Category: Geometry

Question Number 208940 by efronzo1 last updated on 28/Jun/24 Answered by kapoorshah last updated on 28/Jun/24 $$r=OD=\sqrt{10}$$$$BE=\sqrt{15}$$$$BOE\sim BCA$$$$\frac{BO}{BE}=\frac{BC}{AB}$$$$\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{15}}}\:=\:\frac{{BC}}{\mathrm{2}\sqrt{\mathrm{10}}}\:\:\Rightarrow\:{BC}\:=\:\frac{\mathrm{20}}{\:\sqrt{\mathrm{15}}}\:=\:\mathrm{5}.\mathrm{164}…

Question Number 208915 by Tawa11 last updated on 27/Jun/24 Answered by mr W last updated on 27/Jun/24 Commented by mr W last updated on 27/Jun/24…

Question Number 208828 by efronzo1 last updated on 24/Jun/24 Answered by mr W last updated on 24/Jun/24 $$BM=MC=a,say$$$$AM=DM=2a$$$$\frac{BM}{6}=\frac{MQ}{4}\Rightarrow MQ=\frac{2a}{3}\Rightarrow QD=\frac{4a}{3}$$$$\left(\mathrm{6}+\mathrm{4}\right)^{\mathrm{2}} ={a}^{\mathrm{2}}…

Question Number 208772 by Jubr last updated on 22/Jun/24 Answered by mr W last updated on 22/Jun/24 Commented by mr W last updated on 22/Jun/24…

Question Number 208690 by Tawa11 last updated on 21/Jun/24 Answered by mr W last updated on 21/Jun/24 $$(2r-3)\times 3=2^2$$$$\Rightarrow r=\frac{13}{6}$$$$shadedarea=\frac{\pi }{2}{\left(\frac{13}{6}\right)}^2-\left(\frac{13}{6}\right)\times 2\approx 3.04$$…