesh123 last updated on 23/Feb/23 Answered by a.lgnaoui last updated on 24/Feb/23 $${O}_{\mathrm{1}} :{centre}\:{du}\:{cercle}\left({C}_{\mathrm{1}} \right)=\:\:\:{origine}\:{Ref} \\ $$$${Cercles}\left(\:{C}_{\mathrm{1}} \:\right){et}\left({C}_{\mathrm{2}} \right){d}\:{equations}\: \\ $$$$\begin{cases}{{x}^{\mathrm{2}}…