Consider a segment of length 10. Points A and B are chosen randomly such that A and B divide the segment into three smaller segments. What is the probability that the three smaller segments could form the sides of a triangle?

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∻Solution by Steps∻ ‖ step 1 ⋮ Consider a segment of length 10. To form a triangle, the sum of the lengths of any two segments must be greater than the length of the third segment. This is known as the triangle inequality. ‖ ‖ step 2 ⋮ Let the first point A divide the segment into two parts of lengths x and 10 - x. The second point B will divide one of these parts further. Without loss of generality, assume B divides the segment of length x into parts of lengths y and x - y. ‖ ‖ step 3 ⋮ The three segments formed are of lengths y, x - y, and 10 - x. The triangle inequalities that must be satisfied are: y + (x - y) > 10 - x, y + (10 - x) > x - y, and (x - y) + (10 - x) > y. ‖ ‖ step 4 ⋮ Simplifying the inequalities, we get: x > 5, 10 - x > 5, and x - 2y > 0. ‖ ‖ step 5 ⋮ The inequalities imply that x must be between 5 and 10, and y must be between 0 and x/2. ‖ ‖ step 6 ⋮ The geometric probability is the area of the region satisfying the triangle inequalities divided by the total area of the possible region where points A and B can be chosen. ‖ ‖ step 7 ⋮ The total possible area is a triangle with base 10 and height 5, which corresponds to the case where 0 < x < 10 and 0 < y < x. ‖ ‖ step 8 ⋮ The area satisfying the triangle inequalities is a triangle with base 5 and height 2.5. ‖ ‖ step 9 ⋮ Calculate the area of the triangle representing all possible outcomes: A_{total} = \frac{1}{2} \times 10 \times 5 = 25. ‖ ‖ step 10 ⋮ Calculate the area of the triangle representing the successful outcomes: A_{success} = \frac{1}{2} \times 5 \times 2.5 = 6.25. ‖ ‖ step 11 ⋮ The probability is the ratio of the successful area to the total area: P = \frac{A_{success}}{A_{total}} = \frac{6.25}{25}. ‖ ‖ step 12 ⋮ Simplify the probability: P = \frac{6.25}{25} = \frac{1}{4}. ‖ ∻Answer∻ ⚹ The probability that the three smaller segments could form the sides of a triangle is $\frac{1}{4}$. ⚹ ∻Key Concept∻ ⚹Geometric Probability and Triangle Inequality⚹ ∻Explanation∻ ⚹The problem is solved using geometric probability, considering the area of successful outcomes over the total possible area, while ensuring the triangle inequalities are satisfied.⚹◊What is the condition for three segments to form a triangle when dividing a segment into three smaller segments?⍭ Generate me a similar question◊

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