Question

triangle 1 is composed of line segments \\(\overline{ab}\\), \\(\overline{bc}\\), and \\(\overline{ca}\\). triangle 2 is composed of line segments \\(de\\), \\(ef\\), and \\(fd\\). the triangles are similar with a scale of 5:2. select the true statement about the proportions of the sides.\\(\frac{a}{e} = 1\\)\\(\frac{m\overline{ab}}{m\overline{de}} = \frac{2}{5}\\)\\(\frac{m\overline{ac}}{m\overline{df}} = \frac{5}{2}\\)\\(\frac{e}{f} = \frac{5}{2}\\)

Explanation:

Step1: Recall Similar Triangles Property

For similar triangles, the ratio of corresponding sides is equal to the scale factor. The scale of Triangle 1 to Triangle 2 is \(5:2\) (or Triangle 2 to Triangle 1 is \(2:5\)). So, \(\frac{\text{Side of Triangle 1}}{\text{Corresponding Side of Triangle 2}}=\frac{5}{2}\) and \(\frac{\text{Side of Triangle 2}}{\text{Corresponding Side of Triangle 1}}=\frac{2}{5}\).

Step2: Analyze Each Option

• Option 1 (\(\frac{a}{e} = 1\)): This implies \(a = e\), but since triangles are similar with scale \(5:2\), corresponding sides can't be equal. Eliminate.
• Option 2 (\(\frac{m\overline{AB}}{m\overline{DE}}=\frac{2}{5}\)): If scale from Triangle 1 to 2 is \(5:2\), then \(\frac{m\overline{AB}}{m\overline{DE}}=\frac{5}{2}\) (since \(AB\) is from Triangle 1, \(DE\) from Triangle 2). So this is incorrect (it's reversed). Eliminate.
• Option 3 (\(\frac{m\overline{AC}}{m\overline{DF}}=\frac{5}{2}\)): \(AC\) is a side of Triangle 1, \(DF\) is the corresponding side of Triangle 2. By similar triangles, \(\frac{\text{Triangle 1 side}}{\text{Triangle 2 side}}=\frac{5}{2}\). This matches.
• Option 4 (\(\frac{e}{f}=\frac{5}{2}\)): Need to check corresponding sides. \(e\) and \(f\) are angles? Wait, no, labels: Wait, the sides: Wait, maybe mislabel. But from the scale, the ratio of Triangle 1 to 2 is \(5:2\). The third option with \(AC\) (Triangle 1) and \(DF\) (Triangle 2) has ratio \(\frac{5}{2}\), which is correct.

Answer:

\(\frac{m\overline{AC}}{m\overline{DF}}=\frac{5}{2}\) (the option with \(\frac{m\overline{AC}}{m\overline{DF}}=\frac{5}{2}\))