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Category: Mensuration

Question-188378

Question Number 188378 by Mingma last updated on 28/Feb/23 Answered by HeferH last updated on 09/Mar/23 $${let}\:{x}\:{be}\:{the}\:{side}\:{of}\:{the}\:{red}\:{triangle}\:{and}\:{y}\:{be}\:{the} \\ $$$${side}\:{of}\:{the}\:{blue}\:{one} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{18}°\right)\:=\:\frac{{y}}{{x}}\: \\ $$$$\frac{{A}_{{r}} }{{A}_{{b}} }\:=\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{18}°\right)}\right)^{\mathrm{2}}…

Question-188250

Question Number 188250 by Rupesh123 last updated on 27/Feb/23 Answered by HeferH last updated on 27/Feb/23 $$\left(\mathrm{1}\:−\:\frac{{h}}{\mathrm{2}}\right)\frac{{b}}{\mathrm{2}}\:=\:{b}\centerdot{h} \\ $$$$\:\frac{{b}}{\mathrm{2}}−\frac{{bh}}{\mathrm{4}}\:=\:{bh} \\ $$$$\:\mathrm{2}\:=\:\mathrm{5}{h} \\ $$$$\:{h}\:=\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$…

Question-188078

Question Number 188078 by Rupesh123 last updated on 25/Feb/23 Answered by a.lgnaoui last updated on 25/Feb/23 $${from}\:{Graphe}:\bigtriangleup{ABC}\:\:{Equilaterale} \\ $$$${big}\:{circle}\left({radius}\:{OF}=\sqrt{\mathrm{4}+\frac{{EF}^{\mathrm{2}} }{\mathrm{4}}\:}\:\right) \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}}{{OF}}\Rightarrow\left({OF}=\mathrm{4};{EF}=\mathrm{4}\sqrt{\mathrm{3}.}\:\:\right) \\ $$$${Area}\left({big}\:{circleC}_{{B}} =\mathrm{16}\pi\right)…