Category: Geometry

Question Number 196225 by mr W last updated on 20/Aug/23 Commented by mr W last updated on 20/Aug/23 $${find}\:{the}\:{area}\:{of}\:{red}\:{part}\:{in}\:{the} \\ $$$${parallelogram}. \\ $$ Answered by…

Question Number 195896 by AROUNAMoussa last updated on 12/Aug/23 Answered by HeferH last updated on 13/Aug/23 Commented by HeferH last updated on 13/Aug/23 $$\mathrm{180}°−\mathrm{4}{x}\:+\:\mathrm{40}°+\mathrm{2}{x}\:=\:\mathrm{180}° \\…

Question Number 195880 by aawb_247 last updated on 12/Aug/23 Commented by aawb_247 last updated on 12/Aug/23 $${anyone}\:{can}\:{help}..\:{what}\:{a}\:{value}\:{of}\:{x}? \\ $$ Answered by mr W last updated…

Question Number 195740 by universe last updated on 09/Aug/23 Answered by Frix last updated on 09/Aug/23 $$\mathrm{Triangle}\:\Rightarrow\:{a}+{b}>{c}\wedge{a}+{c}>{b}\wedge{b}+{c}>{a} \\ $$$$\mathrm{Let}\:{b}=\left({u}−{v}\right){a}\wedge{c}=\left({u}+{v}\right){a} \\ $$$$\Rightarrow\:{u}>\frac{\mathrm{1}}{\mathrm{2}}\wedge−\frac{\mathrm{1}}{\mathrm{2}}<{v}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left({u},\:{v}\right)=\frac{\mathrm{1}}{\mathrm{2}{u}}+\frac{{u}−{v}}{{u}+{v}+\mathrm{1}}+\frac{{u}+{v}}{{u}−{v}+\mathrm{1}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant{f}\left({u},\:{v}\right)<\mathrm{2}…

Question Number 195344 by Mingma last updated on 31/Jul/23 Answered by kapoorshah last updated on 31/Jul/23 $${cos}\:\alpha\:=\:{cos}\:\alpha \\ $$$$\frac{{a}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{a}.\mathrm{6}}\:=\:\frac{{b}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{b}.\mathrm{6}}…