Question Number 103489 by Study last updated on 15/Jul/20 | ||
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Answered by bobhans last updated on 15/Jul/20 | ||
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$${AB}=\:\sqrt{\mathrm{6}}\:+\:\mathrm{2}\sqrt{\mathrm{6}}\:=\:\mathrm{3}\sqrt{\mathrm{6}}\:\oplus \\ $$ | ||
Answered by bramlex last updated on 15/Jul/20 | ||
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Commented by bramlex last updated on 15/Jul/20 | ||
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$$\left(\mathrm{1}\right)\:\mathrm{AC}\:=\:\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{24}}\:=\:\mathrm{2}\sqrt{\mathrm{6}}\: \\ $$$$\left(\mathrm{2}\right)\:\mathrm{SC}\:=\:\sqrt{\mathrm{3}.\mathrm{5}^{\mathrm{2}} −\mathrm{2}.\mathrm{5}^{\mathrm{2}} }=\sqrt{\mathrm{6}×\mathrm{1}}\:=\:\sqrt{\mathrm{6}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{AB}\:=\:\mathrm{AP}\:+\:\mathrm{PB}\:=\:\mathrm{3}\sqrt{\mathrm{6}}\:\mathrm{cm} \\ $$ | ||