https://file.aiquickdraw.com/mathbot/file/1727361144199vfzyayna.jpg

Let's break down the problem based on the image you've provided.

You are working with a right-angled triangle \( ABC \), where:
- \( \angle A = 90^\circ \)
- \( AB = 10 \, \text{cm} \)
- \( AC = 20 \, \text{cm} \)

There are two points, \( P \) and \( Q \), that move as described:
- \( P \) starts at \( A \) and moves along side \( AB \) towards \( B \), at a speed of 2 cm per second. Once \( P \) reaches \( B \), it turns back and moves towards \( A \) at the same speed.
- \( Q \) starts at \( A \) and moves along side \( AC \) towards \( C \), also at a speed of 2 cm per second.

You are asked about the area of triangle \( APQ \) after \( x \) seconds under two conditions:
1. \( 0 \leq x \leq 5 \)
2. \( 5 \leq x \leq 10 \)

Let’s solve these one by one.

### 1. Calculation of the area of \( \triangle APQ \)

For the triangle \( APQ \), the area can be expressed as:
\[
\text{Area of } \triangle APQ = \frac{1}{2} \times AP \times AQ
\]

- The length of \( AP \) is based on the motion of point \( P \), which moves along \( AB \).
- The length of \( AQ \) is based on the motion of point \( Q \), which moves along \( AC \).

#### Case 1: \( 0 \leq x \leq 5 \)

- \( P \) is moving along \( AB \), and its distance from \( A \) after \( x \) seconds is \( AP = 2x \) (since \( P \) moves 2 cm per second).
- \( Q \) is moving along \( AC \), and its distance from \( A \) after \( x \) seconds is \( AQ = 2x \).

Thus, the area of \( \triangle APQ \) after \( x \) seconds is:
\[
\text{Area of } \triangle APQ = \frac{1}{2} \times (2x) \times (2x) = 2x^2
\]

#### Case 2: \( 5 \leq x \leq 10 \)

For this case, \( P \) reaches \( B \) at \( x = 5 \) seconds and then starts moving back towards \( A \). So for \( x \geq 5 \), we need to adjust the distance \( AP \):
- \( AP = 10 - 2(x - 5) = 20 - 2x \)

Meanwhile, \( Q \) continues to move along \( AC \), so:
- \( AQ = 2x \)

Thus, the area of \( \triangle APQ \) after \( x \) seconds (when \( 5 \leq x \leq 10 \)) is:
\[
\text{Area of } \triangle APQ = \frac{1}{2} \times (20 - 2x) \times (2x) = (20 - 2x) \times x = 20x - 2x^2
\]

### 2. When the area of \( \triangle PBQ \) is half of the area of \( \triangle ABC \)

We are also asked to find the time when the area of \( \triangle PBQ \) is half the area of \( \triangle ABC \).

The area of \( \triangle ABC \) is:
\[
\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 10 \times 20 = 100 \, \text{cm}^2
\]

The area of \( \triangle PBQ \) needs to be \( \frac{100}{2} = 50 \, \text{cm}^2 \).

This area can be expressed as:
\[
\text{Area of } \triangle PBQ = \frac{1}{2} \times PB \times BQ
\]
Where:
- \( PB = 10 - AP \)
- \( BQ = AQ \)

Now, let me know if you would like to proceed further into calculations or clarifications.

---

### Related Questions:
1. What happens to the area of \( \triangle APQ \) after 10 seconds?
2. How would the result change if point \( P \) and \( Q \) moved at different speeds?
3. Can you generalize the area expression for a time greater than 10 seconds?
4. What if the lengths of \( AB \) and \( AC \) were different?
5. How does the path of \( P \) affect the shape of \( \triangle APQ \)?

### Tip:
Always visualize the motion of the points and their positions at specific times to help break down the problem step-by-step.

Understand how to calculate the area of triangle APQ based on time, given a right-angled triangle ABC with points P and Q moving at constant speeds. This problem involves the motion of points along triangle sides and applies area formulas for dynamic geometries. Suitable for Grades 9-10 students.

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