Question

Answer 1

a: Ta có: BI+IC=BC

=>\(CI=BC-BI=BC-\frac13BC=\frac23BC\)

=>\(S_{AIC}=\frac23\cdot S_{ABC}=\frac23\cdot180=120\left(\operatorname{cm}^2\right)\)

Ta có: NA+NC=AC

=>\(AC=NC+\frac12NC=\frac32NC\)

=>\(CN=\frac23CA\)

=>\(S_{CNI}=\frac23\cdot S_{AIC}=\frac23\cdot120=80\left(\operatorname{cm}^2\right)\)

Ta có: \(NA=\frac12NC\)

=>\(S_{ANI}=\frac12\cdot S_{INC}=\frac12\cdot80=40\left(\operatorname{cm}^2\right)\)

b: Ta có: \(S_{ANI}=\frac12\cdot S_{NIC}\)

=>\(S_{ANI}