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Let's solve the problem step by step.

### Given:
- The boy and the girl will arrive between 5 p.m. and 6 p.m. (i.e., within a 60-minute interval).
- They do not wait for more than 20 minutes for the other person.
- The arrivals are uniformly distributed between 5 p.m. and 6 p.m.

We need to find the probability that they will meet.

### 1. Representation of the problem:
Let \( x \) be the arrival time of the boy (in minutes after 5 p.m.) and \( y \) be the arrival time of the girl (in minutes after 5 p.m.). Both \( x \) and \( y \) are uniformly distributed between 0 and 60 minutes.

The two will meet if the difference between their arrival times is less than or equal to 20 minutes, i.e.,
\[
|x - y| \leq 20.
\]

### 2. Visualization:
This can be visualized as a geometric problem in the coordinate plane, where \( x \) and \( y \) range from 0 to 60. The region where they meet is bounded by the lines:
\[
y = x + 20 \quad \text{and} \quad y = x - 20.
\]

The total possible outcomes are the area of the square defined by \( 0 \leq x \leq 60 \) and \( 0 \leq y \leq 60 \). The area of this square is:
\[
A_{\text{total}} = 60 \times 60 = 3600.
\]

### 3. Meeting Region:
The region where \( |x - y| \leq 20 \) forms a band around the line \( y = x \). The boundaries of this band are the lines \( y = x + 20 \) and \( y = x - 20 \).

We need to calculate the area of this region. The region consists of two triangular areas that fall outside the main diagonal region where they meet.

Each triangular area has a base and height of \( 40 \) (since the meeting condition applies only within a 20-minute window on each side of the diagonal), and the area of each triangle is:
\[
A_{\text{triangle}} = \frac{1}{2} \times 40 \times 40 = 800.
\]

Thus, the total area outside the meeting region is \( 2 \times 800 = 1600 \).

### 4. Probability of Meeting:
The area where they will meet is the total area minus the area outside the meeting region:
\[
A_{\text{meet}} = 3600 - 1600 = 2000.
\]

The probability that they will meet is the ratio of the area where they meet to the total area:
\[
P(\text{meet}) = \frac{2000}{3600} = \frac{5}{9}.
\]

### Final Answer:
The probability that the boy and girl will meet is \( \frac{5}{9} \).

---

Would you like more details on any part of the solution? Here are 5 follow-up questions you might find helpful:
1. How does uniform distribution affect the setup of the problem?
2. What would happen to the probability if the waiting time were reduced to 15 minutes?
3. Can we generalize this problem for different meeting intervals or waiting times?
4. How is the total area of the square relevant in determining the probability?
5. How could we solve this problem using a different method, such as integration?

**Tip:** Visualizing problems involving uniform distributions and time constraints as geometric areas can make them easier to solve.

A boy and a girl decide to meet between 5 and 6 p.m. in a park. They decide not to wait for the other for more than 20 minutes. Assuming arrivals to be independent and uniformly distributed, find the probability that they will meet.

Solve a probability problem where a boy and a girl decide to meet between 5 and 6 p.m. but will only wait 20 minutes for each other. Explore the solution using uniform distribution and geometric probability concepts. Suitable for high school students in Grades 10-12.

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