Category: Geometry

Question Number 180365 by Acem last updated on 11/Nov/22 Answered by greougoury555 last updated on 11/Nov/22 $${let}\:\angle\:{ABC}\:=\:\beta \\ $$$$\angle\:{AOC}=\:\mathrm{360}°−\mathrm{2}\beta \\ $$$$\:\mathrm{2}\alpha\:+\mathrm{360}°−\mathrm{2}\beta\:=\:\mathrm{180} \\ $$$$\:\Leftrightarrow\:\beta−\alpha\:=\:\frac{\mathrm{360}°−\mathrm{180}°}{\mathrm{2}}\:=\mathrm{90}° \\ $$…

Question Number 49280 by mr W last updated on 05/Dec/18 Commented by mr W last updated on 05/Dec/18 $${The}\:{distances}\:{from}\:{its}\:{centroid}\:{to} \\ $$$${its}\:{vertices}\:{are}\:{given}.\:{Find}\:{the}\:{area} \\ $$$${of}\:{the}\:{triangle}. \\ $$…

Question Number 180300 by mnjuly1970 last updated on 10/Nov/22 Answered by Acem last updated on 11/Nov/22 Commented by Acem last updated on 11/Nov/22 $${Sorry}\:{i}\:{haven}'{t}\:{a}\:{compass} \\…

Question Number 180257 by mr W last updated on 09/Nov/22 Commented by HeferH last updated on 09/Nov/22 Answered by HeferH last updated on 09/Nov/22 $${x}\:−\:\mathrm{60}°\:=\:\mathrm{45}°…

Question Number 180258 by mr W last updated on 09/Nov/22 Commented by mr W last updated on 09/Nov/22 $${find}\:{the}\:{area}\:{of}\:{square}. \\ $$ Answered by HeferH last…