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Category: Geometry

In-AB-C-B-90-o-BB-CC-BB-and-CC-are-medians-m-c-m-a-note-CC-m-c-AA-m-a-

Question Number 207980 by mnjuly1970 last updated on 01/Jun/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{In}\:\:\:{A}\overset{\Delta} {{B}C}\::\:\:{B}=\:\mathrm{90}^{\:\mathrm{o}} \: \\ $$$$\:\mathrm{BB}'\:\:\bot\:\mathrm{CC}'\:\left(\:\mathrm{BB}'\:{and}\:\mathrm{CC}'\:{are}\:{medians}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\frac{\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\boldsymbol{{m}}_{\boldsymbol{{a}}} \:}\:=\:? \\ $$$$\boldsymbol{{note}}:\:\:\mid\:\mathrm{CC}'\:\mid\:=\:\boldsymbol{{m}}_{\boldsymbol{{c}}} \:\:,\:\mid\mathrm{AA}'\mid=\:\boldsymbol{{m}}_{\boldsymbol{{a}}}…

Question-207764

Question Number 207764 by mr W last updated on 25/May/24 Answered by som(math1967) last updated on 25/May/24 $$\:{R}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}^{\mathrm{2}} +{R}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{R}^{\mathrm{2}} =\mathrm{16}−\mathrm{8}…