Question Number 214350 by ajfour last updated on 06/Dec/24 Answered by mr W last updated on 06/Dec/24 $$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\mathrm{5}}−\mathrm{2} \\ $$$$\mathrm{2}+{r}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\left({r}+\mathrm{1}\right)^{\mathrm{2}} −\left({r}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}+\left(\sqrt{\mathrm{5}}−\mathrm{2}\right){r}=\mathrm{2}\sqrt{{r}} \\…
Question Number 214372 by ajfour last updated on 06/Dec/24 Commented by ajfour last updated on 06/Dec/24 $${Find}\:{maximum}\:{r}.\:{Outer}\:{figure}\:{is}\:{a} \\ $$$${rhombus}.{Circle}\:{is}\:{inscribed}\:{in}\:{an}\: \\ $$$${isosceles}\:{triangle}\:{as}\:{shown}\:{above}. \\ $$ Answered by…
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Question Number 214297 by ajfour last updated on 04/Dec/24 Commented by ajfour last updated on 04/Dec/24 $${the}\:{tangents}\:{are}\:{at}\:{right}\:{angles}\:{to} \\ $$$${one}\:{another}\:{and}\:{corner}\:{maynot}\:{be} \\ $$$${at}\:{semicircle}\:{centre}. \\ $$ Commented by…
Question Number 214228 by mr W last updated on 02/Dec/24 Commented by mr W last updated on 02/Dec/24 $${find}\:{area}\:{of}\:{the}\:{square} \\ $$ Answered by aleks041103 last…
Question Number 214220 by Ari last updated on 01/Dec/24 Answered by mr W last updated on 01/Dec/24 $${say}\:{a}>{b} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${a}−{b}=\mathrm{6}\:\:\:…\left({ii}\right)…
Question Number 214199 by mr W last updated on 01/Dec/24 Commented by mr W last updated on 01/Dec/24 $$\left[{see}\:{Q}\mathrm{214176}\right] \\ $$$${find}\:{the}\:{radius}\:{of}\:{the}\:{maximal} \\ $$$${circle}\:{inscribed}\:{between}\:{the}\:{curves} \\ $$$${f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{as}\:{shown}.…
Question Number 214100 by ajfour last updated on 28/Nov/24 Answered by ajfour last updated on 29/Nov/24 Commented by ajfour last updated on 29/Nov/24 $$\mathrm{sin}\:\alpha=\frac{{a}}{\mathrm{2}{b}+{a}}=\frac{\mathrm{1}}{\mathrm{2}{s}+\mathrm{1}}\:\:\:\forall\:\:{s}=\frac{{b}}{{a}},\:{t}=\frac{{r}}{{a}} \\…
Question Number 214040 by ajfour last updated on 25/Nov/24 Commented by ajfour last updated on 25/Nov/24 $${The}\:\bigtriangleup\:{is}\:{equilateral}.\:{Find}\:{its}\:{side}\:\boldsymbol{{s}}. \\ $$ Answered by mr W last updated…
Question Number 214012 by ajfour last updated on 24/Nov/24 Commented by ajfour last updated on 24/Nov/24 $${Outer}\:{circle}\:{radius}\:{is}\:{R}.\:{Circle}\:{with} \\ $$$${center}\:{A}\:{has}\:{radius}\:{r}={R}/\mathrm{2}. \\ $$$${If}\:\bigtriangleup{ABC}\:{is}\:{equilateral},\:{find}\:{its} \\ $$$${edge}\:{length}\:\left({say}\:{a}\right). \\ $$…