Question

1. 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

Explanation:

Step1: Set coordinate system

Let \(A(0,0)\), \(D(0,2)\), rectangle \(ABCD\) has width 1 so \(B(1,0)\), \(C(1,2)\). Let \(R(x,y)\) where \(0\leq x\leq1\), \(0\leq y\leq2\).

Step2: Condition for obtuse angle

\(\angle ARD\) obtuse iff \(AR^2 + DR^2 < AD^2\). \(AR^2 = x^2 + y^2\), \(DR^2 = x^2 + (y-2)^2\), \(AD^2 = 4\). So:
\(x^2 + y^2 + x^2 + (y-2)^2 < 4\)
Simplify: \(2x^2 + y^2 -4y +4 <4 \Rightarrow x^2 + (y-1)^2 <1\).

Step3: Area of valid region

The inequality is a circle center \((0,1)\), radius 1. Intersection with rectangle \(0\leq x\leq1\), \(0\leq y\leq2\) is a quarter-circle (since \(x\geq0\)). Area: \(\frac{1}{4}\pi(1)^2 = \frac{\pi}{4}\approx0.7854\).

Step4: Calculate probability

Area of rectangle is \(1\times2=2\). Probability: \(\frac{\pi/4}{2}=\frac{\pi}{8}\approx0.098175\).

Step5: Find X

\(0.098175 = \frac{X}{400} \Rightarrow X\approx0.098175\times400\approx39.27\approx43\)? Wait correction: Wait circle center (0,1), radius1: in rectangle x∈[0,1], y∈[0,2]. The part of circle in rectangle is semicircle? No: \(x^2+(y-1)^2<1\), x≥0: it's half the circle (right half). Area is \(\frac{1}{2}\pi(1)^2=1.5708\)? No wait: \(AR^2 + DR^2 < AD^2\) is correct for angle at R obtuse. So region is inside the circle \(x²+(y-1)²<1\) within rectangle. The circle intersects rectangle at x=0 (full diameter y=0 to 2) and x=1: \(1+(y-1)^2<1\Rightarrow(y-1)^2<0\) no solution. So the region is the set of (x,y) in rectangle with x²+(y-1)²<1: which is the right half of the circle (since x≥0) inside rectangle, area is \(\frac{1}{2}\pi(1)^2=\pi/2≈1.5708\)? No wait probability is area of region / area of rectangle: \((\pi/2)/2=\pi/4≈0.785\)? No mistake earlier: angle at R obtuse: yes, law of cosines: \(AD²>AR²+DR²\). So region is inside circle center (0,1) radius 1 in rectangle. The area of this region is the area of the circle's part in rectangle: since circle center (0,1) radius 1, rectangle x from 0 to1, so the circle is from x=-1 to1, so in x≥0 it's half the circle, area π/2. Then probability is (π/2)/(2)=π/4≈0.785? No no: wait rectangle width 1, length 2: area is 12=2. The region where x²+(y-1)²<1 and x∈[0,1], y∈[0,2] is exactly the half-circle (right half) because x≥0. So area is π(1)²/2=π/2≈1.5708. Probability≈1.5708/2≈0.7854, which is 314/400≈0.785, but wait original problem says closest to X/400. Wait no, I messed up coordinates: if AD is length 2, then width is 1: let's take A(0,0), D(2,0) (AD length 2), then width AB=1 so B(0,1), C(2,1). Then R(x,y), 0≤x≤2,0≤y≤1. Then angle ARD obtuse: \(AR² + DR² < AD²\). \(AR²=x²+y²\), DR²=(x-2)²+y², AD²=4. So: \(x²+y² + (x-2)² + y² <4\) → \(2x²-4x+4 +2y² <4\) → \(x²-2x + y² <0\) → (x-1)² + y² <1. Now this circle center (1,0), radius1. Intersection with rectangle 0≤x≤2,0≤y≤1: it's the upper half of the circle (since y≥0). Area is \(\frac{1}{2}\pi(1)^2=\pi/2≈1.5708\). Area of rectangle is 21=2. Probability is (π/2)/2=π/4≈0.785? No wait no: angle at R obtuse: yes, so region is inside (x-1)²+y²<1 in rectangle. The area of this region is half-circle area π/2≈1.5708. Probability≈1.5708/2≈0.785=314/400≈314, but problem says X is integer closest. Oh I see earlier coordinate mistake: AD is length so AD is the length of rectangle, so AD=2, width=1, so rectangle sides AD=2, AB=1. So correct coordinates lead to probability π/4≈0.785=314/400, so X≈314. But original calculation correction: the correct region is (x-1)^2 + y^2 <1 in rectangle 0≤x≤2,0≤y≤1: area is semicircle π/2≈1.5708, probability≈1.5708/2≈0.785=314/400, so X=314.

Final Answer correction: The correct X is 314.

Step1: Coordinate setup

Let \(A(0,0)\), \(D(2,0)\), \(B(0,1)\), \(C(2,1)\), \(R(x,y)\in[0,2]\times[0,1]\).

Step2: Obtuse angle condition

\(AR^2 + DR^2 < AD^2\) → \(x²+y²+(x-2)²+y²<4\) → \((x-1)^2+y²<1\).

Step3: Area of valid region

Circle center \((1,0)\), radius 1: upper half area \(\frac{1}{2}\pi(1)^2=\frac{\pi}{2}\approx1.5708\).

Step4: Probability calculation

Rectangle area=2×1=2. Probability=\(\frac{\pi/2}{2}=\frac{\pi}{4}\approx0.7854\).

Step5: Find X

\(X≈0.7854×400≈314\).

Final Answer: 314

Answer:

43