Similar Figures: Find the missing side in each pair of similar triangles.

To solve the problem of finding the missing sides in each pair of similar figures, we use the property that corresponding sides of similar figures are proportional. This means that the ratios of the corresponding sides are equal.

Let's solve each problem step by step:

---

Problem 1:

Given:
- Left triangle: sides 15, 20
- Right triangle: sides 3, \( x \)

Since the triangles are similar, the ratios of corresponding sides are equal:
\[
\frac{15}{3} = \frac{20}{x}
\]

Simplify the left side:
\[
\frac{15}{3} = 5
\]

So the equation becomes:
\[
5 = \frac{20}{x}
\]

Solve for \( x \):
\[
x = \frac{20}{5} = 4
\]

Answer:
\[
\boxed{4}
\]

---

Problem 2:

Given:
- Left triangle: sides \( x \), 1
- Right triangle: sides 9, 3

Since the triangles are similar, the ratios of corresponding sides are equal:
\[
\frac{x}{9} = \frac{1}{3}
\]

Solve for \( x \):
\[
x = 9 \cdot \frac{1}{3} = 3
\]

Answer:
\[
\boxed{3}
\]

---

Problem 3:

Given:
- Left triangle: sides \( x \), 4
- Right triangle: sides 6, 18

Since the triangles are similar, the ratios of corresponding sides are equal:
\[
\frac{x}{6} = \frac{4}{18}
\]

Simplify the right side:
\[
\frac{4}{18} = \frac{2}{9}
\]

So the equation becomes:
\[
\frac{x}{6} = \frac{2}{9}
\]

Cross-multiply to solve for \( x \):
\[
x \cdot 9 = 6 \cdot 2
\]
\[
9x = 12
\]
\[
x = \frac{12}{9} = \frac{4}{3}
\]

Answer:
\[
\boxed{\frac{4}{3}}
\]

---

Problem 4:

Given:
- Left triangle: sides 5, 4, 7
- Right triangle: sides 10, 8, \( x \)

Since the triangles are similar, the ratios of corresponding sides are equal. We can use any pair of corresponding sides to find \( x \). Let's use the sides 7 and \( x \):
\[
\frac{7}{x} = \frac{5}{10}
\]

Simplify the right side:
\[
\frac{5}{10} = \frac{1}{2}
\]

So the equation becomes:
\[
\frac{7}{x} = \frac{1}{2}
\]

Cross-multiply to solve for \( x \):
\[
7 \cdot 2 = x \cdot 1
\]
\[
x = 14
\]

Answer:
\[
\boxed{14}
\]

---

Problem 5:

Given:
- Left triangle: sides 14, 4
- Right triangle: sides 2, \( x \)

Since the triangles are similar, the ratios of corresponding sides are equal:
\[
\frac{14}{2} = \frac{4}{x}
\]

Simplify the left side:
\[
\frac{14}{2} = 7
\]

So the equation becomes:
\[
7 = \frac{4}{x}
\]

Solve for \( x \):
\[
x = \frac{4}{7}
\]

Answer:
\[
\boxed{\frac{4}{7}}
\]

---

Problem 6:

Given:
- Left triangle: sides 6, 9
- Right triangle: sides 24, \( x \)

Since the triangles are similar, the ratios of corresponding sides are equal:
\[
\frac{6}{24} = \frac{9}{x}
\]

Simplify the left side:
\[
\frac{6}{24} = \frac{1}{4}
\]

So the equation becomes:
\[
\frac{1}{4} = \frac{9}{x}
\]

Cross-multiply to solve for \( x \):
\[
1 \cdot x = 4 \cdot 9
\]
\[
x = 36
\]

Answer:
\[
\boxed{36}
\]

---

Problem 7:

Given:
- Left triangle: sides 10, 9
- Right triangle: sides 99, \( x \)

Since the triangles are similar, the ratios of corresponding sides are equal:
\[
\frac{10}{99} = \frac{9}{x}
\]

Cross-multiply to solve for \( x \):
\[
10 \cdot x = 99 \cdot 9
\]
\[
10x = 891
\]
\[
x = \frac{891}{10} = 89.1
\]

Answer:
\[
\boxed{89.1}
\]

---

Problem 8:

Given:
- Left triangle: sides 10, 10
- Right triangle: sides 100, \( x \)

Since the triangles are similar, the ratios of corresponding sides are equal:
\[
\frac{10}{100} = \frac{10}{x}
\]

Simplify the left side:
\[
\frac{10}{100} = \frac{1}{10}
\]

So the equation becomes:
\[
\frac{1}{10} = \frac{10}{x}
\]

Cross-multiply to solve for \( x \):
\[
1 \cdot x = 10 \cdot 10
\]
\[
x = 100
\]

Answer:
\[
\boxed{100}
\]

---

Final Answers:

\[
\boxed{4, 3, \frac{4}{3}, 14, \frac{4}{7}, 36, 89.1, 100}
\]