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Question-217617




Question Number 217617 by chidera last updated on 16/Mar/25
Answered by som(math1967) last updated on 17/Mar/25
 Probability a point lies △ADM   =((Ar△ADM)/(ABCD))=(1/4)  △ADM but not △ADN  =((△DOM)/(ABCD))=(1/8)  neither △ADM nor △ADN  =((Ar MONBC)/(ABCD))=1−(3/8)=(5/8)
$$\:{Probability}\:{a}\:{point}\:{lies}\:\bigtriangleup{ADM} \\ $$$$\:=\frac{{Ar}\bigtriangleup{ADM}}{{ABCD}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\bigtriangleup{ADM}\:{but}\:{not}\:\bigtriangleup{ADN} \\ $$$$=\frac{\bigtriangleup{DOM}}{{ABCD}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${neither}\:\bigtriangleup{ADM}\:{nor}\:\bigtriangleup{ADN} \\ $$$$=\frac{{Ar}\:{MONBC}}{{ABCD}}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$
Commented by som(math1967) last updated on 17/Mar/25

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