Question Number 187810 by Rupesh123 last updated on 22/Feb/23 Commented by a.lgnaoui last updated on 23/Feb/23 $$\left.\mathrm{E}\right)\mathrm{30} \\ $$ Commented by mr W last updated…
Question Number 187721 by Mingma last updated on 20/Feb/23 Answered by mr W last updated on 21/Feb/23 Commented by mr W last updated on 21/Feb/23…
Question Number 187701 by Mingma last updated on 20/Feb/23 Answered by HeferH last updated on 20/Feb/23 $${let}\:“{h}''\:{be}\:{the}\:{height}\:{of}\:{the}\:{top}\:{triangle}\:{and}\:“{a}'' \\ $$$${the}\:{base} \\ $$$$\:\frac{\mathrm{7}{ah}}{\mathrm{2}}\:=\:\mathrm{28} \\ $$$$\:{ah}\:=\:\frac{\mathrm{28}\centerdot\mathrm{2}}{\mathrm{7}}\:=\:\mathrm{8}\: \\ $$$$\:{Total}\:{area}\:=\:\mathrm{4}{a}\centerdot\mathrm{4}{h}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{8}{ah}\:=\:\mathrm{64}{u}^{\mathrm{2}}…
Question Number 187689 by Mingma last updated on 20/Feb/23 Answered by mr W last updated on 21/Feb/23 Commented by mr W last updated on 26/Feb/23…
Question Number 187589 by Mingma last updated on 19/Feb/23 Commented by Mingma last updated on 19/Feb/23 Blue area=? Answered by cortano12 last updated on 19/Feb/23 $$\mathrm{cos}\:\alpha=\frac{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 187587 by Mingma last updated on 19/Feb/23 Commented by Mingma last updated on 19/Feb/23 Blue area=? Answered by mr W last updated on 19/Feb/23…
Question Number 121793 by ajfour last updated on 11/Nov/20 Commented by ajfour last updated on 13/Nov/20 $${If}\:{O}\overset{\frown} {{DE}}\:\:\:{is}\:{a}\:{sector},\:{and}\:\bigtriangleup{ABC}\: \\ $$$${is}\:{equilateral}\:{with}\:{minimum}\:{area} \\ $$$${has}\:{an}\:{area}\:{equal}\:{to}\:\bigtriangleup{OEF}\:,\:{find}\:\theta. \\ $$$$ \\…
Question Number 187301 by Rupesh123 last updated on 15/Feb/23 Answered by HeferH last updated on 16/Feb/23 $$\:{chord}\:{theorem}: \\ $$$$\mathrm{5}^{\mathrm{2}} \:=\:{x}\left(\mathrm{5}\:+\:\mathrm{3}\right) \\ $$$$\:\frac{\mathrm{25}}{\mathrm{8}}\:=\:{x} \\ $$$$\:\frac{\mathrm{25}}{\mathrm{8}}\:+\:\mathrm{8}\:=\:\mathrm{2}{r} \\…
Question Number 187284 by Rupesh123 last updated on 15/Feb/23 Commented by mr W last updated on 15/Feb/23 $${i}\:{think}\:{the}\:{green}\:{area}\:{is}\:{not}\:{unique}. \\ $$ Commented by mr W last…
Question Number 187253 by Rupesh123 last updated on 15/Feb/23 Answered by MikeH last updated on 15/Feb/23 $$\left(\mathrm{i}\right)\:\mathrm{Area}\:\mathrm{of}\:\bigtriangleup{BCD}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right)\left(\mathrm{4}\right)\:\mathrm{sin}\:{x}\:=\:\mathrm{8}\:\mathrm{sin}\:{x} \\ $$$$\:\:\:\:\:\:\:\mathrm{Area}\:\mathrm{of}\:\Delta{ABD}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}\right)\left(\mathrm{4}\right)\mathrm{sin}\:\left(\mathrm{90}−{x}\right)\:=\:\mathrm{6}\:\mathrm{cos}\:{x} \\ $$$$\mathrm{total}\:\mathrm{area}\:=\:\mathrm{area}\:\mathrm{of}\:\Delta{BCD}\:+\:\mathrm{area}\:\mathrm{of}\:\Delta{ABD} \\ $$$$\Rightarrow\:{A}\:=\:\mathrm{6}\:\mathrm{cos}\:{x}\:+\:\mathrm{8}\:\mathrm{sin}\:{x}\:\blacksquare \\ $$…