Cho tam giác ABC có AB=16,2cm, BC=24,3cm, AC=32,7cm. Tính độ dài các cạnh của tam giác A'B'C'

* Hướng dẫn giải

$\begin{array}{l}a)A\text{'}B\text{'}=\mathrm{16,2}+\mathrm{10,8}=27(cm).Do\Delta ABC~\Delta A\text{'}B\text{'}C\text{'}\\ \Rightarrow \frac{B\text{'}C\text{'}}{BC}=\frac{A\text{'}C\text{'}}{AC}=\frac{27}{\mathrm{16,2}}=\frac{5}{3}\Rightarrow \left\{\begin{array}{l}B\text{'}C\text{'}=\frac{5.BC}{3}=\mathrm{40,5}\\ A\text{'}C\text{'}=\frac{5.AC}{3}=\mathrm{54,5}\end{array}\right.\end{array}$

Vậy $A\text{'}B\text{'}=27cm,B\text{'}C\text{'}=\mathrm{40,5}cm,A\text{'}C\text{'}=\mathrm{54,5}cm$

$\begin{array}{l}b)A\text{'}B\text{'}=\mathrm{16,2}-\mathrm{5,4}=\mathrm{10,8.}Do\Delta ABC~\Delta A\text{'}B\text{'}C\text{'}\\ \Rightarrow \frac{B\text{'}C\text{'}}{BC}=\frac{A\text{'}C\text{'}}{AC}=\frac{\mathrm{10,8}}{\mathrm{16,2}}=\frac{2}{3}\Rightarrow \left\{\begin{array}{l}B\text{'}C\text{'}=\frac{2BC}{3}=\mathrm{16,2}\left(cm\right)\\ A\text{'}C\text{'}=\frac{2AC}{3}=\mathrm{21,8}\left(cm\right)\end{array}\right.\end{array}$

Vậy $A\text{'}B\text{'}=\mathrm{10,8}cm,B\text{'}C\text{'}=\mathrm{16,2}cm,A\text{'}C\text{'}=\mathrm{21,8}cm$