Geometric Probability in a Triangle
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Q5: A triangle has sides of lengths 3, 7, and 8. What is the probability of the randomly chosen point in the triangle has less distance than 1 to at least one of it's corners? A) $\frac{\pi\sqrt{2}}{36}$ B) $\frac{\pi\sqrt{3}}{36}$ C) $\frac{\pi}{36}$ D) $\frac{1}{2}$ E) $\frac{3}{4}$
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Hi Serkan, let's solve this geometric probability problem together. We need to find the probability that a point chosen inside a specific triangle is close to one of its corners.
Problem Analysis
A random point chosen in a region follows a uniform distribution, so the probability is simply the ratio of the favorable area to the total area of the triangle.
Let's start by finding the total area of the triangle with sides 3, 7, and 8. We can use Heron's formula for this.
1. Area of the Triangle
First, we calculate the semi-perimeter, s, which is the sum of the sides divided by two.
Now we plug the values into the area formula: 9 times 9 minus 3, times 9 minus 7, times 9 minus 8.
This simplifies to the square root of 9 times 6 times 2 times 1, which equals the square root of 108.
Next, let's identify the favorable region. These are points with a distance less than 1 from at least one corner.
2. Favorable Region
This region consists of three circular sectors centered at the vertices, each with a radius of 1. Since the shortest side is 3, these sectors do not overlap.
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