Category: Mensuration

Question Number 124743 by ajfour last updated on 05/Dec/20 Commented by ajfour last updated on 05/Dec/20 $${Given}\:\:{arc}\:{length}\:{AB}=\mathrm{1},\:{find} \\ $$$${maximum}\:{radius}\:{of}\:{smaller} \\ $$$${circles}. \\ $$ Answered by…

Question Number 189971 by Rupesh123 last updated on 25/Mar/23 Answered by ajfour last updated on 25/Mar/23 $${y}=\frac{\mathrm{3}}{\mathrm{2}}{x} \\ $$$${s}=\sqrt{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{{s}}{\:\sqrt{\mathrm{2}}}\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{13}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}} \\ $$$${q}=\frac{\mathrm{3}}{\mathrm{4}}−\frac{{s}}{\:\sqrt{\mathrm{2}}}\mathrm{sin}\:\theta=\frac{\mathrm{3}}{\mathrm{4}}−\frac{\sqrt{\mathrm{13}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}}{\mathrm{3}}…

Question Number 124367 by ajfour last updated on 02/Dec/20 Commented by ajfour last updated on 02/Dec/20 $${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{ABC}. \\ $$ Commented by mr W last updated…

Question Number 189898 by Mingma last updated on 23/Mar/23 Answered by HeferH last updated on 24/Mar/23 Commented by HeferH last updated on 24/Mar/23 $$\mathrm{HD}\:=\:\mathrm{diameter}\Rightarrow \\…

Question Number 189860 by Rupesh123 last updated on 23/Mar/23 Commented by HeferH last updated on 24/Mar/23 $$\mathrm{is}\:\mathrm{ABC}\:\mathrm{isosceles}? \\ $$ Terms of Service Privacy Policy Contact:…