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Category: Geometry

Question-189608

Question Number 189608 by mr W last updated on 19/Mar/23 Commented by mr W last updated on 19/Mar/23 $${a}\:{bin}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{frustum} \\ $$$${as}\:{shown}.\:{an}\:{ant}\:{climbs}\:{on}\:{the}\:{outer} \\ $$$${surface}\:{of}\:{the}\:{bin}\:{from}\:{point} \\ $$$${A}\:{to}\:{point}\:{B}\:{following}\:{a}\:{path}\:{with}…

Question-189605

Question Number 189605 by normans last updated on 19/Mar/23 Commented by mr W last updated on 19/Mar/23 $$\frac{{a}/\mathrm{4}}{\mathrm{7}}=\frac{\mathrm{9}−{a}/\mathrm{4}}{\mathrm{9}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}×\mathrm{7}×\mathrm{9}}{\mathrm{7}+\mathrm{9}}=\frac{\mathrm{63}}{\mathrm{4}} \\ $$ Commented by normans…

Question-189603

Question Number 189603 by normans last updated on 19/Mar/23 Answered by a.lgnaoui last updated on 19/Mar/23 $$\bigtriangleup{ADF}:\:\:\:\:\:\measuredangle{ADF}=\mathrm{90}−\mathrm{18}=\mathrm{72} \\ $$$$\:\frac{\mathrm{sin}\:\mathrm{18}}{\mathrm{DF}}=\frac{\mathrm{sin}\:\mathrm{72}}{\mathrm{AF}}\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\:\bigtriangleup{ABF}\::\:\:\measuredangle{BAF}=\mathrm{48}+\mathrm{18}=\mathrm{66};\:\:\measuredangle{ABF}=\mathrm{90}−\mathrm{66}=\mathrm{24} \\ $$$$\frac{\mathrm{sin}\:\mathrm{24}}{\mathrm{AF}}=\frac{\mathrm{sin}\:\mathrm{66}}{\mathrm{BF}}\:\:\:\:\mathrm{BF}=\mathrm{BD}+\mathrm{DF}=\mathrm{6}+\mathrm{DF} \\ $$$$\frac{\mathrm{sin}\:\mathrm{24}}{\mathrm{AF}}=\frac{\mathrm{sin}\:\mathrm{66}}{\mathrm{6}+\mathrm{DF}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)…

Question-58512

Question Number 58512 by ajfour last updated on 24/Apr/19 Commented by ajfour last updated on 24/Apr/19 $$\mathrm{Find}\:\:\:\:\:\frac{\mathrm{Area}\left(\bigtriangleup\mathrm{PQR}\right)}{\mathrm{Area}\left(\bigtriangleup\mathrm{ABC}\right)}\:\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}. \\ $$ Answered by mr W…

Question-58493

Question Number 58493 by Tawa1 last updated on 23/Apr/19 Answered by math1967 last updated on 24/Apr/19 $${At}\:\mathrm{2}.\mathrm{00}{pm}\:{distance}\:{between}\:\mathrm{2}{cars} \\ $$$$=\mathrm{915}\:−\mathrm{75}=\mathrm{840}{km} \\ $$$${time}\:{taken}\:{both}\:{meet}=\frac{\mathrm{840}}{\mathrm{75}+\mathrm{100}}=\mathrm{4}\frac{\mathrm{4}}{\mathrm{5}}{hr} \\ $$$$\mathrm{2}{pm}+\mathrm{4}{hr}\mathrm{48}{m}=\mathrm{6}:\mathrm{48}{pm} \\ $$…