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Category: Geometry

Question-223962

Question Number 223962 by behi834171 last updated on 11/Aug/25 Commented by behi834171 last updated on 11/Aug/25 $$\boldsymbol{{Area}}\:\boldsymbol{{of}}\:\boldsymbol{{circles}}: \\ $$$$\begin{cases}{\boldsymbol{{yellow}}=\mathrm{1}}\\{\boldsymbol{{blue}}=\sqrt{\mathrm{2}}}\\{\boldsymbol{{green}}=\sqrt{\mathrm{3}}}\end{cases}\:\:\:\:\:\:\left(\boldsymbol{{any}}\:\boldsymbol{{unit}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{Area}}\:\boldsymbol{{of}}\:\boldsymbol{{gray}}\left(\boldsymbol{{big}}\right)\:\:\boldsymbol{{circle}}=? \\ $$$$ \\ $$$$\boldsymbol{{circles}}\:\boldsymbol{{and}}\:\boldsymbol{{lines}}\:\boldsymbol{{are}}\:\boldsymbol{{tangent}}\:\boldsymbol{{to}}…

Question-223908

Question Number 223908 by yerlow3 last updated on 09/Aug/25 Answered by dionigi last updated on 10/Aug/25 $$ \\ $$$${ABCD}\:{should}\:{be}\:{a}\:{regular}\:{hemi}−{hexagon} \\ $$$${r}\:=\:{PA}\:=\:{PB}\:=\:{PC}\:=\:{PD} \\ $$$${r}\:=\:{DA}\:=\:{AB}\:=\:{BC} \\ $$$${and}\:{P}\:{the}\:{center}\:{of}\:{thr}\:{circle}…

Question-223865

Question Number 223865 by ajfour last updated on 07/Aug/25 Commented by ajfour last updated on 07/Aug/25 $${Contrary}\:{to}\:{diagram}\:{labelling},\:{lets} \\ $$$${have}\:{a}=\mathrm{1},\:{b}={b}.\:\angle{CAE}=\pi/\mathrm{6}. \\ $$$${If}\:{APC}\:{is}\:{a}\:{straight}\:{line},\:{P}\:{being}\:{a} \\ $$$${point}\:{of}\:{tangency},\:{then}\:{find}\:{b},\:{r},\:{R}. \\ $$…

Question-223836

Question Number 223836 by Tawa11 last updated on 06/Aug/25 Answered by som(math1967) last updated on 06/Aug/25 $$\bigtriangleup{ABG}\sim\bigtriangleup{HFG} \\ $$$${let}\:{AB}={x} \\ $$$$\:\:\therefore\frac{{x}}{{x}+\mathrm{105}}=\frac{{x}−\mathrm{112}}{{x}−\mathrm{105}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{112}{x}+\mathrm{105}{x}−\mathrm{112}×\mathrm{105} \\…