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Category: Geometry

Question-38032

Question Number 38032 by behi83417@gmail.com last updated on 20/Jun/18 Commented by behi83417@gmail.com last updated on 21/Jun/18 $$\boldsymbol{{in}}\:\boldsymbol{{triangle}}\:\boldsymbol{{A}}\overset{\blacktriangle} {\boldsymbol{{B}C}}:\angle\boldsymbol{{A}}=\mathrm{72}^{\bullet} ,\boldsymbol{{BC}}=\mathrm{10}. \\ $$$$\boldsymbol{{D}}\:\boldsymbol{{and}}\:\boldsymbol{{E}},\boldsymbol{{can}}\:\boldsymbol{{move}}\:\boldsymbol{{on}}\:\boldsymbol{{BC}},\boldsymbol{{but}}\:\boldsymbol{{such}} \\ $$$$\boldsymbol{{that}},\boldsymbol{{alwyes}}:\boldsymbol{{AD}}=\boldsymbol{{AE}}. \\ $$$$\left.\mathrm{1}\right)\boldsymbol{{evaluate}}\:\angle\boldsymbol{{DAE}},\boldsymbol{{when}}\:\boldsymbol{{area}}\:\boldsymbol{{of}}\:…

Question-37804

Question Number 37804 by ajfour last updated on 17/Jun/18 Commented by ajfour last updated on 17/Jun/18 $${Find}\:{maximum}\:{area}\:{of} \\ $$$${quadrilateral}\:{MNOP}\:{in}\:{terms} \\ $$$${of}\:{radius}\:\boldsymbol{{R}}\:{of}\:{circle}. \\ $$$$\left({O}\:{is}\:{the}\:{center}\:{of}\:{circle}\right). \\ $$…

Question-103338

Question Number 103338 by I want to learn more last updated on 14/Jul/20 Commented by som(math1967) last updated on 14/Jul/20 $$\mathrm{OC}=\mathrm{14}−\mathrm{r} \\ $$$$\mathrm{again}\:\mathrm{OC}=\mathrm{r}\sqrt{\mathrm{2}} \\ $$$$\therefore\mathrm{r}\sqrt{\mathrm{2}}=\mathrm{14}−\mathrm{r}…

Question-103318

Question Number 103318 by I want to learn more last updated on 14/Jul/20 Commented by som(math1967) last updated on 14/Jul/20 $$\frac{\mathrm{AT}}{\mathrm{AB}}=\mathrm{cos}\angle\mathrm{BAT}=\mathrm{cos30}° \\ $$$$\mathrm{AT}=\mathrm{15}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cm} \\ $$$$\therefore\mathrm{BP}=\mathrm{QD}=\left(\mathrm{15}−\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{cm}…