Category: Mensuration

Question Number 125108 by peter frank last updated on 08/Dec/20 Commented by mr W last updated on 08/Dec/20 $${where}\:{did}\:{you}\:{get}\:{these}\:{questions}? \\ $$$${they}\:{are}\:{wrong},\:{i}.{e}.\:{there}\:{is}\:{no} \\ $$$${solution}. \\ $$$${the}\:{last}\:{digit}\:{of}\:\mathrm{2}^{{any}\:{integer}}…

Question Number 125021 by Ggjj last updated on 07/Dec/20 Answered by mathmax by abdo last updated on 07/Dec/20 $$\mathrm{s}\Rightarrow\begin{cases}{\mid\mathrm{x}−\mathrm{1}\mid+\mid\mathrm{y}−\mathrm{5}\mid=\mathrm{1}}\\{\mid\mathrm{x}−\mathrm{1}\mid−\left(\mathrm{y}−\mathrm{5}\right)=\mathrm{0}\:\:\:\:\mathrm{due}\:\mathrm{to}\:\mid\mathrm{x}−\mathrm{1}\mid\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{y}\geqslant\mathrm{5}\:\Rightarrow}\end{cases} \\ $$$$\mathrm{s}\Rightarrow\begin{cases}{\mid\mathrm{x}−\mathrm{1}\mid+\mathrm{y}−\mathrm{5}=\mathrm{1}}\\{\mid\mathrm{x}−\mathrm{1}\mid−\left(\mathrm{y}−\mathrm{5}\right)=\mathrm{0}\:\Rightarrow\mathrm{2}\mid\mathrm{x}−\mathrm{1}\mid=\mathrm{1}\:\Rightarrow\mid\mathrm{x}−\mathrm{1}\mid=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow\mathrm{x}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{or}\:\mathrm{x}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\mathrm{y}=\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{2}}\:\Rightarrow\mathrm{x}+\mathrm{y}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{11}}{\mathrm{2}}=\mathrm{7}…

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…