Final answer:
The task involves partitioning a line segment internally and externally in a 3:2 ratio. For internal division, the point B is such that AB:BC is 3:5:2:5. For external division, if AC is the unit segment, the point D is such that AC:CD is 1:2/3.
Step-by-step explanation:
The subject of the question is partitioning a segment in a given ratio, which is a common task in Mathematics, specifically in the area of geometry. When a student is given a unit segment, say AC, and is asked to partition it internally and externally in the ratio of 3 to 2, they need to find points B and D such that AB:BC = 3:2 for internal division, and AC:CD = 3:2 for external division. To label the internal point B and the external point D, one must calculate the segment lengths based on the given ratios.
For internal division, let AB = 3x and BC = 2x, where AC = AB + BC = 3x + 2x = 5x. Since AC is a unit segment, 5x = 1, so x = 1/5. Thus, AB = 3(1/5) = 3/5 and BC = 2/5. Hence, the internal proportion is AB:BC = 3/5:2/5.
For external division, let AC = 3y and CD = 2y, where AD = AC + CD = 3y + 2y = 5y. Since AC is a unit segment, 3y = 1, so y = 1/3. Thus, CD = 2(1/3) = 2/3 and the external proportion is AC:CD = 1:2/3.