Question Number 124353 by ajfour last updated on 02/Dec/20 Commented by ajfour last updated on 02/Dec/20 $${Find}\:{maximum}\:{radius}\:{of}\:{the} \\ $$$${blue}\:{circles}. \\ $$ Answered by ajfour last…
Question Number 189864 by normans last updated on 23/Mar/23 Answered by HeferH last updated on 23/Mar/23 Commented by HeferH last updated on 23/Mar/23 $$\mathrm{2x}\:+\:\mathrm{12}°\:=\:\mathrm{180}° \\…
Question Number 124329 by ajfour last updated on 02/Dec/20 Commented by ajfour last updated on 02/Dec/20 $${Find}\:{maximum}\:{of}\:{shaded}\:{area}. \\ $$ Commented by mr W last updated…
Question Number 189843 by normans last updated on 22/Mar/23 Answered by HeferH last updated on 22/Mar/23 $$\mathrm{2}\alpha\:+\:\mathrm{x}\:=\:\mathrm{180}° \\ $$$$\mathrm{62}°\:+\:\mathrm{56}°\:+\:\alpha\:=\:\mathrm{180}° \\ $$$$\alpha\:=\:\mathrm{180}°\:−\mathrm{118}°=\mathrm{62}° \\ $$$$\:\mathrm{x}\:=\:\mathrm{180}°\:−\:\mathrm{124}°\:=\:\mathrm{56}° \\ $$…
Question Number 58756 by Tawa1 last updated on 29/Apr/19 Answered by Kunal12588 last updated on 29/Apr/19 Commented by Kunal12588 last updated on 29/Apr/19 $${PR}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}}…
Question Number 124291 by bramlexs22 last updated on 02/Dec/20 Answered by liberty last updated on 02/Dec/20 $${The}\:{triangle}\:{area}\:{satisfies}\:{L}_{\Delta} =\:\frac{{r}}{\mathrm{2}}\left({a}+{b}+{c}\right) \\ $$$${and}\:{we}\:{have}\:{r}\:=\:\frac{\mathrm{2}{L}\Delta}{{a}+{b}+{c}} \\ $$$${so}\:{the}\:{area}\:{of}\:{the}\:{largest}\:{circle}\:{that}\:{can}\:{be} \\ $$$${cut}\:{from}\:{triangle}\:{is}\:\frac{\mathrm{4}\pi{L}_{\Delta} ^{\mathrm{2}}…
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Question Number 189808 by normans last updated on 22/Mar/23 Answered by mr W last updated on 22/Mar/23 Commented by mr W last updated on 22/Mar/23…
Question Number 189807 by normans last updated on 22/Mar/23 Answered by a.lgnaoui last updated on 22/Mar/23 $$\: \\ $$$${Angle}\:{entre}\:{verticale}\:{et}\:{p}\:{q}\left(\alpha={arc}\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi}{\mathrm{6}}\right) \\ $$$${Angle}\:{entre}\:{Eq}\:{et}\:{qr}\:{est}\:\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\pi}{\mathrm{6}}+{X}+\frac{\pi}{\mathrm{4}}=\pi\Rightarrow\:\:{X}=\frac{\mathrm{7}\pi}{\mathrm{12}} \\ $$…
Question Number 189792 by normans last updated on 21/Mar/23 Answered by HeferH last updated on 22/Mar/23 Commented by HeferH last updated on 22/Mar/23 $$\frac{\mathrm{a}}{\mathrm{5}−\mathrm{a}}\:=\:\frac{\mathrm{9}−\mathrm{a}}{\mathrm{a}};\:\: \\…