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.

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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.