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Category: Geometry

Question-222673

Question Number 222673 by ajfour last updated on 04/Jul/25 Commented by ajfour last updated on 04/Jul/25 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Find}\:{small}\:{circle}\:{radius}\:\boldsymbol{{a}}. \\ $$$${Oh}!\:{i}\:{forgot}\:{to}\:{say}\:{almost}-\:{ABCD}\:{is}\:{sq}. \\ $$ Answered by mr W…

Question-222646

Question Number 222646 by Tawa11 last updated on 03/Jul/25 Answered by som(math1967) last updated on 03/Jul/25 $$\:{AB}={BC}−{d},\:{AC}={BC}+{d} \\ $$$$\:\:{AC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} \\ $$$$\left({BC}+{d}\right)^{\mathrm{2}} =\left({BC}−{d}\right)^{\mathrm{2}} +{BC}^{\mathrm{2}}…

Question-222636

Question Number 222636 by Mingma last updated on 02/Jul/25 Answered by mr W last updated on 04/Jul/25 $${d}=\sqrt{{a}^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }+{x}+\sqrt{{b}^{\mathrm{2}} −\left({b}−{x}\right)^{\mathrm{2}} } \\ $$$$\:\:=\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }+{x}+\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}}…