A point P on the ground is on the same line as the bases and the tips of a pair of trees A and B such that P is on the left of both these trees. The ratio of heights of A and B is 1:3. If the distance between P and A is a metre, then the distance between A and B, in metre, is

The problem describes a scenario involving collinear points and similar triangles. Point P, the bases, and the tips of trees A and B are on the same line, implying the tips of the trees and point P form a single line of sight (angle of elevation). Let the height of tree A be 'h' and tree B be '3h' based on the 1:3 ratio. Let the distance from P to A be 'a' and the distance from A to B be 'x'. Using the property of similar triangles, the ratio of heights must equal the ratio of their horizontal distances from point P. Thus, Height(A)/Height(B) = Distance(PA)/Distance(PB). Substituting the values: h/3h = a/(a + x). This simplifies to 1/3 = a/(a + x), which leads to a + x = 3a. Solving for x gives x = 2a. Therefore, the distance between tree A and tree B is 2a.