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Question-208327




Question Number 208327 by efronzo1 last updated on 12/Jun/24
Answered by A5T last updated on 12/Jun/24
DE=5x,DF=12x  Let the perpendicular from A to BC meet it at H;  AH×BC=3×4=12⇒AH=((12)/5)  ⇒BH=(√(9−((144)/(25))))=(9/5)  ((AD)/(AB))=((DE)/(BC))=x⇒AD=3x⇒AE=4x  BF×BC=BD×BA⇒BF=((3×(3−3x))/5)...(i)  ((BF)/(BH))=((DF)/(AH))⇒BF=(((9×12x)/5)/((12)/5))=9x...(ii)  (i)=(ii)⇒3−3x=15x⇒x=(1/6)⇒DF=12x=2
$${DE}=\mathrm{5}{x},{DF}=\mathrm{12}{x} \\ $$$${Let}\:{the}\:{perpendicular}\:{from}\:{A}\:{to}\:{BC}\:{meet}\:{it}\:{at}\:{H}; \\ $$$${AH}×{BC}=\mathrm{3}×\mathrm{4}=\mathrm{12}\Rightarrow{AH}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\Rightarrow{BH}=\sqrt{\mathrm{9}−\frac{\mathrm{144}}{\mathrm{25}}}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\frac{{AD}}{{AB}}=\frac{{DE}}{{BC}}={x}\Rightarrow{AD}=\mathrm{3}{x}\Rightarrow{AE}=\mathrm{4}{x} \\ $$$${BF}×{BC}={BD}×{BA}\Rightarrow{BF}=\frac{\mathrm{3}×\left(\mathrm{3}−\mathrm{3}{x}\right)}{\mathrm{5}}…\left({i}\right) \\ $$$$\frac{{BF}}{{BH}}=\frac{{DF}}{{AH}}\Rightarrow{BF}=\frac{\frac{\mathrm{9}×\mathrm{12}{x}}{\mathrm{5}}}{\frac{\mathrm{12}}{\mathrm{5}}}=\mathrm{9}{x}…\left({ii}\right) \\ $$$$\left({i}\right)=\left({ii}\right)\Rightarrow\mathrm{3}−\mathrm{3}{x}=\mathrm{15}{x}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{6}}\Rightarrow{DF}=\mathrm{12}{x}=\mathrm{2} \\ $$

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