Question

fie punctele e si f pe segmentul ab astfel incat ae/eb =5/11 si af/fb=3/5.valoarea raportului ef/be este egala cu.....

Answer 1

VARIANTA 1:
$\frac{AE}{BE} = \frac{5}{11} => \frac{AE+BE}{BE}= \frac{5+11}{11} <=> \frac{AB}{BE}= \frac{16}{11}$ $=> BE= \frac{11AB}{16}$

$\frac{AF}{BF}= \frac{3}{5} => \frac{AF+BF}{BF}= \frac{5+3}{5} <=> \frac{AB}{BF}= \frac{8}{5} => BF= \frac{5AB}{8}$

$\frac{BF}{BE} = \frac{ \frac{5AB}{8} }{ \frac{11AB}{16} }= \frac{5AB}{8}* \frac{16}{11AB} = \frac{10}{11}$

$\frac{EF}{BE} = \frac{BE-BF}{BE}= 1- \frac{BF}{BE}= 1-\frac{10}{11}= \frac{1}{11}$

VARIANTA 2:
$\frac{AE}{BE}= \frac{5}{11}=>AE=5a; BE=11a$

$\frac{AF}{BF}= \frac{3}{5} => AF=3b; BF=5b$

[tex]AB=AE+BE=5a+11a=16a \\ AB=AF+BF=3b+5b=8b [/tex]
Din cele doua relatii de mai sus, rezulta: 16a=8b => 2a=b.

EF=BE-BF=11a-5b=11a-10a=a

$\frac{EF}{BE}= \frac{a}{11a}= \frac{1}{11}$