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Category: Geometry

Question-218739

Question Number 218739 by Spillover last updated on 14/Apr/25 Answered by A5T last updated on 15/Apr/25 $$\mathrm{AB}=\mathrm{c}\:;\:\mathrm{BC}=\mathrm{a}\:;\:\mathrm{CA}\:=\mathrm{b}\:;\:\mathrm{AF}=\mathrm{h}_{\mathrm{a}} \\ $$$$\mathrm{AF}\bot\mathrm{BC}\:;\:\mathrm{BG}\bot\mathrm{DC}\:;\:\mathrm{DH}\bot\mathrm{BC}\: \\ $$$$\mathrm{S}_{\bigtriangleup\mathrm{ABC}} =\frac{\mathrm{ah}_{\mathrm{a}} }{\mathrm{2}}\Rightarrow\mathrm{h}_{\mathrm{a}} =\frac{\mathrm{2S}_{\bigtriangleup\mathrm{ABC}} }{\mathrm{a}}…

Question-218733

Question Number 218733 by Spillover last updated on 14/Apr/25 Commented by A5T last updated on 15/Apr/25 $$\mathrm{This}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{contradiction}\left(\mathrm{if}\:\mathrm{the}\:\mathrm{other}\:\mathrm{chord}\right. \\ $$$$\left.\mathrm{were}\:\mathrm{also}\:\mathrm{a}\:\mathrm{diameter}\right). \\ $$ Answered by mr W…

Question-218735

Question Number 218735 by Spillover last updated on 14/Apr/25 Answered by som(math1967) last updated on 15/Apr/25 $${let}\:{rad}\:{of}\:{large}\:{circle}\:={R} \\ $$$${rad}.{of}\:{small}\:{circle}={r} \\ $$$${AD}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}×{R}×\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{2} \\ $$$$\:\Rightarrow{R}=\mathrm{2}−\sqrt{\mathrm{2}}…