ТУ задача

[antoniy]

Screenshot_46.jpg (51.43 KiB) Прегледано 413 пъти

[ptj]

[tex]( \frac{d_1}{2})^2+ (\frac{d_2}{2})^2=a^2[/tex] (Питагорова теорема)

[tex]( \frac{d_1}{2})^2+ (\frac{ \lambda . d_1}{2})^2=( \frac{P}{4} )^2[/tex]

[tex]4(1+ \lambda ^2)d_1^2=P^2[/tex]

[tex]d_1^2= \frac{P^2}{4(1+ \lambda^2) }[/tex]

[tex]S= \frac{d^1.d^2}{2}= \frac{ \lambda .d_1^2}{2}= \frac{ \lambda.P^2 }{8.(1+ \lambda^2 )}[/tex]

[Darina73]

Знаем P и [tex]\lambda[/tex]
S=?
Нека [tex]\angle[/tex]BAD=2[tex]\alpha[/tex] ,BD=[tex]d_{1 }[/tex] и AC=[tex]d_{2 }[/tex] , което значи ,че
tg[tex]\alpha= \frac{ d_{1 } }{ d_{2 } }[/tex] т.е. tg[tex]\alpha= \lambda[/tex]

S=AB.AD.sin2[tex]\alpha[/tex]= [tex]\frac{P}{4} . \frac{P}{4}[/tex].[tex]\frac{2tg \alpha }{1+ tg^{2 } \alpha }[/tex]= [tex]\frac{ P^{2 } \lambda }{8(1+ \lambda ^{2 }) }[/tex]