Jeśli trójkąt  $\Delta A_1{B}_1{C}_1$  jest podobny do trójkąta  $\Delta ABC,$  to skalą podobieństwa nazywamy liczbę  $k,$  

$k=\frac{\left|A_1{B}_1\right|}{\left|AB\right|}=\frac{\left|B_1{C}_1\right|}{\left|BC\right|}=\frac{\left|A_1{C}_1\right|}{\left|AC\right|}.$  

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Z treści zadania wiemy, że:

$\frac{\left|A_1{B}_1\right|}{\left|AB\right|}=0,25$  

$\left|A_1{B}_1\right|=\frac{1}{4}\left|AB\right|$  

oraz   

$\frac{\left|B_1{C}_1\right|}{\left|BC\right|}=0,25$  

$\left|B_1{C}_1\right|=\frac{1}{4}\left|BC\right|$  

oraz

$\frac{\left|A_1{C}_1\right|}{\left|AC\right|}=0,25$  

$\left|A_1{C}_1\right|=\frac{1}{4}\left|AC\right|$  

 

Z treści zadania wiemy, że:

$L_{A_1{B}_1{C}_1}+12=L_{ABC}$  

$\left|A_1{B}_1\right|+\left|B_1{C}_1\right|+\left|A_1{C}_1\right|+12=\left|AB\right|+\left|BC\right|+\left|AC\right|$  

$\frac{1}{4}\left|AB\right|+\frac{1}{4}\left|BC\right|+\frac{1}{4}\left|AC\right|+12=\left|AB\right|+\left|BC\right|+\left|AC\right|\left|-\frac{1}{4}\right|AB\left|-\frac{1}{4}\right|BC\left|-\frac{1}{4}\right|AC\mid$  

$12=\frac{3}{4}\left|AB\right|+\frac{3}{4}\left|BC\right|+\frac{3}{4}\left|AC\right|\mid :\frac{3}{4}$  

$12\cdot \frac{4}{3}=\left|AB\right|+\left|BC\right|+\left|AC\right|$  

$16\left[\text{cm}\right]=L_{ABC}$

 

Wyznaczmy obwód trójkąta A1B1C1.

$L_{A_1{B}_1{C}_1}+12=16$  

$L_{A_1{B}_1{C}_1}=4\left[\text{cm}\right]$