in a rectangle ABCD of width 1 and length 2 suppose AD is a length let R be a point selected at random from within the rectangle. the probability that < ARD is obtuse is closest to X/400 where X is an interger. find X

To solve this problem, we need to find the probability that \(\angle ARD\) in rectangle \(ABCD\) is obtuse, where \(A(0, 0)\), \(B(2, 0)\), \(C(2, 1)\), and \(D(0, 1)\). The rectangle has width 1 and length 2, and point \(R\) is randomly selected from within the rectangle.

### Step 1: Understanding when \(\angle ARD\) is obtuse
An angle is obtuse if the dot product between two vectors corresponding to the sides of the angle is negative. In this case, we want the angle \(\angle ARD\) to be obtuse, meaning that the dot product of vectors \(\vec{AR}\) and \(\vec{RD}\) must be negative.

Let \(R = (x, y)\) be any point inside the rectangle. The vectors \(\vec{AR}\) and \(\vec{RD}\) are:
\[
\vec{AR} = (x - 0, y - 0) = (x, y)
\]
\[
\vec{RD} = (x - 0, y - 1) = (x, y - 1)
\]
The dot product of \(\vec{AR}\) and \(\vec{RD}\) is:
\[
\vec{AR} \cdot \vec{RD} = x(x) + y(y - 1) = x^2 + y(y - 1) = x^2 + y^2 - y
\]
For \(\angle ARD\) to be obtuse, we require:
\[
x^2 + y^2 - y < 0
\]
This simplifies to:
\[
x^2 + y^2 < y
\]
This inequality describes the region inside the rectangle where the angle \(\angle ARD\) is obtuse.

### Step 2: Analyzing the inequality
We need to find the area of the region where \(x^2 + y^2 < y\) inside the rectangle \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\).

Rearrange the inequality:
\[
x^2 + (y - \frac{1}{2})^2 < \frac{1}{4}
\]
This represents a circle centered at \((0, \frac{1}{2})\) with radius \(\frac{1}{2}\).

### Step 3: Finding the area where the inequality holds
The region defined by the inequality is a portion of the circle with radius \(\frac{1}{2}\) centered at \((0, \frac{1}{2})\) that lies inside the rectangle. The area of this circle is:
\[
\text{Area of the full circle} = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}
\]
However, we are only interested in the part of the circle that lies within the rectangle, specifically for \(x \in [0, 2]\) and \(y \in [0, 1]\).

Because the circle is tangent to the top and bottom of the rectangle, we approximate that about half of the circle lies within the rectangle. Thus, the area of the region where \(\angle ARD\) is obtuse is approximately \(\frac{\pi}{8}\).

### Step 4: Calculating the probability
The total area of the rectangle is \(2 \times 1 = 2\).

Therefore, the probability that \(\angle ARD\) is obtuse is approximately:
\[
\frac{\text{Area of the obtuse region}}{\text{Total area of the rectangle}} = \frac{\frac{\pi}{8}}{2} = \frac{\pi}{16}
\]
Using \(\pi \approx 3.1416\), we get:
\[
\frac{\pi}{16} \approx \frac{3.1416}{16} \approx 0.19635
\]
Now, the problem states that this probability is approximately \(\frac{X}{400}\), so:
\[
\frac{X}{400} \approx 0.19635
\]
Solving for \(X\):
\[
X = 0.19635 \times 400 \approx 78.54
\]
Rounding to the nearest integer, \(X = 79\).

Thus, the value of \(X\) is \( \boxed{79} \).

Would you like further details or clarification on any steps?

Here are 5 related questions:
1. How is the concept of a dot product used to determine if an angle is obtuse?
2. What is the significance of the circle equation derived in the solution?
3. How would the result change if the dimensions of the rectangle were different?
4. Can the same approach be applied to other polygons to find obtuse angles?
5. Why is it important to approximate the area of the circle inside the rectangle?

**Tip:** When solving probability problems with geometric regions, visualizing the areas involved can simplify your approach significantly.

In a rectangle ABCD of width 1 and length 2, suppose AD is a length. Let R be a point selected at random from within the rectangle. The probability that ∠ARD is obtuse is closest to X/400, where X is an integer. Find X.

Solve a geometric probability problem involving a rectangle ABCD. Find the probability that the angle ∠ARD is obtuse, using dot product and area of a circle. A step-by-step explanation is provided with key formulas and theorems. Suitable for Grades 11-12.

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