Question Number 228013 by Mingma last updated on 09/Mar/26 Answered by A5T last updated on 14/Mar/26 $$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{AB}=\mathrm{1}\:\mathrm{and}\:\mathrm{BC}=\mathrm{x}\:;\:\mathrm{then}\:\mathrm{AC}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\bigtriangleup\mathrm{CBD}\approx\bigtriangleup\mathrm{CAB}\Rightarrow\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{DC}}{\mathrm{BC}}\:\Rightarrow\mathrm{DC}=\frac{\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{BD}=\sqrt{\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{x}^{\mathrm{4}}…
Question Number 228009 by Mingma last updated on 09/Mar/26 Answered by fantastic2 last updated on 09/Mar/26 Commented by fantastic2 last updated on 09/Mar/26 $$\angle{OAB}=\mathrm{20}^{\mathrm{0}} \Rightarrow\angle{AOB}=\mathrm{140}^{\mathrm{0}}…
Question Number 226626 by sankarsanmahanta097 last updated on 07/Dec/25 Terms of Service Privacy Policy Contact: [email protected]
Question Number 224608 by fantastic last updated on 21/Sep/25 Commented by fantastic last updated on 21/Sep/25 $$\measuredangle{EFD}=? \\ $$ Answered by som(math1967) last updated on…
Question Number 224324 by fantastic last updated on 03/Sep/25 Answered by fantastic last updated on 03/Sep/25 Commented by fantastic last updated on 05/Sep/25 $${Area}\:{EFHJ}={AGHJE}−{AGHF}−{AEF} \\…
Question Number 224282 by fantastic last updated on 31/Aug/25 Commented by fantastic last updated on 31/Aug/25 $${ABCD}\:{is}\:{a}\:{square}. \\ $$$${What}\:{is}\:{the}\:{total}\:{area}\:{of}\: \\ $$$${two}\:{semicircles} \\ $$ Answered by…
Question Number 224255 by fantastic last updated on 30/Aug/25 Commented by fantastic last updated on 30/Aug/25 $$\bigtriangleup{ABC}=? \\ $$ Commented by fantastic last updated on…
Question Number 224176 by fantastic last updated on 23/Aug/25 Commented by fantastic last updated on 23/Aug/25 $${whats}\:{the}\:{green}\:{area}? \\ $$$$ \\ $$$${please}\:{give}\:{me}\:{a}\:{detailed}\:{solution}\: \\ $$ Answered by…
Question Number 224121 by fantastic last updated on 21/Aug/25 Answered by mr W last updated on 21/Aug/25 $${R}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{10}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{10}}}−\frac{\mathrm{1}}{\mathrm{10}}−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{15} \\ $$$$\pi{R}^{\mathrm{2}} =\mathrm{225}\pi\:\checkmark \\ $$ Commented by…
Question Number 224122 by fantastic last updated on 21/Aug/25 Commented by fantastic last updated on 21/Aug/25 $${Three}\:{circles}\:{in}\:{a}\:{big}\:{circle} \\ $$$${A}\:{pointed}\:{circle}\:{r}_{\mathrm{1}} \\ $$$${B}\:{pointed}\:{circle}\:{r}_{\mathrm{2}} \\ $$$${C}\:{pointed}\:{circle}\:{r}_{\mathrm{3}} \\ $$$${R}=?…