To solve this problem, let's break down the scenario step by step.

We have a rectangle $ABCD$ with width 1 and length 2. Suppose point $R$ is randomly selected from within the rectangle. We're interested in the probability that the angle $\angle ADR$ is obtuse.

Step 1: Understanding the problem

The angle $\angle ADR$ will be obtuse if the vector $\overrightarrow{AD}$ and the vector $\overrightarrow{AR}$ form an angle greater than 90°.

• The rectangle's dimensions give us that $A = (0, 0)$, $B = (2, 0)$, $C = (2, 1)$, and $D = (0, 1)$.
• The vector $\overrightarrow{AD}$ is simply the vector from $A = (0, 0)$ to $D = (0, 1)$, which is the vertical vector $\overrightarrow{AD} = (0, 1)$.
• Point $R = (x, y)$ is randomly chosen inside the rectangle. So, $x$ is between 0 and 2, and $y$ is between 0 and 1.

Step 2: Condition for obtuse angle

The angle between two vectors is obtuse if their dot product is negative. The vectors involved here are:

• $\overrightarrow{AD} = (0, 1)$
• $\overrightarrow{AR} = (x, y)$

The dot product of $\overrightarrow{AD}$ and $\overrightarrow{AR}$ is: $\overrightarrow{AD} \cdot \overrightarrow{AR} = (0)(x) + (1)(y) = y$

For the angle $\angle ADR$ to be obtuse, we require: $y < 0$

However, since $y$ represents the $y$-coordinate of the point $R$ in the rectangle, and $y$ ranges from 0 to 1, this condition is never satisfied. Therefore, there's no point within the rectangle where the angle $\angle ADR$ is obtuse.

Step 3: Probability

Since the angle $\angle ADR$ cannot be obtuse for any point inside the rectangle, the probability of selecting a point such that $\angle ADR$ is obtuse is zero.

Therefore, the probability that the angle is obtuse is $0$. If the probability is expressed as $\frac{x}{400}$, then $x = 0$.

Final Answer:

$x = 0$

Let me know if you'd like further clarification or have additional questions! Here are a few related questions you might find interesting:

1. How would the probability change if we were looking for an acute angle instead of an obtuse one?
2. What happens if we change the rectangle's dimensions, say to a square with side length 1?
3. How do we compute the probability that an angle between two vectors in general is obtuse or acute?
4. How would the problem change if the point $R$ were constrained to lie on one of the sides of the rectangle?
5. Can this method be generalized to other polygons, like triangles or circles?

Tip: When dealing with random points within a geometric shape, always visualize or sketch the region of interest and identify how geometric properties influence the desired condition (like obtuse or acute angles).