Question

the polygons in each pair are similar. find the scale factor of the smaller figure to the larger figure.
3)
4)
the polygons in each pair are similar. find the missing side length.
5)
6)

Explanation:

Step1: Find the scale factor for similar polygons

For similar polygons, the scale factor \(k\) is the ratio of corresponding side lengths. For problem 3, the smaller figure has a side length of \(12\) and the larger has \(30\). The scale factor of the smaller to the larger is \(k=\frac{12}{30}\).

$$ \frac{12}{30}=\frac{2}{5} $$

For problem 4, using the corresponding sides \(25\) (smaller) and \(30\) (larger). The scale factor \(k = \frac{25}{30}=\frac{5}{6}\) (wait, no, actually, if we take another pair of corresponding sides: \(15\) and \(18\), \(\frac{15}{18}=\frac{5}{6}\), also \(11.5\) and \(\text{missing}\) (but using \(25\) and \(30\) is wrong, correct: take \(25\) (smaller trapezoid's top - base) and \(42\) (larger's bottom - base? No, no, for similar trapezoids, we use the ratio of corresponding sides. Let's use \(15\) (smaller's non - parallel side) and \(18\) (larger's non - parallel side). Scale factor \(k=\frac{15}{18}=\frac{5}{6}\).

Step2: Find the missing side length for similar polygons

For problem 5, since the triangles are similar (isosceles, as two sides of the first triangle are \(7\) and the base of the second is \(14\), scale factor \(k=\frac{7}{14}=\frac{1}{2}\). Let the missing side be \(x\). Using the ratio \(\frac{x}{12}=\frac{1}{2}\), then \(x = 6\).
For problem 6, the rectangles are similar. The ratio of the widths is \(\frac{3}{9}=\frac{1}{3}\). Let the missing length be \(y\). Using the ratio \(\frac{y}{24}=\frac{1}{3}\), then \(y = 8\).

Answer:

1. \(\frac{2}{5}\); 4) \(\frac{5}{6}\); 5) \(6\); 6) \(8\)