Answer:
Explanation:
We will begin by drawing a diagram to better understand the problem:
P
/ \
/ \
/ \
/ \
/ \
/ \
Q-------------R
O
Let the side length of equilateral triangle PQR be denoted by "s", and let the distance from the center O to any of the vertices P, Q, or R be denoted by "r". It is known that r = s/√3.
To find the probability that X is closer to O than to any of the sides, we need to find the ratio of the area of the region of the triangle that satisfies this condition to the total area of the triangle.
Let us consider the following two cases:
Case 1: X lies inside the circle with center O and radius r.
In this case, any point inside the circle is closer to O than to any of the sides. The area of this circle can be found using the formula for the area of a circle, A = πr^2.
Case 2: X lies outside the circle with center O and radius r, but inside the triangle PQR.
In this case, X is closer to O than to any of the sides if and only if it lies inside the equilateral triangle that is formed by connecting the midpoints of the sides of PQR. This is because any point inside this triangle is closer to O than to any of the sides, and any point outside this triangle is closer to one of the sides than to O.
The area of this equilateral triangle can be found by dividing PQR into four congruent smaller triangles and using the Pythagorean theorem to find the length of the sides of the smaller triangles. The side length of the smaller triangles is s/2, and the height is s√3/2. Therefore, the area of each smaller triangle is (1/2) * (s/2) * (s√3/2) = s^2/8√3. The area of the equilateral triangle is therefore 4 times this value, or s^2/2√3.
The probability that X is closer to O than to any of the sides is therefore given by:
Area of circle with center O and radius r
-----------------------------------------
Total area of triangle PQR
πr^2
------------
s^2√3/4
=----------
s^2√3
= π/3
----
3
+
Area of equilateral triangle with side s/2
-------------------------------------------
Total area of triangle PQR
s^2/2√3
---------
s^2√3/4
---------
s^2√3
= 1/3 - (√3)/4
Therefore, the probability that X is closer to O than to any of the sides is π/9 - (√3)/12 or approximately 0.047.