△ABC ∽ △LMN。在△ABC中,AB = 5.5 cm,BC = 6 cm,CA = 4.5 cm。构造△ABC和△LMN,使得$\frac{BC}{MN} \ =\ \frac{5}{4}$。
已知
∆ABC ~ ∆LMN 且
$\frac{BC}{MN} \ =\ \frac{5}{4}$ ....(i)
求解
我们需要构造∆ABC和∆LMN。
解:
现在,我们知道:
相似三角形的对应边成比例。所以,
$\frac{AB}{LM} \ =\ \frac{BC}{MN} \ =\ \frac{AC}{LN}$ ....(ii)
由方程(i)和(ii)
$\frac{AB}{LM} \ =\ \frac{BC}{MN} \ =\ \frac{AC}{LN} \ =\ \frac{5}{4}$ ...(iii)
因此,
$ \begin{array}{l} \frac{AB}{LM} \ =\ \frac{5}{4}\ \ \ LM\ =\ AB\ \times \ \frac{4}{5}\ \ \ LM\ =\ 5.5\ \times \ \frac{4}{5}\ \ \ LM\ =\ 4.4\ cm \end{array}$
此外,
$ \begin{array}{l} \frac{BC}{MN} \ =\ \frac{5}{4}\ \ \ MN\ =\ BC\ \times \ \frac{4}{5}\ \ \ MN\ =\ 6\ \times \ \frac{4}{5}\ \ \ MN\ =\ 4.8\ cm \end{array}$
并且,
$ \begin{array}{l} \frac{AC}{LN} \ =\ \frac{5}{4}\ \ \ LN\ =\ AC\ \times \ \frac{4}{5}\ \ \ LN\ =\ 4.5\ \times \ \frac{4}{5}\ \ \ LN\ =\ 3.6\ cm \end{array}$
所以,所需的三角形如下