A man wanted to plant 4 trees, but all 4 had to be equal distances from each other. Then how he has to arrange those plants
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square vertices
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equilateral triangle vertices and centroid
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Both 1 and 2
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None of the above
A square has equal side lengths but its diagonal is longer than the sides. An equilateral triangle with its centroid places the centroid at a shorter distance from each vertex than the vertices are from each other. Four points all mutually equidistant is impossible in 2D geometry; it requires a 3D tetrahedron (triangular pyramid) arrangement. Since the question implies planting trees on ground (2D), no 2D arrangement works.
To answer this question, we need to consider the placement of the trees and the condition that they must be equal distances from each other.
Option A) Square vertices - If we place the trees at the vertices of a square, then the distances between the trees will not be equal. The distance between adjacent trees will be equal to the length of a side of the square, but the distance between non-adjacent trees will be greater.
Option B) Equilateral triangle vertices and centroid - If we place the trees at the vertices of an equilateral triangle, then the distances between the trees will also not be equal. The distance between adjacent trees will be equal to the length of a side of the triangle, but the distance between non-adjacent trees will be greater.
Therefore, neither option A nor option B satisfies the condition that all 4 trees must be equal distances from each other.
Since none of the given options meet the requirements, the correct answer is D) None of the above.