Question

Answer 1

a: \(CE=\frac14\times CB\)

=>\(CB=4\times CE\)

=>\(S_{CFB}=4\times S_{CEF}=4\times2=8\left(\operatorname{cm}^2\right)\)

Ta có: \(CF=\frac13\times CA\)

=>\(CA=3\times CF\)

=>\(S_{CAB}=3\times S_{CFB}=3\times8=24\left(\operatorname{cm}^2\right)\)

b: CE+EB=CB

=>\(EB=CB-CE=CB-\frac14\times CB=\frac34\times CB\)

=>BE=3EC

=>\(S_{FEB}=3\times S_{FEC};S_{DEB}=3\times S_{DEC}\)

=>\(S_{DEB}-S_{FEB}=3\times\left(S_{DEC}-S_{FEC}\right)\)

=>\(S_{DFB}=3\times S_{DFC}\)