Category: Geometry

Question Number 45448 by MrW3 last updated on 13/Oct/18 Commented by MrW3 last updated on 13/Oct/18 $${When}\:{an}\:{ellipse}\:{is}\:{rotated}\:{by}\:{an} \\ $$$${angle}\:\alpha,\:{the}\:{five}\:{partial}\:{areas}\:{are}\:{equal}. \\ $$$${Find}\:{the}\:{equation}\:{of}\:{the}\:{ellipse} \\ $$$${for}\:\alpha=\mathrm{90}°\:{and}\:\mathrm{45}°\:{respectively}. \\ $$…

Question Number 45381 by behi83417@gmail.com last updated on 12/Oct/18 Commented by behi83417@gmail.com last updated on 12/Oct/18 $$\angle{ABN}=\angle{NBO}=\angle{OBC}=\frac{\angle{ABC}}{\mathrm{3}} \\ $$$$\angle{BAN}=\angle{NAC},\angle{ACO}=\angle{OCB} \\ $$$${AB}=\mathrm{12},{BC}=\mathrm{8},{AC}=\mathrm{10} \\ $$$${Q}#\mathrm{1}={radius}\:{of}\:{red}\:\:{circle}. \\ $$$${Q}#\mathrm{2}={area}\:{of}\:\:{N}\overset{\lozenge}…

Question Number 176437 by cherokeesay last updated on 19/Sep/22 Answered by HeferH last updated on 19/Sep/22 $$\:{by}\:{similar}\:{triangles}: \\ $$$$\:{b}\:={AD}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{5}}}\:\:=\:\frac{\left(\mathrm{4}\sqrt{\mathrm{5}}\right)}{\left(\mathrm{1}\:+\:\sqrt{\mathrm{5}}\right)} \\ $$$$\:{h}\:=\:\mathrm{4} \\ $$$$\:{A}_{\bigtriangleup{ABD}} \:=\:\frac{{bh}}{\mathrm{2}}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\:+\:\mathrm{1}}\:=\:\mathrm{10}\:−\:\mathrm{2}\sqrt{\mathrm{5}\:} \\…

Question Number 110848 by ajfour last updated on 31/Aug/20 Commented by ajfour last updated on 31/Aug/20 $${Find}\:{c}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}. \\ $$ Commented by bemath last updated on…

Question Number 110782 by Aina Samuel Temidayo last updated on 30/Aug/20 $$\mathrm{A}\:\mathrm{triangle}\:\mathrm{has}\:\mathrm{area}\:\mathrm{15}\:\mathrm{and} \\ $$$$\mathrm{circumradius}\:\mathrm{12}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of} \\ $$$$\mathrm{its}\:\mathrm{heights}. \\ $$ Answered by som(math1967) last updated on 30/Aug/20…